Set Equivalence Less One Element

Theorem
Let $S$ and $T$ be sets such that $S \sim T$, i.e. they are equivalent.

Let $a \in S$ and $b \in T$.

Then:
 * $S \setminus \left\{{a}\right\} \sim T \setminus \left\{{b}\right\}$

where $\setminus$ denotes set difference.

Proof
As $S \sim T$, there exists a bijection $f: S \to T$.

We define the mapping $g: \left({S \setminus \left\{{a}\right\}}\right) \to \left({T \setminus \left\{{b}\right\}}\right)$ as follows:

$\forall x \in S \setminus \left\{{a}\right\}: g \left({x}\right) = \begin{cases} f \left({x}\right): f \left({x}\right) \ne b \\ f \left({a}\right): f \left({x}\right) = b \end{cases}$

We first show that $g$ is an injection.

Let $x, y \in S \setminus \left\{{a}\right\}$ and $x \ne y$.


 * Case $1$. $f \left({x}\right) \ne b$ and $f \left({y}\right) \ne b$

By the definition of $g$, $f \left({x}\right) = g \left({x}\right)$ and $f \left({y}\right) = g \left({y}\right)$.

It follows from the injectivity of $f$ that $g \left({x}\right) \ne g \left({y}\right)$.


 * Case $2$. $f \left({x}\right) = b$

By the definition of $g$, $g \left({x}\right) = f \left({a}\right)$.

Now, $y \ne a$ because $y \in S \setminus \left\{{a}\right\}$.

So $f \left({y}\right) \ne f \left({a}\right) = g \left({x}\right)$ by the injectivity of $f$.

Also $f \left({y}\right) \ne b$, again by the injectivity of $f$.

Thus, by the definition of $g$, $g \left({y}\right) = f \left({y}\right) \ne f \left({a}\right) = g \left({x}\right)$.


 * Case $3$. $f \left({y}\right) = b$

The proof is the same as that of case $2$, with the roles of $x$ and $y$ reversed.

Now, we show that $g$ is a surjection.

Let $y \in T \setminus \left\{{b}\right\}$.

By the surjectivity of $f$, there exists a $x \in S$ such that $f \left({x}\right) = y$.

We wish to show that there exists an element $x' \ne a$ of $S$ such that $f \left({x'}\right) = y$.


 * Case $1$. $f \left({a}\right) = b$

By hypothesis, $f \left({x}\right) = y \ne b$.

Then $x \ne a$ by the rule of transposition, so we let $x' = x$.


 * Case $2$. $f \left({a}\right) \ne b$

By the rule of transposition, $f^{-1} \left({b}\right) \ne a$. Here, $f^{-1} \left({b}\right)$ denotes the preimage of $b$ under $f$.


 * Case $2.1$. $f \left({a}\right) = y$

It follows by the definition of $g$ that $g \left({f^{-1} \left({b}\right)}\right) = y$, so we let $x' = f^{-1} \left({b}\right)$.


 * Case $2.2$. $f \left({a}\right) \ne y$

By hypothesis, $f \left({x}\right) = y \ne f \left({a}\right)$.

Then $x \ne a$ by the rule of transposition, so we let $x' = x$.

So $g$ is both an injection and a surjection, and is hence a bijection.