# 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:

$S \setminus \set a \sim T \setminus \set b$

where $\setminus$ denotes set difference.

## Proof

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

Let $g: \paren {S \setminus \set a} \to \paren {T \setminus \set b}$ be the mapping defined as follows:

$\forall x \in S \setminus \set a: \map g x = \begin{cases} \map f x: \map f x \ne b \\ \map f a: \map f x = b \end{cases}$

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

Let $x, y \in S \setminus \set a$ and $x \ne y$.

The following cases apply:

$(1): \quad \map f x \ne b$ and $\map f y \ne b$

By definition of $g$:

$\map f x = \map g x$

and:

$\map f y = \map g y$

It follows from the injectivity of $f$ that:

$\map g x \ne \map g y$

$(2): \quad \map f x = b$

By definition of $g$:

$\map g x = \map f a$

We have that:

$y \in S \setminus \set a$

Therefore by definition of set difference:

$y \ne a$

So by the injectivity of $f$:

$\map f y \ne \map f a = \map g x$

Also by the injectivity of $f$:

$\map f y \ne b$

Thus, by the definition of $g$:

$\map g y = \map f y \ne \map f a = \map g x$

$(2): \quad \map f y = b$

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

$\Box$

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

Let $y \in T \setminus \set b$.

By definition of surjectivity of $f$:

$\exists x \in S: \map f x = y$

It is to be shown that:

$\exists x' \in S, x \ne a: \map f {x'} = y$

$(1): \quad \map f a = b$

$\map f x = y \ne b$

Then by the rule of transposition:

$x \ne a$

so let:

$x' = x$

$(2): \quad \map f a \ne b$

By the rule of transposition:

$\map {f^{-1} } b \ne a$

where $\map {f^{-1} } b$ denotes the preimage of $b$ under $f$.

Case $(2.1): \quad \map f a = y$

By the definition of $g$:

$\map g {\map {f^{-1} } b} = y$

so let:

$x' = \map {f^{-1} } b$

Case $(2.2): \quad \map f a \ne y$
$\map f x = y \ne \map f a$

Then by the rule of transposition:

$x \ne a$

so let:

$x' = x$

Thus:

$\forall y \in T \setminus \set b: \exists x \in S \setminus \set a: \map g x = y$

and so $g$ is a surjection.

$\Box$

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

$\blacksquare$