Fall 2023
MATH 312: Proofs in Elementary Analysis

Branko Ćurgus

Monday, November 20, 2023

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• Today and tomorrow (Tuesday) we will discuss Section 3.14. We did Exercise 3.14.2., Exercise 3.14.7 and a part of Exercise 3.14.5.

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Sunday, November 19, 2023

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• I got somewhat confused on Friday. Here is clearer presentation.
• We want to use the concept of the infimum of a nonempty set to prove \[ \exists \alpha \in \mathbb{R} \quad \text{such that} \quad \alpha \gt 0 \land \alpha^2 = 2. \]

• For all $s \in \mathbb{R}$ the following two implications hold: \begin{align*} 0 \lt s \ \land \ 2 \lt s^2 \quad & \Rightarrow \quad \frac{s}{2} + \frac{1}{s} \lt s,\\ 0 \lt s \ \land \ s^2 \neq 2 \quad & \Rightarrow \quad 2 \lt \left( \frac{s}{2} + \frac{1}{s} \right)^2. \end{align*}

  Let $s \in \mathbb{R}$ be arbitrary and assume $0 \lt s$ and $2 \lt s^2$. Since $0 \lt 2s$, we have that $0 \lt 1/(2s)$. By Axiom OM multiplying $2 \lt s^2$ by $0 \lt 1/(2s)$ yields \[ \frac{1}{s} \lt \frac{s}{2}. \]

  Adding $s/2$, Axiom OA, yields \[ \frac{s}{2} + \frac{1}{s} \lt \frac{s}{2} + \frac{s}{2} = s. \] This proves the first implication.

  To prove the second implication, assume $0 \lt s$ and $2 \neq s^2$. The following identity holds \begin{align*} \left( \frac{s}{2} + \frac{1}{s} \right)^2 & = \left( \frac{s}{2}\right)^2 + 1 + \left( \frac{1}{s} \right)^2 \\ & = \left( \frac{s}{2}\right)^2 - 1 + \left( \frac{1}{s} \right)^2 + 2 \\ & = \left( \frac{s}{2} - \frac{1}{s} \right)^2 + 2. \end{align*} Notice that \[ 2 \neq s^2 \quad \Leftrightarrow \quad \frac{s}{2} - \frac{1}{s} \neq 0. \] Since we assume $2 \neq s^2$, we have $\frac{s}{2} - \frac{1}{s} \neq 0$, and therefore \[ \left( \frac{s}{2} - \frac{1}{s} \right)^2 \gt 0. \] Therefore \[ \left( \frac{s}{2} + \frac{1}{s} \right)^2 = \left( \frac{s}{2} - \frac{1}{s} \right)^2 + 2 \gt 2. \]

  This proves the second implication.

• The set \[ B = \bigl\{ x \in \mathbb{R} : 0 \lt x \land 2 \lt x^2 \bigr\} \] is nonempty and bounded below. The real number \[ \alpha = \inf \bigl\{ x \in \mathbb{R} : 0 \lt x \land 2 \lt x^2 \bigr\} \] has the properties $0 \lt \alpha$ and $\alpha^2 = 2.$

    Part I. We proved that $0 \lt 2$ and $2 \lt 4 = 2^2$. Therefore, $2 \in B$. Thus, $B \neq \emptyset$. We will prove that $1$ is a lower bound for $B$. Let $x \in B$ be arbitrary. Then $0 \lt x$ and $2 \lt x^2$. Since $1 \lt 2$ we have that $1 \lt x^2$. By an exercise in the class notes, the inequality $1 \lt x^2$ is equivalent to $1 \lt x$. Thus, for all $x \in B$ we have $1 \lt x$. Hence, $1$ is a lower bound for $B$ and $B$ is bounded below. By the Existence of Infimum Theorem, every nonempty bounded below set has an infimum. Thus, the set $B$ has an infimum. Set \[ \alpha = \inf B. \] We proved that $1$ is a lower bound of $B$. Since $\alpha$ is the maximum lower bound of $B$, we deduce that $1 \leq \alpha$. Therefore, $0 \lt \alpha$.

  Part II. Next we will prove that the set $B$ does not have a minimum. Let $x \in B$ be arbitrary. Then $0 \lt x$ and $2 \lt x^2$. By the first implication in the Lemma, we deduce that \[ y = \frac{x}{2} + \frac{1}{x} \lt x. \] By the second implication in the Lemma we have \[ 2 \lt \left( \frac{x}{2} + \frac{1}{x} \right)^2 = y^2. \] Since clearly $y \gt 0$, we proved that $y\in B$. Thus \[ \forall \mkern1px x \in B \quad \exists \mkern2px y \in B \quad \text{such that} \quad y \lt x. \] Thus, $B$ does not have a minimum. Therefore, $\alpha \not\in B$. Since $0 \leq \alpha$ and $\alpha\not\in B$, we deduce that $\alpha^2 \leq 2$.

  Part III. Next we prove that $\alpha^2 \leq 2.$ We use the following property of the infimum: \[ \forall\mkern1px \varepsilon \gt 0 \quad \exists\mkern1px x \in B \quad \text{such that} \quad x \lt \alpha + \varepsilon. \] Consequently, \[ \forall\mkern1px \varepsilon \gt 0 \quad \exists\mkern1px x \in B \quad \text{such that} \quad x^2 \lt (\alpha+ \varepsilon)^2. \] Since, $2 \lt x^2$, we have \[ \forall\mkern1px \varepsilon \gt 0 \quad 2 \lt (\alpha+ \varepsilon)^2 = \alpha^2 + 2\alpha \varepsilon + \varepsilon^2. \] Now we assume that $\varepsilon \lt 1$. Then $\varepsilon^2 \lt \varepsilon$ and therefore \begin{align*} 2 \lt \alpha^2 + 2\alpha \varepsilon + \varepsilon^2 \quad &\Rightarrow \quad 2 \lt \alpha^2 + 2\alpha \varepsilon + \varepsilon \\ &\Rightarrow \quad 2 - \alpha^2 \lt (2\alpha + 1) \varepsilon \\ &\Rightarrow \quad \frac{2 - \alpha^2}{2\alpha + 1} \lt \varepsilon. \end{align*} In the last implication we used the fact that $2\alpha + 1 \gt 0$.

  In the last paragraph we proved that \[ \forall\mkern1px \varepsilon \in (0,1) \quad \text{we have} \quad \frac{2 - \alpha^2}{2\alpha + 1} \lt \varepsilon. \] The preceding statement implies that \[ \frac{2 - \alpha^2}{2\alpha + 1} \leq 0. \] Since $2\alpha + 1 \gt 0$, Axiom OM implies \[ 2 - \alpha^2 \leq 0. \] That is, $2 \leq \alpha^2$.

  In conclusion, we have proved $\alpha^2 \leq 2$ and $2 \leq \alpha^2$. Therefore $\alpha^2 = 2$.

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Tuesday, November 14, 2023

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• Thank you Bennett for the pictures.
• In this item we proved a lemma that will be needed in the following theorem.

  Let $u, v \in\mathbb{R}$ be such that $u \lt v$. Then \[ u \lt \frac{2}{3} u + \frac{1}{3} v \lt \frac{1}{3} u + \frac{2}{3} v \lt v. \]

  Assume $u \lt v$. We use Axiom OA, and add $2u$, then we add $u+v$ and then we add $2v$ to the inequality $u \lt v$. In this way we get three inequalities \[ 3 u \lt 2 u + v, \quad 2 u + v \lt u + 2 v, \quad u + 2 v \lt 3 v. \] The preceding three inequalities and Axiom OT yield \[ 3 u \lt 2 u + v \lt u + 2 v \lt 3 v. \] Since $1/3 \gt 0$, multiplying the preceding inequalities by $1/3$, Axiom OM yields the inequalities in the Lemma.

• The set of real numbers $\mathbb{R}$ is uncountable.

  To prove that $\mathbb{R}$ is uncountable we have to prove that every function $\varphi: \mathbb{N} \to \mathbb{R}$ is not a bijection. We will prove that for arbitrary $\varphi: \mathbb{N} \to \mathbb{R}$ we have that $\varphi$ is not a surjection.

  Let $\varphi: \mathbb{N} \to \mathbb{R}$ be an arbitrary function. We will prove that $\varphi$ is not a surjection. That is we will prove that there exists $c \in \mathbb{R}$ such that \[ \forall n\in \mathbb{N} \quad \varphi_n \neq c. \]

  The idea of the proof is to construct a sequence of "nested" closed intervals \[ \bigl\{ [ \alpha_n , \beta_n ] : n \in \mathbb{N} \bigr\} \] with the following properties:

  •   (I) For all $n \in {\mathbb N}$ we have $\alpha_n \lt \alpha_{n+1} \lt \beta_{n+1} \lt \beta_n$.
  •  (II) For all $k,m \in {\mathbb N}$ we have $\alpha_k \lt \beta_m$.
  • (III) For all $n \in {\mathbb N}$ we have $\varphi_n \not\in \bigl[\alpha_n,\beta_n\bigr]$.

  We define the sequence of "nested" closed intervals \[ \bigl\{ [ \alpha_n , \beta_n ] : n \in \mathbb{N} \bigr\} \] recursively:

  Step 1. Define $\alpha_1 = \varphi_1 + 1$ $\beta_1 = \varphi_1 + 2$. Then \[ \alpha_1 \lt \beta_1 \quad \text{and} \quad and, since $\varphi_1 \not\in [\alpha_1, \beta_1]. \] That is (I) and (III) are satisfied for $n = 1$.

  Step 2. Here we define $\alpha_2$ and $\beta_2$. We set:

      Case 1. $\varphi_{2} \not\in (\alpha_1, \beta_1)$. Set \[ \alpha_{2} = \frac{2}{3} \alpha_1 + \frac{1}{3} \beta_1, \qquad \beta_{2} = \frac{1}{3} \alpha_1 + \frac{2}{3} \beta_1 \] By the Lemma above we have \[ \alpha_1 \lt \alpha_2 \lt \beta_2 \lt \beta_1, \] and, since $\varphi_{2} \not\in (\alpha_1, \beta_1)$, we have \[ \varphi_2 \not\in [\alpha_2,\beta_2]. \] That is (I) and (III) are satisfied for $n = 2$ in this case.

      Case 2. $\varphi_{2} \in (\alpha_1, \beta_1)$. Then \[ \alpha_1 \lt \varphi_2 \lt \beta_1. \] Set \[ \alpha_{2} = \frac{2}{3} \varphi_2 + \frac{1}{3} \beta_1, \qquad \beta_{2} = \frac{1}{3} \varphi_2 + \frac{2}{3} \beta_1 \] By the Lemma above we have \[ \varphi_2 \lt \alpha_2 \lt \beta_{2} \lt \beta_{1}. \] Since $\alpha_1 \lt \varphi_2 \lt \beta_1$, by Axiom OT we deduce \[ \alpha_1 \lt \alpha_2 \lt \beta_2 \lt \beta_1, \] and, since $\varphi_{2} \lt \alpha_2 \lt \beta_2$, we have \[ \varphi_2 \not\in [\alpha_2,\beta_2]. \] That is (I) and (III) are satisfied for $n = 2$ in this case as well.

  Recursive Step. Assume that $n \in \mathbb{N}$ is arbitrary and that $\alpha_k$ and $\beta_k$ had been defined for all $k \in \{1,\ldots,n\}$ such that \[ \alpha_1 \lt \alpha_2 \lt \cdots \lt \alpha_n \lt \beta_n \lt \cdots \lt \beta_2 \lt \beta_1, \] and \[ \forall n \in \{1,\ldots,n\} \quad \text{we have} \quad \varphi_k \not\in [\alpha_k , \beta_k]. \] Now we define $\alpha_{n+1}$ and $\beta_{n+1}$. We set:

      Case 1. $\varphi_{n+1} \not\in (\alpha_n, \beta_n)$. Set \[ \alpha_{n+1} = \frac{2}{3} \alpha_n + \frac{1}{3} \beta_n, \qquad \beta_{n+1} = \frac{1}{3} \alpha_n + \frac{2}{3} \beta_n \] By the Lemma above we have \[ \alpha_n \lt \alpha_{n+1} \lt \beta_{n+1} \lt \beta_n, \] and, since $\varphi_{n+1} \not\in (\alpha_n, \beta_n)$, we have \[ \varphi_{n+1} \not\in [\alpha_{n+1},\beta_{n+1}]. \] That is (I) and (III) are satisfied for $n+1$ in this case.

      Case 2. $\varphi_{n+1} \in (\alpha_n, \beta_n)$. Then \[ \alpha_n \lt \varphi_{n+1} \lt \beta_n. \] Set \[ \alpha_{n+1} = \frac{2}{3} \varphi_{n+1} + \frac{1}{3} \beta_n, \qquad \beta_{n+1} = \frac{1}{3} \varphi_{n+1} + \frac{2}{3} \beta_n \] By the Lemma above we have \[ \varphi_{n+1} \lt \alpha_{n+1} \lt \beta_{n+1} \lt \beta_n. \] Since $\alpha_n \lt \varphi_{n+1} \lt \beta_n$, by Axiom OT we deduce \[ \alpha_n \lt \alpha_{n+1} \lt \beta_{n+1} \lt \beta_n, \] and, since $\varphi_{n+1} \lt \alpha_{n+1} \lt \beta_{n+1}$, we have \[ \varphi_{n+1} \not\in [\alpha_{n+1},\beta_{n+1}]. \] That is (I) and (III) are satisfied for $n+1$ in this case as well.

  By the recursive construction, the sequence of nested intervals \[ \bigl\{ [ \alpha_n , \beta_n ] : n \in \mathbb{N} \bigr\} \] satisfies

  •   (I) For all $n \in {\mathbb N}$ we have $\alpha_n \lt \alpha_{n+1} \lt \beta_{n+1} \lt \beta_n$.
  • (III) For all $n \in {\mathbb N}$ we have $\varphi_n \not\in \bigl[\alpha_n,\beta_n\bigr]$.

  We need to prove (II): \[ \forall k \in {\mathbb N} \ \ \forall m \in {\mathbb N} \quad \text{we have} \quad \alpha_k \lt \beta_m. \] Let $k, m \in {\mathbb N}$ be arbitrary. If $k=m$, then by (I) we have $\alpha_k \lt \beta_m$. If $k \lt m$, then by construction we have that $\alpha_k \lt \alpha_m$ and by (I) we have that $\alpha_m \lt \beta_m$. By Axiom OT, $\alpha_k \lt \beta_m$. If $k \gt m$, the reasoning is similar.

  Set \[ A = \bigl\{ \alpha_k : k \in \mathbb{N} \bigr\}, \qquad B = \bigl\{ \beta_k : k \in \mathbb{N} \bigr\}. \] These sets are clearly nonempty. By property (II), these two sets satisfy \[ \forall a\in A \ \ \forall b \in B \quad a \lt b. \] Thus, the Completeness Axiom applies. Thus, there exists $c \in \mathbb{R}$ such that \[ \forall k \in {\mathbb N} \ \ \forall m \in {\mathbb N} \quad \text{we have} \quad \alpha_k \leq c \leq \beta_m. \] Hence, \[ \forall n \in {\mathbb N} \qquad c \in [\alpha_n,\beta_n]. \] Since \[ \forall n \in {\mathbb N} \qquad \varphi_n \not\in \bigl[\alpha_n,\beta_n\bigr], \] from the last two displayed statements we deduce that \[ \forall n \in {\mathbb N} \qquad \varphi_n \neq c. \] Consequently $\varphi : \mathbb{N} \to \mathbb{R}$ is not a surjection.

  The preceding theorem is Theorem 3.11.4 in the class notes.

  

• Let $A$ be a nonempty subset of $\mathbb{R}$. A number $w \in \mathbb{R}$ is a supremum (or a least upper bound) of $A$ if

  1. $w$ is an upper bound for $A,$ and
  2. if $v$ is an upper bound for $A$ and $w \neq v$, then $w \lt v.$

  The notation for the supremum of $A$ is \[ w = \sup A. \]

  If $A$ is nonempty bounded above subset of $\mathbb{R}$, then $A$ has a supremum.

  

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Monday, November 13, 2023

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• On Thursday we started the discussion of the role of the Completeness Axiom of Real Numbers. My approach to the Completeness Axiom is unconventional. To make this unique perspective available to all, I have created a dedicated webpage detailing my approach. While currently a work in progress, it will soon be fully developed. You can access the evolving content centered around Completeness Axiom here.

• Let $A \neq \emptyset$, $B \neq \emptyset$, $A \subset \mathbb{R}$, and $B \subset \mathbb{R}$. If $\forall a \in A$ and $\forall\mkern 1mu b \in B$ we have $a \leq b$, then $\exists\mkern 1.5mu c \in \mathbb{R}$ such that $\forall a \in A$ and $\forall\mkern 1mu b \in B$ we have $a \leq c \leq b$. Briefly, \[ \forall a \in A \ \ \forall\mkern 1px b \in B \quad a \leq b \quad \Rightarrow \quad \exists\mkern 1.5px c \in \mathbb{R} \ \ \text{such that} \ \ \forall a \in A \ \ \forall\mkern 1px b \in B \quad a \leq c \leq b \]

• The emphasis in the Completeness Axiom is that it applies to two nonempty sets $A$ and $B$ of real numbers which are loosely speaking separated; where separated means \[ \forall a \in A \ \ \forall\mkern 1mu b \in B \quad \text{we have} \quad a \leq b. \] Than, if we have two such sets $A$ and $B$, the Completeness Axiom establishes the existence of a real number $c \in \mathbb{R}$ such that \[ \forall a \in A \ \ \forall\mkern 1px b \in B \quad \text{we have} \quad a \leq c \leq b. \]
• Few blackboards: (thank you Calvin)

  • The Completeness Axiom and four properties of $\mathbb{R}$ that cannot be proved without it.

  • A short review of the proof \[ \exists \mkern 2px \alpha \in \mathbb{R} \quad \text{such that} \quad \alpha \gt 0 \land \alpha^2 = 2. \] A proof by contradiction of the Archimedean Property: \[ \forall \mkern 1px x \in \mathbb{R} \ \ \exists \mkern 1.5px n \in \mathbb{N} \quad \text{such that} \quad x \leq n. \]

  • As a consequence of the Archimedean Property we can define the floor and ceiling function.

    The floor function is defined as follows: For $x \in \mathbb{R}$ we set \[ \lfloor x \rfloor = \max \bigl\{ k \in \mathbb{Z} : k \leq x \bigr\}. \] In words: The floor function of a real number $x$ is defined as the largest integer less than or equal to $x$. This means it rounds $x$ down to the nearest integer. For example, $\lfloor \pi \rfloor = 3$, $\lfloor -e \rfloor = -3$.

    The ceiling function is defined as follows: For $x \in \mathbb{R}$ we set \[ \lceil x \rceil = \min \bigl\{ k \in \mathbb{Z} : x \leq k \bigr\}. \] In words: The ceiling function of a real number $x$ is the smallest integer greater than or equal to $x$. This means it rounds $x$ up to the nearest integer. For example, $\lceil \pi \rceil = 4$, $\lceil -e \rceil = -2$.

  • The basic properties of the floor and ceiling functions are as follows: For all $x \in \mathbb{R}$ we have \begin{alignat*}{2} \lfloor x \rfloor \leq & \mkern 8px x \lt \lfloor x \rfloor + 1 & \qquad \quad \lceil x \rceil - 1 &\lt \mkern 8px x \leq \lceil x \rceil \\ x-1 \lt & \lfloor x \rfloor \leq x & \qquad x &\leq \lceil x \rceil \lt x +1 \\ \end{alignat*}

  • Let $\alpha, \beta \in \mathbb{R}$. If $\beta - \alpha \geq 1$, then \[ \alpha \lt \frac{\lceil \alpha \rceil + \lfloor \beta \rfloor }{2} \lt \beta. \]

    Let $\alpha, \beta \in \mathbb{R}$ be such that $\beta - \alpha \geq 1$. By the basic properties of the floor and ceiling function we have \begin{alignat*}{2} \alpha \leq & \ \lceil \alpha \rceil &&\lt \alpha +1 \\ \beta - 1 \lt & \ \lfloor \beta \rfloor && \leq \beta. \end{alignat*} Since we assume that \[ \alpha + 1 \leq \beta \qquad \text{and} \qquad \alpha \leq \beta - 1, \] Axiom OT and the preceding inequalities yield: \begin{alignat*}{2} \alpha \leq & \ \lceil \alpha \rceil &&\lt \alpha +1 \leq \beta \\ \alpha \leq \beta - 1 \lt & \ \lfloor \beta \rfloor && \leq \beta. \end{alignat*} Or, briefly, \begin{alignat*}{2} \alpha \leq & \ \lceil \alpha \rceil &&\lt \beta \\ \alpha \lt & \ \lfloor \beta \rfloor && \leq \beta. \end{alignat*} Adding the preceding inequalities yield \[ 2 \alpha \lt \lceil \alpha \rceil + \lfloor \beta \rfloor \lt 2 \beta. \] Hence, \[ \alpha \lt \frac{\lceil \alpha \rceil + \lfloor \beta \rfloor}{2} \lt \beta. \]

    Let $a, b \in \mathbb{R}$ be such that $a \lt b$. Then \[ a \lt \frac{\left\lceil \left\lceil \dfrac{1}{b-a} \right\rceil a \right\rceil + \left\lfloor \left\lceil \dfrac{1}{b-a} \right\rceil b \right\rfloor}{2 \, \left\lceil \dfrac{1}{b-a} \right\rceil} \lt b. \]

    Let $a, b \in \mathbb{R}$ be such that $a \lt b$. Thus, $b - a \gt 0$. Can we multiply $b - a \gt 0$ with a large number and make the produce $\geq 1$, so that we can use the preceding Lemma? Yes. The point here is that $b-a \gt 0$ is tiny. So, the real number \[ \frac{1}{b-a} \] is a huge positive real number. Since $1/(b-a) \gt 0$, notice that \[ \left\lceil \dfrac{1}{b-a} \right\rceil \geq 1. \] Basic property of the ceiling function yields: \[ \frac{1}{b-a} \leq \left\lceil \dfrac{1}{b-a} \right\rceil. \] Consequently. \[ 1 \leq \left\lceil \dfrac{1}{b-a} \right\rceil (b-a) = \left\lceil \dfrac{1}{b-a} \right\rceil b - \left\lceil \dfrac{1}{b-a} \right\rceil a. \] Now we can set \[ \alpha = \left\lceil \dfrac{1}{b-a} \right\rceil a, \qquad \beta = \left\lceil \dfrac{1}{b-a} \right\rceil b, \] and since $\beta - \alpha \geq 1$, we can apply the preceding Lemma to deduce \[ \left\lceil \dfrac{1}{b-a} \right\rceil a \lt \frac{ \left\lceil \left\lceil \dfrac{1}{b-a} \right\rceil a \right\rceil + \left\lfloor \left\lceil \dfrac{1}{b-a} \right\rceil b \right\rfloor}{2} \lt \left\lceil \dfrac{1}{b-a} \right\rceil b. \] Divide the preceding inequalities by \[ \left\lceil \dfrac{1}{b-a} \right\rceil \] to get \[ a \lt \frac{ \left\lceil \left\lceil \dfrac{1}{b-a} \right\rceil a \right\rceil + \left\lfloor \left\lceil \dfrac{1}{b-a} \right\rceil b \right\rfloor}{2\left\lceil \dfrac{1}{b-a} \right\rceil} \lt b. \]

    Let $a, b \in \mathbb{R}$ be such that $a \lt b$. Then there exists $r \in \mathbb{Q}$ such that \[ a \lt r \lt b. \]

    In the preceding Theorem we have \[ \left\lceil \left\lceil \dfrac{1}{b-a} \right\rceil a \right\rceil + \left\lfloor \left\lceil \dfrac{1}{b-a} \right\rceil b \right\rfloor \in \mathbb{Z}, \] and \[ 2 \, \left\lceil \dfrac{1}{b-a} \right\rceil \in \mathbb{N}. \] Therefore, \[ r = \frac{ \left\lceil \left\lceil \dfrac{1}{b-a} \right\rceil a \right\rceil + \left\lfloor \left\lceil \dfrac{1}{b-a} \right\rceil b \right\rfloor}{2\left\lceil \dfrac{1}{b-a} \right\rceil} \in \mathbb{Q}. \] Thus, the claim of the Corollary follows from the preceding Theorem.
    

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Monday, November 6, 2023

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• In the table below, I list the first twenty one members of $\mathbb{N}\times\mathbb{N}$ following a particular pattern. To each ordered pair we associate a positive integer. The table below suggest the following function \[ A(s,t) = \frac{1}{2}(s+t-1)(s+t) - t + 1, \qquad (s,t) \in \mathbb{N}\times\mathbb{N}. \]
• The function $A$ defined in the preceding item is clearly a function defined on $\mathbb{N}\times\mathbb{N}$ with the values in $\mathbb{N}$. That is \[ A : \mathbb{N}\times\mathbb{N} \to \mathbb{N}. \] I claim that $A$ is a bijection. To prove this claim, I will find the inverse function of $A$. (To be updated here, see Subsection 3.9.1 in the class notes.)

• In this item I introduce the sequence of triangular numbers. That is the function $T: \mathbb{N} \to \mathbb{N}$ defined given by the following recursive definition: \[ T_1 = 1, \qquad \forall\mkern 1pt n \in \mathbb{N} \quad T_{n+1} = T_n + (n+1). \]

  • The above recursive definition yields: \[ \forall\mkern 1pt n \in \mathbb{N} \qquad T_{n+1} - T_n = n+1. \] It is convenient to set $T_0 = 0.$ Then, the preceding statement can be stated as \[ \forall\mkern 1pt n \in \mathbb{N} \qquad T_{n} - T_{n-1} = n. \] The preceding statement implies that $\forall\mkern 1pt n \in \mathbb{N}$ the following set \[ \bigl\{ k \in \mathbb{N} : T_{n-1} \lt k \ \land \ k \leq T_{n} \bigr\} \] has $n$ elements. That is \[ \forall\mkern 1pt n \in \mathbb{N} \qquad \operatorname{card} \bigl\{ k \in \mathbb{N} : T_{n-1} \lt k \ \land \ k \leq T_{n} \bigr\} = n. \] In words: there are exactly $n$ positive integers which are larger than $T_{n-1}$ and less or equal to $T_n.$

  • It is a nice exercise in Mathematical Induction to prove \[ \forall\mkern 1pt n \in \mathbb{N} \qquad T_n = \frac{1}{2}n(n+1). \]

• In this item we define a sequence that I call the repeat sequence $R: \mathbb{N} \to \mathbb{N}$ such that \begin{align*} R_1 & = 1, \\ R_2 & = 2, \ \ R_3 = 2, \\ R_4 & = 3, \ \ R_5 = 3, \ \ R_6 = 3, \\ R_7 & = 4, \ \ R_8 = 4, \ \ R_9 = 4, \ \ R_{10} = 4, \\ R_{11} &= 5, \ \ R_{12} = 5, \ \ R_{13} = 5, \ \ R_{14} = 5, \ \ R_{15} = 5, \ldots \end{align*} The relationship between the sequence $R$ and the triangular numbers is given in the following table:
  • A formal definition of the repeat sequence is \[ \forall\mkern 1pt n \in \mathbb{N} \qquad R_n = \min \bigl\{ m \in \mathbb{N} : n \leq T_m \bigr\}. \] Clearly $n(n+1) \geq 2n$ for all $n \in \mathbb{N}$. Therefore, $T_n \geq n$ for all $n\in \mathbb{N}$. Consequently the set in the definition of $R_n$ is not empty. We proved in Exercise that every nonempty subset of $\mathbb{N}$ has a minimum, so $R_n$ is well defined.
  • This item is here help the understanding of the repeat sequence. In words, $R_n$ is the smallest positive integer for which the triangular number $T_{R_n}$ has the property \[ n \leq T_{R_n}. \] For example, Let us calculate $R_{34}$. We need to study triangular numbers nearby $34$. Let us find some. For $m = 7$ we have $T_7 = 28$. Thus, $R_{27} \gt 7$. Let us calculate $T_8 = 36$. Since \[ T_7 = 28 \lt 34 \leq 36 = T_8, \] we conclude that $R_{34} = 8$. Based on the preceding inequalities we have \[ R_{28} = 7, \ R_{29} = R_{30} = R_{31} = R_{32} = R_{33} =R_{34} = R_{35} = R_{36} = 8, \ R_{37} = 9. \]
  • To some extend, the repeat sequence is an inverse of the sequence of the triangular numbers.
  • A recursive definition of the repeat sequence is as follows \[ R_1 = 1, \qquad \forall \mkern 1pt n \in \mathbb{N} \quad R_{n+1} = 1 + R_{(n+1-R_{n})}. \]
• With the repeat sequence at our disposal we can define the inverse function of the function $B: \mathbb{N} \to \mathbb{N}\times\mathbb{N}$ which is the inverse of the function $A:\mathbb{N}\times\mathbb{N}\to \mathbb{N}$ defined by \[ A(s,t) = \frac{1}{2}(s+t-1)(s+t) - t + 1 = T_{(s+t-1)} - t+1 , \quad (s,t) \in \mathbb{N}\times\mathbb{N}, \]
• In this item we prove that \[ \forall \, s,t, n \in \mathbb{N} \quad B\bigl(A(s,t)\bigr) = (s,t) \quad \text{and} \quad A\bigl(B(n)\bigr) = n. \]
  • Recall \begin{align*} A(s,t) & = T_{(s+t-1)} - t+1 , \quad (s,t) \in \mathbb{N}\times\mathbb{N}, \\ B(n) & = \Bigl(n - T_{R_n-1}\, , \, T_{R_n} - n + 1 \Bigr) , \quad n \in \mathbb{N}. \end{align*}
  • Let $s,t \in \mathbb{N}$. We evaluate $B\bigl(A(s,t)\bigr)$. For that we need to evaluate $R_{A(s,t)}$ first. Since by the definition of triangular numbers we have \[ T_{(s+t-1)} - T_{(s+t-2)} = (s + t - 1), \] and since $s \gt 0$, we have \[ T_{(s+t-2)} = T_{(s+t-1)} - (s + t - 1) \lt A(s,t) = T_{(s+t-1)} - t+1 \leq T_{(s+t-1)}. % \] Thus, $A(s,t)$ is in a special way between two consecutive triangular numbers: \[ T_{(s+t-2)} \lt A(s,t) \leq T_{(s+t-1)}. \] By the definition of the repeat function the lase inequality yields \[ R_{A(s,t)} = s+t-1. \] Using the preceding equality we have Therefore \begin{align*} B\bigl(A(s,t)\bigr) & = \Bigl(A(s,t) - T_{s+t-1-1}\, , \, - A(s,t) + 1 + T_{s+t-1} \Bigr) \\ & = \Bigl( T_{(s+t-1)} - t+1 - T_{s+t-2}\, , \, - \bigl( T_{(s+t-1)} - t+1 \bigr) + 1 + T_{s+t-1} \Bigr) \\ & = \bigl( s \, , \, t - 1 + 1 \bigr) \\ & = \bigl( s , t \bigr). \end{align*}
  • Let $n\in \mathbb{N}$ be arbitrary. Recall that \[ B(n) = \Bigl(n - T_{R_n-1}\, , \, - n + 1 + T_{R_n} \Bigr). \]

    For the next calculation we need \[ \bigl(n - T_{R_n-1}\bigr) + \bigl( - n + 1 + T_{R_n} \bigr) - 1 = T_{R_n} - T_{R_n-1} = R_n. \] Comment: If $B(n) = (s,t)$, then this is $s+t-1$ which is used in the definition of $A(s,t)$, but now it is $A\bigl(B(n)\bigr)$.

    Now we evaluate $A\bigl(B(n)\bigr)$: \begin{equation*} A\bigl(B(n)\bigr) = T_{R_n} - \bigl( - n + 1 + T_{R_n} \bigr) + 1 = n. \end{equation*}

• I wrote a Mathematica notebook NbijectionNxN.nb in which I experimented with the function $A$ above and its inverse. A pdf printout of this notebook is NbijectionNxN.pdf. You can download the notebook here NbijectionNxN.nb
• Computer algebra system Mathematica is very useful for Mathematical explorations. To get started with Mathematica visit my dedicated Mathematica page. There you will find my brief introduction to Mathematica and a selection of Mathematica videos providing a practical and efficient way to familiarize yourself with incredible Mathematical tool. Mathematica is available in the computer labs located in BH 215 and BH 209, look in the Start Menu under Wolfram Mathematica.

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Friday, November 3, 2023 (updated Sunday, November 5, 2023)

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• On Wednesday, I summarized how we use $\mathbb{N}$ to give a formal definition of a finite set and how special subsets of $\mathbb{N}$ are used for counting. A formal definition of a finite set leads to the definition of an infinite set. Before that we emphasised two other special features of the set $\mathbb{N}$: recursive definitions of functions with the domain $\mathbb{N}$ and Proofs by Mathematical Induction.
• The following five special features of the set $\mathbb{N}$ of natural numbers (positive integers) are worth emphasising in an ordered list:
  • A function with the domain $\mathbb{N}$ can be defined using a Recursive Definition.
  • If $P(n)$ is a statement involving $n \in \mathbb{N}$, than the proposition \[ \forall\mkern 1pt n \in \mathbb{N} \mkern 5pt P(n), \] that is, \[ \text{for all} \ \ n \in \mathbb{N} \ \ \text{we have that} \ \ P(n) \ \ \text{is true}, \] can be proved by Mathematical Induction.
  • The special subsets ${[\mkern -1.4pt [ 1, n ]\mkern -1.4pt]}_{\mathbb{N}}$ of $\mathbb{N}$ are used to give a formal definition of a finite set and for counting.
  • The formal definition of a finite set leads to the definition of an infinite set, and $\mathbb{N}$ is an example of the "simplest" infinite set.
  • The set $\mathbb{N}$ is used to define a countable set: A nonempty set $A$ is countable if there exists a bijection $f:\mathbb{N}\to A$. (In some sense, countable sets are the simplest infinite sets.)
• Infinite sets are scary. We need to handle them with care. And that means with rigor. When handling a specific infinite set, it is essential to recognize if it has a property that resembles a common property of a finite set. The last sentence is vague; it is like a guiding principle. I will emphasize the occurrences of that principle in proofs.

• The set $\mathbb{N}$ is infinite.

  Proof. We proved the following implication: Let $A$ be a nonempty subset of $\mathbb{R}$. If $A$ is finite, then $A$ has a maximum. The contrapositive of the preceding implication is: Let $A$ be a nonempty subset of $\mathbb{R}$. If $A$ does not have a maximum, then $A$ is infinite. Now we prove that $\mathbb{N}$ does not have a maximum. Let $n \in \mathbb{N}$ be arbitrary. By the definition of $\mathbb{N}$, we have $n+1 \in \mathbb{N}$. By Axiom OA, we have $n\lt n+1$. Thus, $\mathbb{N}$ does not have a maximum. Consequently, $\mathbb{N}$ is infinite.

• So, the preceding item tells us $\mathbb{N}$ is scary. The next item will tell us that $\mathbb{N}$ shares a feature with finite sets. Recall that we proved the following implication: Let $A$ be a nonempty subset of $\mathbb{R}$. If $A$ is finite, then $A$ has a minimum. Recall that we also proved: If $B$ is a nonempty subset of a finite set $A$, then $B$ is finite. Together these two implications yield the following implication: Let $A$ be a subset of $\mathbb{R}$. If $B$ is a nonempty subset of a finite set $A$, then $B$ has a minimum. In the next item we will prove that $\mathbb{N}$ has the same property.

• If $A$ is a nonempty subset of $\mathbb{N}$, then $A$ has a minimum.

  Proof. Let $A \neq \emptyset$ and $A \subseteq \mathbb{N}$. Since $A \neq \emptyset$ there exists $n \in A$. Since $A \subseteq \mathbb{N}$. Now consider the set \[ B = A \cap {[\mkern -1.4pt [ 1, n ]\mkern -1.4pt]}_{\mathbb{N}} \] Since $n \in A$ and $n \in {[\mkern -1.4pt [ 1, n ]\mkern -1.4pt]}_{\mathbb{N}}$, we have $n\in B$. Thus $B\neq \emptyset.$ Clearly, $B \subseteq {[\mkern -1.4pt [ 1, n ]\mkern -1.4pt]}_{\mathbb{N}}.$ Since $B$ is nonempty sunset of a finite set ${[\mkern -1.4pt [ 1, n ]\mkern -1.4pt]}_{\mathbb{N}},$ $B$ has a minimum. Set $b = \min B.$ Thus $b \in B$ and for all $x \in B$ we have $b \leq x$. Since $n \in B$, we have $b \leq n$. Since $B\subseteq A$, we have that $b \in A$. Next we will prove that for all $y \in A$ we have $b \leq y$. Let $y \in A$ be arbitrary. By Axiom OE, we have the trichotomy $y \lt n$, $y = n$, or $n \lt y$. We continue the proof with three two cases. Case 1: $y \leq n$. In this case $y \in A$ and $y \in {[\mkern -1.4pt [ 1, n ]\mkern -1.4pt]}_{\mathbb{N}}.$ hence $y \in B$. Since $b = \min B$, we have $b \leq y$. Case 2: $n \lt y$. We proved that $b \leq n$. By Axiom OT, we have $b \lt y$. Thus, in each case, $b \leq y$. Since $y\in A$ was arbitrary, we have proved for all $y \in A$ we have $b \leq y$. Earlier we proved $b \in A$. Consequently $b = \min A$.

• The preceding theorem is used all the time in Number Theory. Sometimes it is used that to prove that certain sets on positive integers are empty. For example, consider the set \[ A = \bigl\{ p \in \mathbb{N} : \exists\mkern 1pt q \in \mathbb{N} \ \ \text{s.t.} \ \ p^2 = 2 q^2 \bigr\}. \] You probably learned in other classes that the above set $A$ is empty. Can you recall how that was proved? Although, I do not like proofs by contradiction, I will use a proof by contradiction here. (In the notes I proved this directly, not using a proof by contradiction.)

  Proof of $A = \emptyset$. Assume that $A \neq \emptyset.$ Next we will prove that $A$ does not have a minimum. Let $a \in A$ be arbitrary. By the definition of $A$, there exists $b \in \mathbb{N}$ such that $a^2 = 2 b^2$. Here we use background knowledge: a positive integer is even if and only if its square is even. Since $a^2 = 2 b^2$, mark style="background-color:#33FF33;">$a^2$ is even. By the background knowledge, it follows that $a$ is even. Hence, there exists $c \in \mathbb{N}$ such that a=2c$. Consequently, substituting a=2c$ in $a^2 = 2 b^2$, we get $4 c^2 = 2 b^2$. Thus, $b^2 = 2 c^2$ with $c \in \mathbb{N}.$ Consequently, $b \in A$. Since $a^2 = 2 b^2 \gt b^2$, by background knowledge: Let $x, y \gt 0$. Then $x \lt y$ if and only if $x^2 \lt y^2$.. Hence, $a^2 = 2 b^2 \gt b^2$, implies $a \gt b$. Since $b \in A$, we proved that $a$ is not a minimum of $A$. This contradicts the theorem in the preceding item.

• Finally, in the next item we prove that each infinite subset of $\mathbb{R}$ has a subset which either does not have a minimum or it does not have a maximum. So, each infinite subset of $\mathbb{R}$ has a subset that genuinely differs from a finite set.

• Let $A \subseteq \mathbb{R}$. If $A$ is infinite, then there exists a nonempty subset $B$ of $A$ such that $B$ does not have a minimum or there exists a nonempty subset $C$ of $A$ such that $C$ does not have a maximum.

  Proof. We will prove the equivalent implication: If $A$ is an infinite subset of $\mathbb{R}$ and each nonempty subset of $A$ has a minimum, then there exist a nonempty subset $C$ of $A$ such that $C$ does not have a maximum.

  Assume that $A$ is an infinite subset of $\mathbb{R}$ and each nonempty subset of $A$ has a minimum. Then, in particular, $\min A$ exists. Let $W$ be the set of all minimums of infinite subsets of $A$. Formally, \begin{equation*} W = \Bigl\{x \in A \ : \ x = \min E \ \ \text{where} \ \ E \subset A \ \ \text{and} \ \ E \ \text{is infinite} \ \Bigr\}. \end{equation*} Clearly $\min A$ is an element in $W$. Hence $W \neq \emptyset$.

  Next we will prove that $W$ does not have a maximum. Let $y \in W$ be arbitrary. Then there exists an infinite subset $F$ of $A$ such that $y = \min F$. Since $F$ is infinite, the set $F\setminus\{y\}$ is also infinite. Since $F\setminus\{y\} \subset A$, by the assumption $z = \min\bigl(F\setminus\{y\}\bigr)$ exists. Therefore, $z \in W$. Since $z \in F\setminus\{y\}$, we have $z \neq y$. Since $z \in F$ and $y = \min F$, we have $z \geq y$. Hence $z > y$. Thus, for each $y \in W$ there exists $z \in W$ such that $z > y$. This proves that $W$ is a nonempty subset of $A$ which does not have a maximum.

• Few blackboards: (thank you Trever)

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Wednesday, November 1, 2023

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• Section 3.8 presents another important application of the natural numbers: counting. In this section I introduce special notation for special subsets of $\mathbb{N}$ which are used for counting. Let $n \in \mathbb{N}$. We set \[ {[\mkern -1.4pt [ 1, n ]\mkern -1.4pt]}_{\mathbb{N}} = \bigl\{ k \in \mathbb{N} \, : \, k \leq n \bigr\}. \] For example \begin{align*} {[\mkern-1.4pt [ 1, 1 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1\} \\ {[\mkern-1.4pt [ 1, 2 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2\} \\ {[\mkern-1.4pt [ 1, 3 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2,3\} \\ {[\mkern-1.4pt [ 1, 4 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2,3,4\} \\ {[\mkern-1.4pt [ 1, 5 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2,3,4,5\} \\ {[\mkern-1.4pt [ 1, 6 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2,3,4,5,6\} \\ {[\mkern-1.4pt [ 1, 7 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2,3,4,5,6,7\} \\ {[\mkern-1.4pt [ 1, 8 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2,3,4,5,6,7,8\} \\ {[\mkern-1.4pt [ 1, 9 ]\mkern-1.4pt]}_{\mathbb{N}} & = \{1,2,3,4,5,6,7,8,9\} \\ & \mkern 10mu \vdots \end{align*}
• Definition. A nonempty set $A$ is finite if there exists $n \in \mathbb{N}$ and a bijection \[ f: {[\mkern-1.4pt [ 1, n ]\mkern-1.4pt]}_{\mathbb{N}} \to A \] In this case we say that $A$ has $n$ elements and write $\operatorname{card}(A) = n$.
• In the notes I state several propositions about finite set which one can consider common sense. However, we now have tools to prove them. Although, for the lack of time we will not do the proofs in class. If you are curious, think about the statements and we can discuss proofs on Canvas. I will state those propositions here:
  • Exercise 3.8.2. If $A$ is finite and $b \not\in A$, then $A\cup \{b\}$ is finite and $\operatorname{card}\bigl( A\cup \{b\} \bigr) = 1 + \operatorname{card}(A)$.
  • Exercise 3.8.3. A nonempty set $A$ is finite if and only if there exists $n \in \mathbb{N}$ and a surjection $f:{[\mkern-1.4pt [ 1, n ]\mkern-1.4pt]}_{\mathbb{N}} \to A$. In this case $\operatorname{card}(A) \leq n$.
  • Exercise 3.8.4. A nonempty subset $B$ of a finite set $A$ is finite and we have $\operatorname{card}(B) \leq \operatorname{card}(A)$.
  • Exercise 3.8.5. A nonempty proper subset $B$ of a finite set $A$ is finite and we have $\operatorname{card}(B) \lt \operatorname{card}(A)$.
  • Exercise 3.8.6. If $A$ is a finite subset of $\mathbb{R}$, then $A$ has a minimum and $A$ has a maximum. In other words, $\min A$ and $\max A$ exist.
  • The last exercise is used very often.
• Having a rigorous definition of a finite set, we can define an infinite set. A nonempty set which is not finite is said to be infinite. But, I do not like definitions that use negation; it leaves to the reader to formulate that negation; and there are several equivalent characterizations of finite sets. The most convenient characterization of an infinite set is obtained by formulating the negation of the characterization of a finite set in Exercise 3.8.3.:
  • A nonempty set $A$ is infinite if and only if for all $n \in \mathbb{N}$ and all functions $f:{[\mkern-1.4pt [ 1, n ]\mkern-1.4pt]}_{\mathbb{N}} \to A$ we have that $f$ is not a surjection.
  • Although, a much easier way to prove that a set is infinite is to use the contrapositive of the implication proved in Exercise 3.8.6. Briefly, if $A$ is finite, then $A$ has a maximum. Formally, \[ A \ \ \text{finite} \quad \Rightarrow \quad \exists \mkern 1mu a \in A \ \ \text{such that} \ \ \forall \mkern 1mu x \in A \ \ \text{we have} \ \ x \leq a. \] The contrapositive is \[ \forall \mkern 1mu x \in A \ \ \exists \mkern 1mu y \in A \ \ \text{such that} \ \ x \lt y \quad \Rightarrow \quad A \ \ \text{infinite}. \] For example, we can use the preceding implication to prove that $\mathbb{N}$ is an infinite set.
• Notice the following trichotomy for sets:
  • the empty set,
  • finite sets,
  • infinite sets.
  For full disclosure, most writings about sets consider the empty set to be a finite set. Then, the number of elements in the empty set is $0$. However, the empty set does not have either the minimum or the maximum. And I consider the existence of the minimum and the maximum the cornerstone properties of finite sets. Thus, to say that every finite set has a minimum and a maximum, I must exclude the empty set from being finite. (I do not want to qualify this statement by saying that every nonempty finite set has a minimum and a maximum.) Another motivation for this trichotomy is the famous trichotomy for real numbers: negative, zero, and positive.

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Monday, October 30, 2023

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• On Friday and today we defined and discussed the set $\mathbb{N}$ of natural numbers (that is, positive integers).
• The most important properties of the set $\mathbb{N}$ of natural numbers in Section 3.6 are Theorem 3.6.4, the Principle of Mathematical Induction, and Proposition 3.6.7, the Principle of Recursive Definition of functions on $\mathbb{N}$.
• All Examples and Exercises in Section 3.7 are important. I hope that some of this content you have seen in other classes.

• In Problem 3 on Homework 3, I ask you to construct a bijection between two intervals. Each student was assigned a different pair of intervals. For the structure of your answer use my posts on Monday, October 23, 2023 and Tuesday, October 24, 2023.

  Notice that my presentation in most of the cases is not complete. In each case, I present a formula for bijection, and the formula for its inverse. But notice that I do not prove that the formula given for the inverse, is truly the inverse.

  For example, in Bijection of kind E I claim: The function defined by \[ x \mapsto \frac{x}{1+|x|}, \qquad x\in \mathbb{R}, \] is a bijection defined on $\mathbb{R}$ with the range $(-1,1)$. The inverse of this bijection is \[ y \mapsto \frac{y}{1-|y|}, \qquad y \in (-1,1). \]

  Next we prove that the given formula is truly the inverse of the given function defined on $\mathbb{R}$.

  Step 1. Let $y \in (-1,1)$ be arbitrary. Then $\dfrac{y}{1-|y|} \in \mathbb{R}$. Notice that $1-|y| \gt 0$ and calculate \[ \frac{\dfrac{y}{1-|y|}}{1+\left|\dfrac{y}{1-|y|}\right|} = \frac{\dfrac{y}{1-|y|}}{1+\dfrac{|y|}{1-|y|}} = \frac{\dfrac{y}{1-|y|}}{\dfrac{1- |y| + |y|}{1-|y|}} = \frac{\dfrac{y}{1-|y|}}{\dfrac{1}{1-|y|}} = y. \] Thus, for arbitrary $y \in (-1,1)$, for $x = \dfrac{y}{1-|y|} \in \mathbb{R}$ we have $\dfrac{x}{1+|x|} = y$. This proves that the range of the given function is $(-1,1)$.

  Step 2. Let $x \in \mathbb{R}$ be arbitrary. Then \[ -|x| \leq x \leq |x|. \] Further, by OA, we get \[ -1 -|x| \lt x \lt 1 + |x|. \] Since $1+|x| \gt 0$, by OM, we get \[ -1 \lt \dfrac{x}{1+|x|} \lt 1. \] That is $\dfrac{x}{1+|x|} \in (-1,1)$. Next we substitute this value in the formula for the proposed inverse and get \[ \dfrac{\dfrac{x}{1+|x|}}{1-\left|\dfrac{x}{1+|x|}\right|} = \dfrac{\dfrac{x}{1+|x|}}{1-\dfrac{|x|}{1+|x|}} = \dfrac{\dfrac{x}{1+|x|}}{\dfrac{1+|x| - |x|}{1+|x|}} = \dfrac{\dfrac{x}{1+|x|}}{\dfrac{1}{1+|x|}} = x. \]

  Step 1 and Step 2 together with Exercise 2.2.3 imply that the given function is a bijection between $\mathbb{R}$ and $(-1,1)$.

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Thursday, October 26, 2023

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• Today in class I illustrated the concepts of a set bounded above and a set bounded below using the set introduced in Exercise 3.5.6. This exercise asks us to prove that the set \[ A = \bigl\{x \in \mathbb{R} : x^2 \lt 2 \bigr\} \] is a bounded set. I proved this in class. But, the argument that I offered used Exercise 3.2.7. This is, of course, valid, but a simpler proof is always desirable. At the end of the class Ugo approached me with a different idea on how to prove the claim in this exercise. Ugo's idea was to use the following statement: \[ \forall \mkern 1mu x \in \mathbb{R} \qquad x \gt 1 \ \ \Rightarrow \ \ x \lt x^2. \] To prove this statement, let $x \in \mathbb{R}$ be arbitrary and assume $x \gt 1$ be arbitrary. Since $1 \gt 0$, we have $x \gt 0$. By Axiom OM, $1 \lt x$ and $0 \lt x$ implies $x \lt x^2$. (Looking carefully at the notes, one can also use Exercise 3.2.8.)

  • Proof. We need to prove that the set \[ A = \bigl\{x \in \mathbb{R} : x^2 \lt 2 \bigr\} \] is bounded above and that it is bounded below. We will prove that $A$ an upper bound for $A$ is $2.$ That is, we need to prove that for all $x \in A$ we have $x\lt 2.$

    Let $x \in A$ be arbitrary. By definition of $A$ we have $x^2 \lt 2.$ Now we distinguish two cases, $1 \lt x$ and $x \leq 1.$ Case 1. Assume $1 \lt x.$ Then, by Ugo's idea, we have $x \lt x^2.$ Now, $x \lt x^2$ and $x^2 \lt 2$, and Axiom OT imply $x \lt 2.$ Case 2. Assume $x \leq 1.$ Now $x \leq 1$ and $1 \lt 2$ and Axiom OT imply $x \lt 2.$

    In each case we have proved that $x\in A$ implies $x^2 \lt 2.$ Thus, $2$ is an upper bound for $A.$

    Next we need to prove that $-2$ is a lower bound for $A.$ That is, we need to prove that for all $x \in A$ we have $-2 \lt x.$ To prove this statement we use Exercise 3.1.2(f). This exercise implies that for all $x \in \mathbb{R}$ we have $(-x)^2 = x^2$.

    Let $x \in A$ be arbitrary. By definition of $A$ we have $x^2 \lt 2.$ Since $(-x)^2 = x^2,$ we have $(-x)^2 \lt 2.$ Therefore $-x \in A.$ We already proved that $2$ is an upper bound for $A.$ Therefore, $-x \lt 2.$ Adding $-2+x$ to both sides of this inequality and Axiom OA, yield $-2 \lt x.$

    In conclusion, we proved that \[ \forall \mkern 2mu x \in A \qquad -2 \lt x \land x \lt 2, \] or, in different notation, \[ A \subseteq (-2,2). \]

• In class I mentioned that it is intuitively clear that the set \[ A = \bigl\{x \in \mathbb{R} : x^2 \lt 2 \bigr\} \] does not have a maximum. However, a proof of this fact requires quite a bit of algebraic fiddling. Please think about it and let me know of your ideas. Recall that we need to prove \[ \forall \mkern 1mu x \in A \quad \exists \mkern 1mu y \in A \quad \text{such that} \quad x \lt y. \] Basically, for a given $x^2 \lt 2$, we need to come up with a formula for $y$ in terms of $x$ such that $x \lt y$ and $y^2 \lt 2.$

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Tuesday, October 24, 2023

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• Thank you Bennett for the pictures. The picture below shows the construction of a bijection \[ f: [0,+\infty) \to (0,+\infty). \] It is convenient here to introduce the notation \[ \mathbb{R}_+ = (0,+\infty). \] The logic of this notation is that we set the subscript $+$ to indicate that in $\mathbb{R}_+$ we include all positive real numbers.

  In the definition of the bijection $f$ we use the set of positive integers \[ \mathbb{N} = \bigl\{1,2,3,4,5,6,7,8,9, \ldots \ \bigr\}. \quad \text{the lower dots indicate that the pattern indicated continues} \] We will give a rigorous definition of the set $\mathbb{N}$ later this week. The definition of $f$ is based on the following disjoint unions \begin{alignat*}{2} [0,+\infty) & = \bigl(\{0\}\cup \mathbb{N}\bigr) \cup \bigl(\mathbb{R}_+ \setminus \mathbb{N}\bigr), && \quad \text{the blue set on the bb} \\ (0,+\infty) & = \mathbb{N} \cup \bigl(\mathbb{R}_+ \setminus \mathbb{N}\bigr). && \quad \text{the yellow set on the bb} \end{alignat*}

  The definition of \[ f: [0,+\infty) \to (0,+\infty) \] is piecewise, one formula is for $x \in \{0\}\cup \mathbb{N}$ and another formula is for $x\in \mathbb{R}_+ \setminus \mathbb{N}$: \[ f(x) = \begin{cases} x+1 & \quad \text{if} \quad x\in \{0\}\cup \mathbb{N}, \\[5pt] x & \quad \text{if} \quad x\in \mathbb{R}_+ \setminus \mathbb{N}. \end{cases} \]

  At the top of the blackboard in the picture above is a composition of three bijections which provides a bijection between the half-open interval $(0,1]$ and the open interval $(0,1)$. \[ (0,1] \mkern 5mu \xrightarrow{\mkern 10mu \dfrac{1}{x}-1 \mkern 10mu} \mkern 5mu [0,+\infty) \mkern 5mu \xrightarrow{\mkern 10mu f \mkern 10mu} \mkern 5mu (0,+\infty) \mkern 5mu \xrightarrow{\mkern 10mu \dfrac{1}{1+x} \mkern 10mu} \mkern 5mu (0,1) \]

  The bijection presented in the preceding displayed formula we denote by \[ h: (0,1] \to (0,1). \]
• On the blackboard below, I used the bijection \[ h: (0,1] \to (0,1). \] from the preceding item to define a bijection \[ k : [-1,1] \to (-1,1) \] as follows \[ k(x) = \begin{cases} -h(-x) & \quad \text{if} \quad x\in [-1,0), \\[5pt] 0 & \quad \text{if} \quad x = 0, \\[5pt] h(x) & \quad \text{if} \quad x\in (0,1]. \end{cases} \] On the blackboard below, the bijection I call $k : [-1,1] \to (-1,1)$ above is called $l(x)$.

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Monday, October 23, 2023

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• Today we talked about bijections between intervals in $\mathbb{R}.$ Here we group the intervals in three classes: Closed Intervals, Open Intervals, and Half-Open Intervals.

• Let $a, b \in \mathbb{R}$ be such that $a \lt b$. By $[a,b]$ we denote the closed interval with the end-points $a$ and $b$. Its definition is \[ [a,b] = \bigl\{x \in \mathbb{R} : a \leq x \land x \leq b \bigr\}. \] The most popular examples are $[-1,1]$ and $[0,1]$.

  Bijections of kind A. For any two closed intervals, there exists a bijection given by a relatively simple formula. Let $a, b, c, d \in \mathbb{R}$ be such that $a \lt b$ and $c \lt d$. The following function \[ x \mapsto \frac{d-c}{b-a}(x-a) + c, \qquad x \in [a,b], \] is a bijection defined on $[a,b]$ with the range $[c,d]$. The inverse of this function is \[ y \mapsto \frac{b-a}{d-c}(y-c) + a, \qquad y \in [c,d]. \]

• Let $a, b \in \mathbb{R}$ be such that $a \lt b$. By $(a,b)$ we denote the open interval with the end-points $a$ and $b$. Its definition is \[ (a,b) = \bigl\{x \in \mathbb{R} : a \lt x \land x \lt b \bigr\}. \] The most popular examples are $(-1,1)$ and $(0,1)$.

  Let $c \in \mathbb{R}$. The following intervals are also open intervals: \begin{align*} (c,+\infty) & = \bigl\{ x \in \mathbb{R} : c \lt x \bigr\}, \\ (-\infty, c) & = \bigl\{ x \in \mathbb{R} : x \lt c \bigr\}. \end{align*} The most popular examples are $(0,+\infty)$ and $(-\infty, 0)$. Since these two intervals appear often, we often use the following notation \[ \mathbb{R}_+ = (0,+\infty), \qquad \mathbb{R}_- = (-\infty, 0). \]

  The set of real numbers $\mathbb{R}$ is also an open interval. Sometimes it is denoted by $(-\infty, + \infty).$

  For any two finite open intervals, there exists a bijection given by a relatively simple formula.

  Bijections of kind B. Let $a, b, c, d \in \mathbb{R}$ be such that $a \lt b$ and $c \lt d$. The following function \[ x \mapsto \frac{d-c}{b-a}(x-a) + c, \qquad x \in (a,b), \] is a bijection defined on $(a,b)$ with the range $(c,d)$. The inverse of this function is \[ y \mapsto \frac{b-a}{d-c}(y-c) + a, \qquad y \in (c,d). \]

  Bijections of kind C. Let $c, d \in \mathbb{R}$. C1. The following function \[ x \mapsto d-c+x, \qquad x \in (c,+\infty), \] is a bijection defined on $(c,+\infty)$ with the range $(d,+\infty)$. The inverse of this function is \[ y \mapsto c-d+x, \qquad y \in (d,+\infty). \] C2. The following function \[ x \mapsto d+c-x, \qquad x \in (c,+\infty), \] is a bijection defined on $(c,+\infty)$ with the range $(-\infty,d)$. The inverse of this function is \[ y \mapsto c+d-x, \qquad y \in (-\infty,d). \] C3. The following function \[ x \mapsto d-c+x, \qquad x \in (-\infty,c), \] is a bijection defined on $(-\infty,c)$ with the range $(-\infty,d)$. The inverse of this function is \[ y \mapsto c-d+x, \qquad y \in (-\infty,d). \]

  Bijections of kind D. A bijection between $(0,1)$ and $(0,+\infty) = \mathbb{R}_+$ is given by \[ x \mapsto \frac{1}{x} - 1, \qquad x\in (0,1). \] The inverse of this bijection is \[ y \mapsto \frac{1}{1+y}, \qquad y \in (0,+\infty) = \mathbb{R}_+. \] D1. Let $a, b, c \in \mathbb{R}$ be such that $a \lt b$. The following function \[ x \mapsto \frac{b-x}{x-a} + c, \qquad x \in (a,b), \] is a bijection defined on $(a,b)$ with the range $(c,+\infty)$. The inverse of this function is \[ y \mapsto \frac{b-a}{1-c + y} + a, \qquad x \in (c,+\infty) \] D2. Let $a, b, c \in \mathbb{R}$ be such that $a \lt b$. The following function \[ x \mapsto -\frac{b-x}{x-a} + c, \qquad x \in (a,b), \] is a bijection defined on $(a,b)$ with the range $(-\infty,c)$. The inverse of this function is \[ y \mapsto \frac{b-a}{1+ c - y} + a, \qquad x \in (-\infty,c) \]

  Bijection of kind E. The function defined by \[ x \mapsto \frac{x}{1+|x|}, \qquad x\in \mathbb{R}, \] is a bijection defined on $\mathbb{R}$ with the range $(-1,1)$. The inverse of this bijection is \[ y \mapsto \frac{y}{1-|y|}, \qquad y \in (-1,1). \] Let $a, b \in \mathbb{R}$ be such that $a \lt b$. We compose the preceding bijection with a bijection of kind B to get a bijection with domain $\mathbb{R}$ and the range $(a,b)$. The following function \[ x \mapsto \frac{b-a}{2}\left(\frac{x}{1+|x|}+1\right) + a, \qquad x \in \mathbb{R}, \] is a bijection defined on $\mathbb{R}$ with the range $(a,b)$. The inverse of this function is \[ y \mapsto \frac{\frac{2(x-a)}{b-a}-1}{1-\left|\frac{2(x-a)}{b-a}-1\right|}, \qquad y \in (a,b). \]

  Bijection of kind F. Let $a\in \mathbb{R}$. F1. The following function \[ x \mapsto ???, \qquad x \in \mathbb{R}, \] is a bijection defined on $\mathbb{R}$ with the range $(a,+\infty)$.

  F2. The following function \[ x \mapsto ???, \qquad x \in \mathbb{R}, \] is a bijection defined on $\mathbb{R}$ with the range $(-\infty,a)$.

• Let $a, b \in \mathbb{R}$ be such that $a \lt b$. The half-open intervals with the end-points $a$ and $b$ are defined as follows: \begin{align*} [a,b) & = \bigl\{ x \in \mathbb{R} : a \leq x \land x \lt b \bigr\}, \\ (a,b] & = \bigl\{ x \in \mathbb{R} : a \lt x \land x \leq b \bigr\}. \end{align*} The most popular examples are $[-1,0)$ and $(0,1]$.

  Let $c \in \mathbb{R}$. The following intervals are also half-open intervals: \begin{align*} [c,+\infty) & = \bigl\{ x \in \mathbb{R} : c \leq x \bigr\}, \\ (-\infty, c] & = \bigl\{ x \in \mathbb{R} : x \leq c \bigr\}. \end{align*}

  Bijection of kind G. Let $a, b, c, d \in \mathbb{R}$ be such that $a \lt b$ and $c\lt d$.

  G1. The following function \[ x \mapsto ???, \qquad x \in [a,b), \] is a bijection defined on $[a,b)$ with the range $[c,d)$.

  G2. The following function \[ x \mapsto ???, \qquad x \in [a,b), \] is a bijection defined on $[a,b)$ with the range $(c,d]$.

  Let $a, b, c \in \mathbb{R}$ be such that $a \lt b$.

  G3. The following function \[ x \mapsto ???, \qquad x \in [a,b), \] is a bijection defined on $[a,b)$ with the range $[c,+\infty)$.

  G4. The following function \[ x \mapsto ???, \qquad x \in [a,b), \] is a bijection defined on $[a,b)$ with the range $(-\infty,c]$.

  G5. The following function \[ x \mapsto ???, \qquad x \in (a,b], \] is a bijection defined on $(a,b]$ with the range $[c,+\infty)$.

  G6. The following function \[ x \mapsto ???, \qquad x \in (a,b], \] is a bijection defined on $(a,b]$ with the range $(-\infty,c]$.

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Sunday, October 22, 2023

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• On Thursday and Friday we talked about intervals in $\mathbb{R}.$ Intervals are special subsets of $\mathbb{R}$. Here I will group the intervals in three classes:

  Let $a, b \in \mathbb{R}$ be such that $a \lt b$. By $[a,b]$ we denote the closed interval with the end-points $a$ and $b$. Its definition is \[ [a,b] = \bigl\{x \in \mathbb{R} : a \leq x \land x \leq b \bigr\}. \] The most popular examples are $[-1,1]$ and $[0,1]$.

  Let $a, b \in \mathbb{R}$ be such that $a \lt b$. By $(a,b)$ we denote the open interval with the end-points $a$ and $b$. Its definition is \[ (a,b) = \bigl\{x \in \mathbb{R} : a \lt x \land x \lt b \bigr\}. \] The most popular examples are $(-1,1)$ and $(0,1)$.

  Let $c \in \mathbb{R}$. The following intervals are also open intervals: \begin{align*} (c,+\infty) & = \bigl\{ x \in \mathbb{R} : c \lt x \bigr\}, \\ (-\infty, c) & = \bigl\{ x \in \mathbb{R} : x \lt c \bigr\}. \end{align*}

  The set of real numbers $\mathbb{R}$ is also an open interval. Sometimes it is denoted by $(-\infty, + \infty).$

  Let $a, b \in \mathbb{R}$ be such that $a \lt b$. The half-open intervals with the end-points $a$ and $b$ are defined as follows: \begin{align*} [a,b) & = \bigl\{ x \in \mathbb{R} : a \leq x \land x \lt b \bigr\}, \\ (a,b] & = \bigl\{ x \in \mathbb{R} : a \lt x \land x \leq b \bigr\}. \end{align*} The most popular examples are $[-1,0)$ and $(0,1]$.

  Let $c \in \mathbb{R}$. The following intervals are also half-open intervals: \begin{align*} [c,+\infty) & = \bigl\{ x \in \mathbb{R} : c \leq x \bigr\}, \\ (-\infty, c] & = \bigl\{ x \in \mathbb{R} : x \leq c \bigr\}. \end{align*}

• On Monday we will look at bijections between intervals. For each of the three classes for any two intervals in that class, there exists a bijection defined by a formula that you might have encountered in Precalculus mathematics.
• On Friday we discussed unions and intersections of infinitely many intervals, see Exercise 3.4.4 through Exercise 3.4.8. Exercise 3.4.6 is solved in the notes.
• Exercise 3.4.4 is as follows: Let $a \in \mathbb{R}$. Prove \[ \bigcap \bigl\{(a-u, a+u) : u \gt 0 \bigr\} = \{a\}. \] What is the meaning of the expression \[ \bigcap \bigl\{(a-u, a+u) : u \gt 0 \bigr\} ? \] Here $\bigl\{(a-u, a+u) : u \gt 0 \bigr\}$ stands for the family of open intervals. For every $u \gt 0$ we are given an interval $(a-u,a+u)$. This is a set of sets, which we often call a family of sets.
• Given any nonempty family of sets $\mathcal{A}$ we write \[ \bigcap \bigl\{A : A \in \mathcal{A} \bigr\}. \] The meaning of this expression is as follows \[ x \in \bigcap \bigl\{A : A \in \mathcal{A} \bigr\} \quad \Leftrightarrow \quad \forall \mkern 0mu A \in \mathcal{A} \ \ \text{we have} \ \ x \in A. \]
• Similarly, given any nonempty family of sets $\mathcal{A}$ we write \[ \bigcup \bigl\{A : A \in \mathcal{A} \bigr\}. \] The meaning of this expression is as follows \[ x \in \bigcup \bigl\{A : A \in \mathcal{A} \bigr\} \quad \Leftrightarrow \quad \exists \mkern 0mu A \in \mathcal{A} \ \ \text{such that} \ \ x \in A. \]
• Thus, to prove \[ \bigcap \bigl\{(a-u, a+u) : u \gt 0 \bigr\} = \{a\}, \] we have to prove the following equivalence \[ x \in \bigcap \bigl\{(a-u, a+u) : u \gt 0 \bigr\} \quad \Leftrightarrow \quad x = a. \] In other words, we have to prove \[ \forall \mkern 1mu u \gt 0 \ \ \text{we have} \ \ x \in (a-u, a+u) \quad \Leftrightarrow \quad x = a. \] Let us prove the preceding equivalence in two steps. First we proved $\Leftarrow$, then we prove $\Rightarrow$.
  • Assume that $x=a$. Let $u \gt 0$ be arbitrary. By Axiom OA, adding $a$ to both sides of $0 \lt u$ we get $a \lt a+u$. Adding $a-u$ to both sides of $0 \lt u$ we get $a-u \lt a$. Hence $x=a \in (a-u, a+u)$. Since $u \gt 0$ was arbitrary we proved $\Leftarrow$.
  • Assume that \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad x \in (a-u, a+u). \] Then \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad a-u \lt x \land x \lt a+u. \] We can rewrite the last quantified statement as two quantified statements \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad a-u \lt x \] and \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad x \lt a+u. \] Applying Axiom OA, from the preceding two statements we obtain \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad a-x \lt u \] and \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad x - a \lt u. \] Now recall Exercise 3.2.11: \[ \forall \mkern 1mu u \gt 0 \ \ \text{we have} \ \ \alpha \lt u \quad \Rightarrow \quad \alpha \leq 0. \] Therefore \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad a-x \lt u \quad \Rightarrow \quad a-x \leq 0 \] and \[ \forall \mkern 1mu u \gt 0 \quad \text{we have} \quad x - a \lt u \quad \Rightarrow \quad x-a \leq 0. \] Thus, we proved $a \leq x$ and $x \leq a$. Consequently, $x = a$. In conclusion, we proved \[ \forall \mkern 1mu u \gt 0 \ \ \text{we have} \ \ x \in (a-u, a+u) \quad \Rightarrow \quad x = a. \]

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Thursday, October 19, 2023 (updated)

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• Updated with a proof of Exercise 3.3.17 and related comments.
• Answering your question about a favorite number, I had to mention the Hall of Fame of Numbers: \[ \bigl\{1, 0, \pi, e, i \bigr\}, \] and the fascinating equality that ties these numbers in one equality: \[ e^{i\pi} + 1 = 0. \] Further, I shared with you a number, that I think should be the official number of Western:
  • I found this interesting number on campus.
    Looking for connections between mathematics and art, I discovered a fascinating fact that the mathematical spirit of Isamu Noguchi's Skyviewing Sculpture, which dominates WWU's Red Square, is the remarkable number My talk
    "Numbers in the Sky(viewing Sculpture)"
    was a celebration of this discovery.
  • The short paper Numbers in the Sky(viewing Sculpture) is based on my talk.
  • The slides from this talk might be interesting.
  • This is a handout that I made for this talk. Print this handout on heavy yellow paper.
  • Another handout for the talk. It explains how to make a model of the Skyviewing Sculpture.
  • The following three pdf files are three animations from the talk. Pdf animations will not play in a browser. If you want to see these animations you can download these pdf files and open them in Adobe Acrobat. I hope that the animations will play. Scene 1   Scene 2   Scene 3
• We proved Exercise 3.3.17 in class: For all $x \in \mathbb{R}$ and all $a\in \mathbb{R}\setminus\{0\}$ the following implication holds \[ |x - a| \lt \frac{|a|}{2} \quad \Rightarrow \quad \frac{|a|}{2} \lt |x|. \]
  • Proof. After some explorations it occurred to me that the Reverse Triangle Inequality can be useful here. Recall the Reverse Triangle Inequality \[ \bigl| |a| - |x| \bigr| \leq |x - a| \] and recall that for all $u \in \mathbb{R}$ we have $u \leq |u|$ yields \[ |a| - |x| \leq \bigl| |x| - |a| \bigr|. \] Thus, a special case of the Reverse Triangle Inequality is \[ |a| - |x| \leq |x - a|. \] Assume that \[ |x - a| \lt \frac{|a|}{2}. \] The last two displayed inequalities and Axiom OT yield \[ |a| - |x| \lt \frac{|a|}{2}. \] Now Axiom OA (we are adding $|x| - |a|/2$) yields \[ \frac{|a|}{2} \lt |x|. \] This completes the proof.

  • Why is the inequality that we proved in the previous item important? The proven inequality is important since we can use it to assess the change in the reciprocal function using the change in the variable. For example, let $x,a \in \mathbb{R}\setminus\{0\}$ and we want to estimate the distance \[ \left| \frac{1}{x} - \frac{1}{a} \right| \] by using $|x - a|$. We think of $a$ here as fixed and we think of $x$ as varying.

    First we simplify the expression \[ \left| \frac{1}{x} - \frac{1}{a} \right| = \left| \frac{a - x}{xa} \right| = \frac{1}{|x| \mkern 2mu |a|} |x-a|. \] We do not like $\displaystyle \frac{1}{|x| \mkern 2mu |a|}$ since $x$ is varying so, the previous equality does not give us an estimate for \[ \left| \frac{1}{x} - \frac{1}{a} \right| \qquad \text{using} \qquad |x-a|. \] But, the proven inequality comes to rescue here. Assume that \[ |x - a| \lt \frac{|a|}{2} \quad \text{then we know that} \quad \frac{|a|}{2} \lt |x|. \] So assuming that \[ |x - a| \lt \frac{|a|}{2}, \] we can use $\displaystyle \frac{|a|}{2} \lt |x|$ and the pizza-party-principle to conclude that \[ \frac{1}{|x| \mkern 2mu |a|} |x-a| \leq \frac{1}{\frac{|a|}{2} \mkern 2mu |a|} |x-a| = \frac{2}{|a|^2} |x-a|. \] Since \[ \left| \frac{1}{x} - \frac{1}{a} \right| = \frac{1}{|x| \mkern 2mu |a|} |x-a|, \] we have proved that \[ |x - a| \lt \frac{|a|}{2} \quad \Rightarrow \quad \left| \frac{1}{x} - \frac{1}{a} \right| \leq \frac{2}{|a|^2} |x-a|. \]

  • For example, if $a=2$ we are guarantied to have that \[ |x - 2| \lt 1 \quad \Rightarrow \quad \left| \frac{1}{x} - \frac{1}{2} \right| \leq \frac{1}{2} |x - 2|. \] Further, we have, just for example, \[ |x - 2| \lt \frac{1}{8} \quad \Rightarrow \quad \left| \frac{1}{x} - \frac{1}{2} \right| \lt \frac{1}{16}. \] In general, for all $\epsilon \in (0,1]$ we have \[ |x - 2| \lt \epsilon \quad \Rightarrow \quad \left| \frac{1}{x} - \frac{1}{2} \right| \lt \frac{\epsilon}{2}. \] So the preceding implication yields that if $x$ is close to $2$, that $1/x$ is even closer to $1/2$ and we know exactly how the "closedness" changes.

    However, the siltation is different for $a \in (0,1).$ Let $a=1/3$. We are guarantied to have that \[ \left|x - \frac{1}{3} \right| \lt \frac{1}{6} \quad \Rightarrow \quad \left| \frac{1}{x} - 3 \right| \leq 18 \left|x - \frac{1}{3} \right|. \] Further, we have, just for example, \[ \left|x - \frac{1}{3} \right| \lt \frac{1}{8} \quad \Rightarrow \quad \left| \frac{1}{x} - 3 \right| \lt \frac{18}{8} = \frac{9}{4}. \] In general, for all $\epsilon \in (0,1/6]$ we have \[ \left|x - \frac{1}{3} \right| \lt \epsilon \quad \Rightarrow \quad \left| \frac{1}{x} - 3 \right| \lt 8\epsilon. \] So the preceding implication yields that if $x$ is close to $1/3$, that $1/x$ is can be close to $3$ but we have to choose very small $\epsilon$.

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Tuesday, October 10, 2023

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• On Monday we discussed Exercise 2.2.13 since it is background knowledge for Exercise 2.3.13. I presented a detailed proof of Exercise 2.3.13. (I hope to write this proof on the website later. It would also be nice to write special cases of the proof for the set of positive integers. We will come back to this later.)
• We ended Monday class by presenting the first five axioms of $\mathbb{R}$. Those are the axioms of a commutative group. On Tuesday we presented all eleven axioms of a field. This is presented in Section 3.1 Axioms of a field. We continued by presenting four axioms of order in a field. Those fifteen axioms define an ordered field. Those are the first fifteen axioms of $\mathbb{R}$.
• We ended Tuesday class by showing how the fifteen presented axioms are used to prove the familiar algebraic properties of the real numbers. We presented proofs of Exercise 3.1.2 (b), (d). Neither of these proofs is in the notes. As an exercise, think of reproducing these proofs.
• Further, we proved that $0 \lt 1$. This is proved in the notes in Exercise 3.2.2 (5). In the context of Exercise 3.2.2, one can give a direct proof of $0 \lt 1$. Today in class, I proved $0 \lt 1$ by contradiction.
• I will write more about proofs by contradiction here later. If I forget, please remind me.

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Sunday, October 8, 2023 (corrected)

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• We covered a lot of material on Friday. I would like to emphasize a subtlety in Exercise 2.2.12 that I did not emphasize sufficiently in class. Exercise 2.2.12 reads as follows: Prove that \[ f:\mathbb{R}\setminus\{-1\} \to f:\mathbb{R}\setminus\{-1\} \] defined by \[ f(x) = \frac{1-x}{1+x} \] is a bijection.
  • Before starting this exercise recall the definition of function, for example at Functions, or in the class lecture notes.
  • It is not clear that the formula given in Exercise 2.2.12 defines a function \[ f:\mathbb{R}\setminus\{-1\} \to f:\mathbb{R}\setminus\{-1\}. \] Recall the definition of a function from Functions:
    Definition. Let $A$ and $B$ be nonempty sets. A function $f$ from $A$ to $B$ is a subset of the Cartesian product $A\!\times\!B$ such that

    • Fun 1.  For every $x \in A$ there exists $y \in B$ such that the pair $(x,y)$ belongs to $f$.
    • Fun 2.  If pairs $(x,y_1)$ and $(x,y_2)$ belong to $f$, then $y_1= y_2$.
  • In Exercise 2.2.12 we have \[ A = \mathbb{R}\setminus\{-1\} \quad \text{and} \quad B = A = \mathbb{R}\setminus\{-1\}. \] It is not clear that with the formula \[ f(x) = \frac{1-x}{1+x} \] the property Fun 1 is satisfied. That is, it is not clear that the following statement is true: For every $x \in \mathbb{R}\setminus\{-1\}$ there exists $y \in \mathbb{R}\setminus\{-1\}$ such that the pair $(x,y)$ belongs to $f$? Since $f$ is given by the displayed formula, we have that $y = \frac{1-x}{1+x}$. Thus, we have to prove the following statement \[ \forall \mkern 1mu x \in \mathbb{R}\setminus\{-1\} \quad \text{we have} \quad \frac{1-x}{1+x} \in \mathbb{R}\setminus\{-1\} \]
  • The proof is interesting. Here it is. Assume that $x \in \mathbb{R}\setminus\{-1\}$. This means that $x \in \mathbb{R}$ and $x\neq -1$. Therefore, $1+x \neq 0.$ From algebra of real numbers we know that the following sequence of implications is true \[ 1 \neq -1 \ \ \Rightarrow 1 - x \neq -1 - x \ \ \Rightarrow 1 - x \neq -(1 + x). \] Since $1+x \neq 0,$ it follows from the algebra of real numbers that \[ 1 - x \neq -(1 + x) \quad \Rightarrow \quad \frac{1-x}{1+x} \neq -1. \] Thus, we proved that \[ x \in \mathbb{R}\setminus\{-1\} \quad \Rightarrow \quad \frac{1-x}{1+x} \neq -1. \]
  • Therefore, in Exercise 2.2.12 we are given a function \[ f:\mathbb{R}\setminus\{-1\} \to f:\mathbb{R}\setminus\{-1\} \]
  • To prove that the function $f$ given in Exercise 2.2.12 is a bijection, for $x \in \mathbb{R}\setminus\{-1\}$ we calculate \[ (f\circ f)(x) = \frac{1-f(x)}{1+f(x)} = \frac{1- \frac{1-x}{1+x}}{1+ \frac{1-x}{1+x}} = \frac{\frac{1+x-(1-x)}{1+x}}{\frac{1+x+1-x}{1+x}} = \frac{2x}{2} = x. \]
  • Thus, for all $x \in \mathbb{R}\setminus\{-1\}$ we have $(f\circ f)(x) = x.$ Now, Exercise 2.2.3 implies that $f$ is a bijection. Please make sure that you understand why this is true.

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Thursday, October 5, 2023

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• I was asked to do Exercise 2.2.6 today. That involved a review of the concepts of an injective function, a surjective function, a bijection, and an inverse function. I explained that this exercise is related to Exercise 2.2.4 (I proved this in class) and Exercise 2.2.5 (it is proved in the notes). Then, I used Exercise 2.2.3 (proved in the notes) to complete a proof of Exercise 2.2.6
• Exercise 2.2.12 is a good inspiration for Problem 2 on Homework 1.
• Exercise 2.2.13 is a good practice for forming a negation of a quantified statement. It is solved in the notes. Think about it before reading the solution.
• Exercise 2.2.13 is a good practice for forming a negation of a quantified statement. It is solved in the notes. Think about it before reading the solution.
• For Friday, read Section 2.3: Cardinality of sets  in the class lecture notes. There are three exercises with statements that you can prove: Exercise 2.3.2, Exercise 2.3.3, and Exercise 2.3.4.

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Tuesday, October 3, 2023

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• Today we talked about Section 2.1: Sets  in the class lecture notes. There are three exercises with statements that you can prove: Exercise 2.1.8, Exercise 2.1.10, and Exercise 2.1.15. I did the first part of Exercise 2.1.8 in class.
• Today we started Section 2.2: Functions. Read this section for Thursday. I think that I gave a better presentation of functions at my webpage about Functions. Read this webpage as well. Reading both, the notes and the webpage might inspire some questions for Thursday.

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Monday, October 2, 2023

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• Homework 1 has been posted today at the class Canvas page in Assignments.
  • We discussed the content of Problem 1 today.
  • We will talk about the content related to Problem 2 tomorrow.
• If you have questions, post them at the class Canvas page in Discussions.
  • Even if you don't have a question right now, keep an open mind; one may soon emerge.
  • Remember, everyone has a unique perspective. Only you can pose questions from your specific viewpoint.
  • I've taught some courses for decades and have consistently been amazed by students' questions – ones I'd never even considered. And I genuinely strive to keep my mind always open to new questions.

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Sunday, October 1, 2023

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• Here are few exercises involving quantified statements with three variables. For each statement formulate the negation and decide which one is true: the original statement or its negation. Provide a proof of your claim.
  • $\forall\, a \ \forall\, b \ \exists\, c \quad cb \gt a$    (the universe of discourse consists of all real numbers)
  • $\forall\, a \ \forall\, b \ \exists\, c \quad cb \gt a$    (the universe of discourse consists of all positive real numbers)
  • $\forall\, x \ \exists\, y \ \forall\, z \quad (z \gt y) \ \Rightarrow \ (zx \gt 1)$    (the universe of discourse consists of all real numbers)
  • $\forall\, x \ \exists\, y \ \forall\, z \quad (z \gt y) \ \Rightarrow \ (zx \gt 1)$    (the universe of discourse consists of all positive real numbers)
  • $\forall\, x \ \exists\, y \ \forall\, z \quad (z \gt y) \ \Rightarrow \ (z^2 \gt x^2)$    (the universe of discourse consists of all real numbers)
  • $\forall\, x \ \exists\, y \ \forall\, z \quad (z \gt y) \ \Rightarrow \ (z^2 \gt x)$    (the universe of discourse consists of all real numbers)
• A good practice with quantified statements is to look at grids consisting of randomly colored black and white boxes.
  • Let $B(j,k)$ be the following propositional function: "The box in the $j$-th row and $k$-th column is colored black". Here we assume that the rows are counted from top to bottom and the columns are counted from left to right.
  • Consider grids which consist of black and white cells. For example consider the following four grids:

    Grid 1	Grid 2	Grid 3	Grid 4

    Further, consider the following four propositions. In each of the propositions below, the Universe of Discourse is the set \(\bigl\{1,2,3,4,5,6,7,8,9,10,11\bigr\}.\)
    • Proposition 1.   $\forall j \, \exists k \ B(j,k)$
    • Proposition 2.   $\exists k \, \forall j \ B(j,k)$
    • Proposition 3.   $\exists j \, \forall k \ \neg B(j,k)$
    • Proposition 4.   $\forall k \, \exists j \ B(j,k)$
    For each grid identify the propositions that are true for that grid. For each grid identify the propositions that are false for that grid. State the negations of each of Propositions 1 through 4.

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Thursday, September 28, 2023

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• We talked about Mathematical Logic today. The relevant reading is in the first three sections (1. Logic, 2. Propositional calculus, 3. Implications) of my essay Brief Review of Mathematical Logic.
• For the next class read Section 4: Propositional functions and Section 5: Examples in my essay Brief Review of Mathematical Logic. As a big example of how to do proofs with quantifiers you can read my webpage Mathematical Rigor in the Context of Quadratic Functions.
• Section 1.3 of the class lecture notes is about Mathematical Logic. You can read it, and compare it with my Brief Review of Mathematical Logic. I am curious about your thoughts on Exercise 1.3.3 in Section 1.3 of the class lecture notes. Think about it after tomorrow's class. You can share your thoughts in Discussions on the class Canvas page.

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Wednesday, September 27, 2023

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• The information sheet
• In this class we present a rigorous approach to basic theorems in calculus. Since rigorous reasoning in Mathematics is based on Mathematical Logic, we will start with a Brief Review of Mathematical Logic. After reading Brief Review of Mathematical Logic you can read my webpage Mathematical Rigor in the Context of Quadratic Functions.
• The class lecture notes.
• Some useful Wikipedia links:
  • Mathematical proof
  • Set
  • The number 1
  • The number 0
  • Integers
  • Natural numbers
  • Rational numbers
  • Real numbers
  • Intervals
  • Sign function
  • Floor and ceiling
• My one-page on the Power of Math.
• My one-page on Rigorous Reasoning.

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