Proper Subset of an Infinite Set is Equinumerous to the Set Containing It

Question

I noticed that there is a question about $S$ being denumerable, which implies $S$ is equinumerous with a proper subset of itself, but what about an infinite set? That is, how to do I prove that every infinite set is equinumerous with a proper subset of itself?

Answer 1

It seems that you know how to do it if $S$ is infinite and countable (denumerable). Otherwise, partition $S$ into a countably infinite set $S_1$ and another set $S_2 = S \setminus S_1$. You know $S_1$ is equinumerous with a proper subset of itself - use this fact to construct a bijection between $S \equiv S_1 \sqcup S_2$ and a proper subset of itself.

ELABORATION: suppose $S_1$ is equinumerous with a proper subset $S_3$ of itself, with bijection $f: S_1 \to S_3$. Define the bijection $g:S \equiv S_1 \sqcup S_2 \to S_3 \sqcup S_2$ by $g(x) = f(x)$ for $x \in S_1$, $g(x) = x$ for $x \in S_2$ (sorry, I always forget how to do \cases). $ S_3 \sqcup S_2$ is a proper subset of $S$.