Set Equivalence Less One Element

Theorem

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

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

Then $S - \left\{{a}\right\} \sim T - \left\{{b}\right\}$

Proof

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

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

$ \forall x \in S - \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} $

It is easily confirmed that this is a bijection:

As $f$ is injective, we have that:

$\forall x_1, x_2 \in S: x_1 \ne x_2 \implies f \left({x_1}\right) \ne f \left({x_2}\right)$

Hence

$\forall x_1, x_2 \in S - \left\{{a}\right\}: x_1 \ne x_2 \implies f \left({x_1}\right) \ne f \left({x_2}\right)$

It follows that:

$\forall x_1, x_2 \in S - \left\{{a}\right\}: x_1 \ne x_2 \implies g \left({x_1}\right) \ne g \left({x_2}\right)$

and so $f$ is injective.

$\forall y \in T: \exists x \in S: f \left({x}\right) = y$

as $f$ is surjective.

So, all elements of $T - \left\{{b}\right\}$ have images under $g$ which are the same as under $f$

Except, that is, for the element whose preimage under $f$ is $a$.

Instead, this element is mapped to from the preimage of $b$ under $f$.

Hence we have that:

$\forall y \in T - \left\{{b}\right\}: \exists x \in S - \left\{{a}\right\}: y = f \left({x}\right)$

Thus $g \left({x}\right)$ is surjective.

So, being both injective and surjective, $g$ is a bijection.

Sources

• Seth Warner: Modern Algebra (1965): $\S 17$: Theorem $17.2$