Denumerability and Equinumerosity Between the Subsets and a Given Set I'm asked to prove that if $S=\{s_1,s_2,s_3,\dots\}$ is denumerable, then $S$ is equinumerous with a proper subset of itself. I think that the sets $K_1=\{S\setminus\{s_1\}\}$, $K_2=\{K_1\setminus \{s_2\}\}, \dots$ Each have the same cardinality as $S$, but I'm not sure how to demonstrate this...
 A: HINT: The most straightforward way to show equinumerosity of sets $A$ and $B$ is to show that there is a bijection between them. The map $n\mapsto n+1$ gives a bijection from $\Bbb Z^+$ to $\Bbb Z^+\setminus\{1\}$; adapt this to show that $S$ and $K_1$ are equinumerous.
More generally, if $m\in\Bbb Z^+$, the map
$$f:\Bbb Z^+\to\Bbb Z^+\setminus\{m\}:k\mapsto\begin{cases}
k,&\text{if }k<m\\
k+1,&\text{if }k\ge m
\end{cases}$$
is a bijection that you can modify similarly to get a bijection from $S$ onto your $K_m$.
A: If a set $S=\{s_1,s_2,s_3,\dots\}$ is denumerable, then there exists a bijection $f:\mathbb{N}\longrightarrow S$, and for $S$ to be equinumerous with a proper subset of itself, say $S'$, then there must exist a bijection $\varphi :S\longrightarrow S'$. Now consider the proper subsets $S'_1=S\setminus\{s_1\}$, $S'_2=S\setminus\{s_1,s_2\}$, and so on. Here we see that for each set $S'_i$ we can construct a function $\varphi : S\rightarrow S'_n$ defined by
$$\varphi(s_i)=s_{i+n},$$
which is the desired mapping that shows existence.
