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.