I updated the notes: The class lecture notes version of May 18, 2024 23:04. I added colored illustrations for the concepts of infimum and supremum in Section 6.3.
I updated the notes: The class lecture notes version of May 18, 2024 23:04. I added colored illustrations for the concepts of infimum and supremum in Section 6.3.
I updated the notes: The class lecture notes version of May 15, 2024 17:37. I added an illustration for Lemma 6.1.6. This lemma could be nicknamed: split in thirds, emphasize the midpoint.
I also wrote a new proof that \(\mathbb{R}\) is not countable, Theorem 6.1.7. I believe that this proof is the most accessible proof based on First Principle Reasoning.
I updated the notes: The class lecture notes version of May 14, 2024 13:08. I wrote Remark 6.1.3 in which I present my reasoning behind the proof that the set \(A\) introduced in the solution for Exercise 6.1.2 does not have a minimum.
I also wrote a new proof that \(\mathbb{R}\) is not countable, Theorem 6.1.6. I believe that this proof is the most accessible proof based on First Principle Reasoning.
I updated the notes: The class lecture notes version of May 14, 2024 00:06. I introduced a new subsection Subsection 5.4.2 The Quadruplicity of Sets. In this section I present the case for my quadruplicity of sets.
I also wrote a new proof that \(\mathbb{R}\) is not countable, Theorem 6.1.6. I believe that this proof is the most accessible proof based on First Principle Reasoning.
I updated the notes: The class lecture notes version of May 8, 2024 21:27. I introduced a new section Section 5.4 Countable Sets. In this section I give detailed proofs of three theorems Theorem 5.4.2, Theorem 5.4.3, Theorem 5.4.4. Each of these theorems is a powerful tool for proving countability when we cannot construct an explicit bijection between \(\mathbb{N}\) and a set that we want to prove is countable. One of these theorems you are using on the assignment.
I also wrote a new proof that \(\mathbb{R}\) is not countable, Theorem 6.1.6. I believe that this proof is the most accessible proof based on First Principle Reasoning.
Inductive step. Let $k\in\mathbb{N}$ be arbitrary. We need to prove \[ P(k) \quad \Rightarrow \quad P(k+1). \] Assume $P(k)$. That is assume \[ \forall \mkern 2mu (b_1,\ldots,b_k) \in \bigl(\mathbb{R}_{\gt 0}\bigr)^k \qquad \prod_{i=1}^k b_i = 1 \quad \Rightarrow \quad k \leq \sum_{i=1}^k b_i. \] The preceding implication is the Inductive hypothesis.
We will prove $P(k+1)$. That is we will prove \[ \forall \mkern 2mu (a_1,\ldots,a_{k+1}) \in \bigl(\mathbb{R}_{\gt 0}\bigr)^{k+1} \qquad \prod_{i=1}^{k+1} a_i = 1 \quad \Rightarrow \quad k+1 \leq \sum_{i=1}^{k+1} a_i. \]
Here is a proof of $P(k+1)$. Let $(a_1,\ldots,a_{k+1}) \in \bigl(\mathbb{R}_{\gt 0}\bigr)^{k+1}$ be arbitrary and assume \[ \prod_{i=1}^{k+1} a_i = 1. \] Since neither product, nor sum depend on the order of the $(k+1)$-tuple, we can also assume that \[ \forall \mkern 1mu i \in \{1,\ldots,k+1\} \quad a_1 \leq a_i \leq a_{k+1}. \] With this order, by Axiom OT, we have \[ a_{k+1} \lt 1 \quad \Rightarrow \quad \forall \mkern 1mu i \in \{1,\ldots,k+1\} \ \ a_i \lt 1 \quad \Rightarrow \quad \prod_{i=1}^{k+1} a_i \lt 1. \] The contrapositive of the implication \[ a_{k+1} \lt 1 \quad \Rightarrow \quad \prod_{i=1}^{k+1} a_i \lt 1 \] is \[ \prod_{i=1}^{k+1} a_i \geq 1 \quad \Rightarrow \quad a_{k+1} \geq 1. \] Since we assume, \[ \prod_{i=1}^{k+1} a_i = 1, \] from the contrapositive, we deduce that $a_{k+1} \geq 1$.
Similarly, the assumed order yields \[ a_{1} \gt 1 \quad \Rightarrow \quad \prod_{i=1}^{k+1} a_i \gt 1. \] The contrapositive of the preceding implication yields that the assumption $\displaystyle \prod_{i=1}^{k+1} a_i = 1$ implies $a_1 \leq 1$.
Notice that \begin{align*} a_1 \leq 1 \ \land \ a_{k+1} \geq 1 &\quad \Rightarrow \quad \bigl(1-a_1\bigr) \bigl(1-a_{k+1}\bigr) \leq 0 \\ &\quad \Rightarrow \quad a_1 a_{k+1} \leq a_1 + a_{k+1} - 1. \end{align*}
Now we use the Inductive hypothesis. To use the Inductive hypothesis we need to define a \(k\)-tuple $(b_1,\ldots,b_k)$. If \(k=1\), we set $(b_1) = (a_1a_{k+1})$. If \(k \gt 1\), let the elements of $(b_1,\ldots,b_k)$ be defined as follows \begin{align*} b_1 & = a_1a_{k+1}, \\ b_i & = a_i \quad \forall \mkern 1mu i \in \{2,\ldots,k\}. \end{align*} Since \[ \prod_{i=1}^k b_i = \prod_{i=1}^{k+1} a_i = 1, \] we can apply the Inductive hypothesis and deduce that \[ k \leq \sum_{i=1}^k b_i. \] Since we proved that \[ b_1 = a_1 a_{k+1} \leq a_1 + a_{k+1} - 1, \] we deduce that \[ k \leq \sum_{i=1}^k b_i = a_1a_{k+1} + \sum_{i=2}^k a_i \leq a_1 + a_{k+1} - 1 + \sum_{i=2}^k a_i. \] Thus, \[ k+1 \leq \sum_{i=1}^{k+1} a_i. \]
In Subsection 4.3.2 Cardinality of intervals we construct a bijection between any two intervals of real numbers. See the details in the lecture notes. A bijection I find particularly cute is the following one: Its graph is symmetric, it is defined by an intriguing formula, and it is accompanied by an animation I created.
The function with domain \(\mathbb{R}_{\gt 0}\) and range \(\mathbb{R}_{\geq 0}\) given by the formula \[ \forall\mkern+0.5mu x \in \mathbb{R}_{\gt 0} \qquad x \mapsto 2 \lceil x \rceil - 1 - x, \] is a bijection. Its inverse is a function with domain \(\mathbb{R}_{\geq 0}\) and range \(\mathbb{R}_{\gt 0}\) is given by the formula: \[ \forall\mkern+0.5mu x \in \mathbb{R}_{\geq 0} \qquad x \mapsto 2 \lfloor x \rfloor + 1 - x. \] The claim that these two functions are inverses of each other is verified using properties of the floor and ceiling.
The animation below illustrates the action of the blue bijection above on the set of positive real numbers \(\mathbb{R}_{\gt 0}\). The set \(\mathbb{R}_{\gt 0}\) is split into half-open intervals \((n-1,n]\), where \(n \in \mathbb{N}\), the set of positive integers. Then, each of these intervals is flipped about its midpoint. That is the meaning of the blue plot above. The animation starts with the domain, \(\mathbb{R}_{\gt 0}\), the point \(0\) is excluded, signified by the circle placed on the number line at point corresponding to the number \(0\). The animation ends with the range, \(\mathbb{R}_{\geq 0}\), the point \(0\) is included, signified by the disk placed on the number line at the point corresponding to the number \(0\).
Place the cursor over the image to start the animation.
In Subsection 4.3.2 Cardinality of intervals we construct a bijection between any two intervals of real numbers. See the details in the lecture notes. A bijection I find particularly cute is the following one: Its graph is symmetric, it is defined by an intriguing formula, and it is accompanied by an animation I created.
On Tuesday I posted about a beautiful bijection with domain \(\mathbb{R}_{\gt 0}\) and range \(\mathbb{R}_{\geq 0}\). The bijection from Tuesday is mentioned in the lecture notes only in Remark 4.3.12. The main bijection in the lecture notes is the bijection \(\phi: \mathbb{R}_{\gt 0} \to \mathbb{R}_{\geq 0}\) given by the following piecewise formula \[ \forall\mkern+0.5mu x \in \mathbb{R}_{\gt 0} \qquad \phi(x) = \begin{cases} x & \text{if} \mkern+20mu x \in \mathbb{R}_{\gt 0} \setminus \mathbb{N}, \\[3pt] x - 1 & \text{if} \mkern+20mu x\in \mathbb{N}. \end{cases} \] Its inverse \[ \forall\mkern+0.5mu x \in \mathbb{R}_{\geq 0} \qquad \phi^{-1}(x) = \begin{cases} x & \text{if} \mkern+20mu x \in \mathbb{R}_{\gt 0} \setminus \mathbb{N}, \\[3pt] x + 1 & \text{if} \mkern+20mu x\in \mathbb{N}\cup\{0\}, \end{cases} \] is a bijection with domain \(\mathbb{R}_{\geq 0}\) and range \(\mathbb{R}_{\gt 0}\). Below are their graphs.
The animation below illustrates the action of the bijection \(\phi\) on the set of positive real numbers \(\mathbb{R}_{\gt 0}\). There is no action on the positive real numbers that are not positive integers. Positive integers are shifted backward. The animation ends with the range, \(\mathbb{R}_{\geq 0}\), the point \(0\) is included, signified by the disk placed on the number line at the point corresponding to the number \(0\).
Place the cursor over the image to start the animation.
In Subsection 4.3.2 Cardinality of intervals we construct a bijection between any two intervals of real numbers. See the details in the lecture notes. A bijection I find particularly cute is the following one: Its graph is symmetric, it is defined by an intriguing formula, and it is accompanied by an animation I created.
The function with domain \(\mathbb{R}_{\gt 0}\) and range \(\mathbb{R}_{\geq 0}\) given by the formula \[ \forall\mkern+0.5mu x \in \mathbb{R}_{\gt 0} \qquad x \mapsto 2 \lceil x \rceil - 1 - x, \] is a bijection. Its inverse is a function with domain \(\mathbb{R}_{\geq 0}\) and range \(\mathbb{R}_{\gt 0}\) is given by the formula: \[ \forall\mkern+0.5mu x \in \mathbb{R}_{\geq 0} \qquad x \mapsto 2 \lfloor x \rfloor + 1 - x. \] The claim that these two functions are inverses of each other is verified using properties of the floor and ceiling.
The animation below illustrates the action of the blue bijection above on the set of positive real numbers \(\mathbb{R}_{\gt 0}\). The set \(\mathbb{R}_{\gt 0}\) is split into half-open intervals \((n-1,n]\), where \(n \in \mathbb{N}\), the set of positive integers. Then, each of these intervals is flipped about its midpoint. That is the meaning of the blue plot above. The animation starts with the domain, \(\mathbb{R}_{\gt 0}\), the point \(0\) is excluded, signified by the circle placed on the number line at point corresponding to the number \(0\). The animation ends with the range, \(\mathbb{R}_{\geq 0}\), the point \(0\) is included, signified by the disk placed on the number line at the point corresponding to the number \(0\).
Place the cursor over the image to start the animation.
In this class, and in all my classes, I emphasize reasoning from First Principles. Solving mathematical problems typically involves utilizing knowledge of related mathematical results. Often, the key to a solution is hidden within these results. I believe that for a solution to be complete, the dependency on related results should be made clear. Effort should be directed towards minimizing the use of related results and replacing them with direct reasoning from First Principles. Discovering First Principles in real-life problems can be challenging; however, in mathematics, they are more accessible and are usually based on Axioms or Fundamental Theorems.
Knowledge built on First Principles is both more robust and more adaptable to new situations. Working from First Principles helps prevent unnecessary complexities in solutions and acts as a safeguard against circular reasoning.
As an illustration of First Principle reasoning, I will prove here that the square of a rational number cannot equal \(2\). You can read the proof given by Student5 in Discussions on Canvas which proves this same theorem using two theorems from number theory. Student7 proved a more general theorem: that for all positive integers \(n\) we have that \(\sqrt{n}\) is rational if and only if \(n\) is a square of an integer. This student used the Fundamental Theorem of Arithmetic in the proof. The proof given by Student5 and my proof below, can be adopted to prove that \(\sqrt{p}\) is irrational for a prime \(p\). I don't think that theorem proven by Student7 can be done without the Fundamental Theorem of Arithmetic.
Step 1. We will prove the claim by a proof by contradiction, see BK5. Assume that the negation of the statement in the theorem is true. That is assume: \[ \exists\mkern+1.5mu r \in \mathbb{Q} \quad r^2 = 2. \] By BK3 there exist \(m \in \mathbb{Z}\) and \(n \in \mathbb{N}\) such that \(r = m/n\). Therefore \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{ \exists\mkern+1.5mu m \in \mathbb{Z} \ \ \exists\mkern+1.5mu n \in \mathbb{N} \qquad m^2 = 2 n^2.} \]
Step 2. Let us now consider the set \[ S = \left\{ q \in \mathbb{N} : \exists\mkern+1.5mu p \in \mathbb{Z} \quad p^2 = 2 q^2 \right\} \] By the green box at the end of Step 1, we deduce that \(n \in S\). Therefore, \(\bbox[5px, #88FF88, border: 1pt solid green]{S\neq \emptyset}\).
Step 3. Now we will prove that \(S\) does not have a minimum. That is we will prove \[ \forall\mkern+0.5mu q \in S \ \ \exists\mkern+1.5mu l \in S \qquad l \lt q. \] Let \(q \in S\) be arbitrary. By the definition of \(S\), there exists \(p \in \mathbb{Z}\) such that \(p^2 = 2 q^2\). The last equality implies that $p^2 \in \mathbb{E}$. By BK2 we have that \(p \in \mathbb{E}\). Since \(p \in \mathbb{E}\), there exists \(k \in \mathbb{Z}\) such that \(p = 2k\). Since \(p^2 = 2 q^2\), we also have \(4 k^2 = 2 q^2\). Consequently, \(q^2 = 2 k^2\). Thus, \(q^2\) and, by BK2, also \(q\) is even. Hence, there exists \(l \in \mathbb{N}\) such that \(q = 2 l\). Since \(q, l \in \mathbb{N}\), we have \(l \lt q\). Since \(q^2 = 2 k^2\), we also have \(4 l^2 = 2 k^2\), that is \( k^2 = 2 l^2\). Since \(l \in\mathbb{N}\), we have \(l \in S\). In conclusion, we proved that \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{ \forall\mkern+0.5mu q \in S \ \ \exists\mkern+1.5mu l \in S \quad l \lt q.} \]
Step 4. In Step 2 we proved that \(S\neq \emptyset\). In Step 3 we proved that \(S\) does not have a minimum. Thus, we have proved the negation of the Well Ordering Axiom. That is, we proved, \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{ \exists\mkern+1.5mu r \in \mathbb{Q} \quad r^2 = 2 \quad \Rightarrow \quad \lnot (\operatorname{WOA}). } \] Thus, the contrapositive of the preceding green box is also true: \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{ \operatorname{WOA} \quad \Rightarrow \quad \forall\mkern+1.5mu r \in \mathbb{Q} \quad r^2 \neq 2 . } \] This completes the proof.
The reason for this notation is that if the set $A$ has $m$ elements and the set $B$ has $n$ elements, then the set $B^A$ has $n^m$ elements. For example, if $A=\{\blacktriangle,\blacksquare\}$ and $B=\{1,2,3\}$ then we can list all the functions from $A$ to $B$ in a table:
$f_1$ | $f_2$ | $f_3$ | $f_4$ | $f_5$ | $f_6$ | $f_7$ | $f_8$ | $f_9$ | |
---|---|---|---|---|---|---|---|---|---|
$\blacktriangle$ | $1$ | $1$ | $1$ | $2$ | $2$ | $2$ | $3$ | $3$ | $3$ |
$\blacksquare$ | $1$ | $2$ | $3$ | $1$ | $2$ | $3$ | $1$ | $2$ | $3$ |
Or, it might be more interesting to make \(B=\{\color{#EE0000}{\mathbf{R}}, \color{#00CC00}{\mathbf{G}}, \color{#0000EE}{\mathbf{B}}\}\), the set of the primary colors, RED, GREEN and BLUE. Then the list of functions is as follows:
$f_1$ | $f_2$ | $f_3$ | $f_4$ | $f_5$ | $f_6$ | $f_7$ | $f_8$ | $f_9$ | |
---|---|---|---|---|---|---|---|---|---|
$\blacktriangle$ | $\color{#EE0000}{\mathbf{R}}$ | $\color{#EE0000}{\mathbf{R}}$ | $\color{#EE0000}{\mathbf{R}}$ | $\color{#00CC00}{\mathbf{G}}$ | $\color{#00CC00}{\mathbf{G}}$ | $\color{#00CC00}{\mathbf{G}}$ | $\color{#0000EE}{\mathbf{B}}$ | $\color{#0000EE}{\mathbf{B}}$ | $\color{#0000EE}{\mathbf{B}}$ |
$\blacksquare$ | $\color{#EE0000}{\mathbf{R}}$ | $\color{#00CC00}{\mathbf{G}}$ | $\color{#0000EE}{\mathbf{B}}$ | $\color{#EE0000}{\mathbf{R}}$ | $\color{#00CC00}{\mathbf{G}}$ | $\color{#0000EE}{\mathbf{B}}$ | $\color{#EE0000}{\mathbf{R}}$ | $\color{#00CC00}{\mathbf{G}}$ | $\color{#0000EE}{\mathbf{B}}$ |
$f_1$ | $f_2$ | $f_3$ | $f_4$ | $f_5$ | $f_6$ | $f_7$ | $f_8$ | $f_9$ | |
---|---|---|---|---|---|---|---|---|---|
$\blacktriangle$ | $\color{#EE0000}{\blacktriangle}$ | $\color{#EE0000}{\blacktriangle}$ | $\color{#EE0000}{\blacktriangle}$ | $\color{#00CC00}{\blacktriangle}$ | $\color{#00CC00}{\blacktriangle}$ | $\color{#00CC00}{\blacktriangle}$ | $\color{#0000EE}{\blacktriangle}$ | $\color{#0000EE}{\blacktriangle}$ | $\color{#0000EE}{\blacktriangle}$ |
$\blacksquare$ | $\color{#EE0000}{\blacksquare}$ | $\color{#00CC00}{\blacksquare}$ | $\color{#0000EE}{\blacksquare}$ | $\color{#EE0000}{\blacksquare}$ | $\color{#00CC00}{\blacksquare}$ | $\color{#0000EE}{\blacksquare}$ | $\color{#EE0000}{\blacksquare}$ | $\color{#00CC00}{\blacksquare}$ | $\color{#0000EE}{\blacksquare}$ |
In the above example, I have $2$ shapes and $3$ colors. So, I can color the first shape in three colors and the second shape in three colors. So, I can color the shapes in $3 \times 3$ ways. If I had $m$ distinct shapes and $n$ distinct colors, then I could color the shapes in $n \times n \times \cdots \times n$ ways; the total of $n^m$ ways. This reasoning is known as the rule of product counting principle. This counting is the reason why the set of all functions $f:A\to B$ with domain $A$ and codomain $B$ is denoted as \[ B^{\Large A} \qquad (n \ \text{colors})^{m \ \text{shapes}} \]
In class I proved
The table below illustrates the proof. Consider $S=\{\blacktriangle,\diamond,\bullet,\star\}$. The table below indicates what the bijection should be. The top part of the table contains sixteen functions in $\{0,1\}^S$; the bottom part contains the subsets of $A$ that are related to the functions above them.
$f_1$ | $f_2$ | $f_3$ | $f_4$ | $f_5$ | $f_6$ | $f_7$ | $f_8$ | $f_9$ | $f_{10}$ | $f_{11}$ | $f_{12}$ | $f_{13}$ | $f_{14}$ | $f_{15}$ | $f_{16}$ | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
$\blacktriangle$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |
$\diamond$ | $0$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$ | $0$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$ |
$\bullet$ | $0$ | $0$ | $1$ | $1$ | $0$ | $0$ | $1$ | $1$ | $0$ | $0$ | $1$ | $1$ | $0$ | $0$ | $1$ | $1$ |
$\star$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ |
s u b s e t |
$\emptyset$ | $\{\star\}$ | $\{\bullet\}$ | $\left\{\begin{array}{c}\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\{\diamond\}$ | $\left\{\begin{array}{c}\!\!\diamond\!\!\\\!\!\star\!\!\end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\diamond\!\!\\\!\!\bullet\!\!\end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\diamond\!\!\\\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\{\blacktriangle\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\star\!\!\end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\bullet\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\!\\\!\!\bullet\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\!\\\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ |
The function \[ \Phi: \mathcal{P}(S) \to \{0,1\}^S \] defined in the proof of Proposition 3.2.17 is given in the columns of the above table. For example, in the above table, if $A = \{\blacktriangle,\bullet,\star\}$, then $\Phi(A) = f_{12}$, that is the function \[ f_{12}(\blacktriangle)=1, \quad f_{12}(\diamond)=0, \quad f_{12}(\bullet)=1, \quad f_{12}(\star)=1. \]
In the proof of this theorem we prove \[ \forall\mkern+1.2mu \Theta: S \to \mathcal{P}(S) \quad \exists\mkern+1.5mu G \subseteq S \quad \forall\mkern+1.2mu x \in S \quad \Theta(x) \neq G. \] To prove the preceding displayed statement, let $\Theta: S \to \mathcal{P}(S)$ be an arbitrary function with domain \(S\) and codomain \(\mathcal{P}(S)\). To prove that $\Theta$ is not a surjection, we need to discover a subset $G$ of $S$ such that \[ \forall\mkern+1.2mu x \in S \quad \Theta(x) \neq G. \] The definition of such subset $G$ is truly ingenious, due to Georg Cantor: \[ G = \bigl\{ s \in S : s \not\in \Theta(s) \bigr\}. \]
I want to illustrate the universality of the definition of \(G\) on the example considered in the preceding item. Let \[ S=\{\blacktriangle,\diamond,\bullet,\star\}. \]
$p$ | $q$ | $p \Rightarrow q$ | $q \Rightarrow p$ | $(p\Rightarrow q) \land (q\Rightarrow p)$ | $p \Leftrightarrow q$ |
---|---|---|---|---|---|
F | F | T | T | T | T |
F | T | T | F | F | F |
T | F | F | T | F | F |
T | T | T | T | T | T |
$p$ | $q$ | $p \oplus q$ | $\lnot(p \oplus q)$ | $p \Leftrightarrow q$ |
---|---|---|---|---|
F | F | F | T | T |
F | T | T | F | F |
T | F | T | F | F |
T | T | F | T | T |
I updated the notes again today: The class lecture notes version of April 8, 2024 20:32. The only new item added is a Venn diagram of a three set symmetric difference, see Figure 10 on page 23. This picture was somewhat challenging to make, so I almost gave up. But, with some "brute force" I succeeded.
I vividly remember the fascination I felt when I was introduced to the set operation of symmetric difference of two sets in an undergraduate class on set theory. Let $A$ and $B$ be two sets. The symmetric difference of two sets $A$ and $B$, denoted by $A\Delta B$, is defined as \[ A\Delta B = ( A \setminus B ) \cup ( B \setminus A ). \] Here $A \setminus B$ and $B \setminus A$ are the set differences defined as \begin{align*} A \setminus B & = \bigl\{ x \in U : x \in A \land x \not\in B \bigr\}, \\ B \setminus A & = \bigl\{ x \in U : x \not\in A \land x \in B \bigr\}. \end{align*} Hence, in formal logical notation the symmetric difference of two sets operation is given as \[ A \Delta B = \Bigl\{ x \in U : \bigl( x \in A \land x \not\in B \bigr) \lor \bigl( x \not\in A \land x \in B \bigr) \Bigr\}. \] A different way of communicating the definition of the symmetric set difference is to give the following truth table. For arbitrary $x \in U$ we have the following four cases:
$x\in A$ | $x\in B$ | $x\in A\Delta B$ | hatched |
---|---|---|---|
F | F | F | none |
F | T | T | slope $1$ |
T | F | T | slope $-1$ |
T | T | F | slopes $-1$ and $1$ |
The above truth table is identical with the truth table of the EXCLUSIVE OR (XOR) compound proposition.
For interesting properties of the symmetric difference see it Wikipedia page.
Below is a Venn diagram of a symmetric difference of two sets $A$ and $B$. In the Venn diagram below, the set $A$ is hatched by thin grey lines with slope $-1$, and the set $B$ is hatched by thin grey lines with slope $1$.
A symmetric difference of two sets in yellow with a teal boundary
Below is a Venn diagram of a symmetric difference of three sets. In the Venn diagram below, the set $A$ is hatched by thin grey lines with slope $-1$, the set $B$ is hatched by thin grey lines with slope $1$, and the set $C$ is hatched by thin grey horizontal lines. The definition of the symmetric difference of three sets rests on the fact that the symmetric difference set operation $\Delta$ is associative: \[ (A \Delta B) \Delta C = A \Delta (B \Delta C) \] One way to prove the preceding equality is to set up the following truth table. For arbitrary $x \in U$ we have the following eight cases:
$x\in A$ | $x\in B$ | $x\in C$ | $x\in A\Delta B$ | $x\in (A\Delta B)\Delta C$ | $x\in B\Delta C$ | $x\in A\Delta (B\Delta C)$ | hatched |
---|---|---|---|---|---|---|---|
F | F | F | F | F | F | F | none |
F | F | T | F | T | T | T | slope $0$ |
F | T | F | T | T | T | T | slope $1$ |
F | T | T | T | F | F | F | slopes $1$, $0$ |
T | F | F | T | T | F | T | slope $-1$ |
T | F | T | T | F | T | F | slopes $-1$, $0$ |
T | T | F | F | F | T | F | slopes $-1$, $1$ |
T | T | T | F | T | F | T | slopes $-1$, $1$, $0$ |
The preceding truth table proves that the following proposition is true \[ \forall\mkern 1.5mu x \in U \qquad \Bigl( x\in (A\Delta B)\Delta C \ \ \Leftrightarrow \ \ x\in A\Delta (B\Delta C) \Bigr). \]
A symmetric difference of three sets in yellow with a teal boundary