Set Equivalence Less One Element

Theorem
Let $S$ and $T$ be sets which are equivalent:


 * $S \sim T$

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 $f: S \to T$.

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

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

First it is shown that $g$ is an injection.

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

The following cases apply:

$(1): \quad f \left({x}\right) \ne b$ and $f \left({y}\right) \ne b$

By 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:
 * $g \left({x}\right) \ne g \left({y}\right)$

$(2): \quad f \left({x}\right) = b$

By definition of $g$:
 * $g \left({x}\right) = f \left({a}\right)$

We have that:
 * $y \in S \setminus \left\{{a}\right\}$

Therefore by definition of set difference:
 * $y \ne a$

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

Also by the injectivity of $f$:
 * $f \left({y}\right) \ne b$

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

$(2): \quad f \left({y}\right) = b$

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

Next it is shown that $g$ is a surjection.

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

By definition of surjectivity of $f$:
 * $\exists x \in S: f \left({x}\right) = y$

It is to be shown that:
 * $\exists x' \in S, x \ne a: f \left({x'}\right) = y$

$(1): \quad f \left({a}\right) = b$

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

Then by the rule of transposition:
 * $x \ne a$

so let:
 * $x' = x$

$(2): \quad f \left({a}\right) \ne b$

By the rule of transposition:
 * $f^{-1} \left({b}\right) \ne a$

where $f^{-1} \left({b}\right)$ denotes the preimage of $b$ under $f$.


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

By the definition of $g$:
 * $g \left({f^{-1} \left({b}\right)}\right) = y$

so let:
 * $x' = f^{-1} \left({b}\right)$


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

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

Then by the rule of transposition:
 * $x \ne a$

so let:
 * $x' = x$

Thus:
 * $\forall y \in T \setminus \left\{{b}\right\}: \exists x \in S \setminus \left\{{a}\right\}: g \left({x}\right) = y$

and so $g$ is a surjection.

Thus $g$ is both an injection and a surjection, and is by definition a bijection.