Tao Analysis I - 3.1.1

Exercise 3.1.1

Let $a$, $b$, $c$, $d$ be objects such that $\{a, b\} = \{c, d\}$. Show that at least one of the two
statements “$a = c$ and $b = d$” and “$a = d$ and $b = c$” hold.

Let's start with the definition of the equality of two sets.

Axiom 3.2 (Equality of sets).Two sets $A$ and $B$ are equal, $A=B$, iff every element of $A$ is an element of $B$ and vice versa.

Let's apply this definition our scenario:

1. $a$ is a member of $\{c,d\}$. That means $(a=c) \lor (a=d)$ is true.
2. $b$ is a member of $\{c,d\}$. That means $(b=c) \lor (b=d)$ is true.
3. $c$ is a member of $\{a,b\}$. That means $(c=a) \lor (c=b)$ is true.
4. $d$ is a member of $\{a,b\}$. That means $(d=a) \lor (d=b)$ is true.

The key to this exercise is recognising that all four statements must be true.

Let's start by considering the case $a=c$, which makes (1) true. It also makes (3) true. How we ensure (2) and (4) are also true? This is only possible when $b=d$. Thus the statement  “$a = c$ and $b = d$” satisfies the definition of set equality.

Let's consider the case $a=d$ which makes (1) true. It also makes (4) true. How do we ensure (2) and (3) are true? This is only possible when $b=c$. Thus the statement  “$a = d$ and $b = c$” satisfies the definition of set equality.

Can both statements be true at the same time? Let's start with $a=c$ from the first statement. The second statement gives us $b=c$. The first statement then gives us $b=d$. That is, $a=b=c=d$ also satisfies the equality of sets.

We have shown that the set equality $\{a, b\} = \{c, d\}$ means at least one of the statements “$a = c$ and $b = d$” and “$a = d$ and $b = c$” holds. $\square$