Infinite Set is Equivalent to Proper Subset/Proof 3

Theorem
A set is infinite if and only if it is equivalent to one of its proper subsets.

Proof
Let $X$ be a set which has a proper subset $Y$ such that:
 * $\operatorname{Card} \left({X}\right) = \operatorname{Card} \left({Y}\right)$

where $\operatorname{Card} \left({X}\right)$ denotes the cardinality of $X$.

Then:
 * $\exists \alpha \in \complement_X \left({Y}\right)$

and
 * $Y \subsetneqq Y \cup \left\{{\alpha}\right\} \subseteq X$

The inclusion mappings:
 * $i_Y: Y \to X: \forall y \in Y: i \left({y}\right) = y$
 * $i_{Y \cup \left\{{\alpha}\right\}}: Y \cup \left\{{\alpha}\right\} \to X: \forall y \in Y: i \left({y}\right) = y$

give:
 * $\operatorname{Card} \left({X}\right) = \operatorname{Card} \left({Y}\right) \le \operatorname{Card} \left({Y}\right) + \mathbf 1 \le \operatorname{Card} \left({X}\right)$

from which:
 * $\operatorname{Card} \left({X}\right) = \operatorname{Card} \left({Y}\right) + \mathbf 1 = \operatorname{Card} \left({X}\right) + \mathbf 1$

So by definition $X$ is infinite.

Now suppose $X$ is infinite.

That is:
 * $\operatorname{Card} \left({X}\right) = \operatorname{Card} \left({X}\right) + \mathbf 1$

Let $\alpha$ be any object such that $\alpha \notin X$.

Then there is a bijection $f: X \cup \left\{{\alpha}\right\} \to X$.

Let $f_{\restriction X}$ be the restriction of $f$ to $X$.

Then from Injection to Image is Bijection:
 * $\operatorname{Im} \left({f_{\restriction X}}\right) = X \setminus \left\{{f \left({\alpha}\right)}\right\}$

which is a proper subset of $X$ which is equivalent to $X$.