Infinite Set is Equivalent to Proper Subset

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

Proof
Let $S = \left\{{a_1, a_2, a_3, \ldots}\right\}$ be a countably infinite subset of an infinite set $T$. Such a subset can always be constructed by Infinite Set has Countable Subset.

Partition $S$ into $S_1 = \left\{{a_1, a_3, a_5, \ldots}\right\}, S_2 = \left\{{a_2, a_4, a_6, \ldots}\right\}$.

We can establish a bijection between $S$ and $S_1$, by letting $a_n \leftrightarrow a_{2n-1}$.

We can extend this to a bijection between $S \cup \left({T \setminus S}\right) = T$ and $S_1 \cup \left({T \setminus S}\right) = T \setminus S_2$ by assigning each element in $T \setminus S$ to itself.

So we have demonstrated a bijection between $T$ and one of its proper subsets $T \setminus S_2$, which shows that if $T$ is infinite, it is equivalent to one of its proper subsets.

Now, let $T_0\subsetneq T$ be a proper subset of $T$, and $f:T\to T_0$ be a bijection. It follows from Subset of Finite Set No Bijection that $T$ must be infinite.

Comment
This is one of the ways that infinite sets can be defined, i.e. $S$ is infinite iff it is equivalent to a proper subset of itself.