how is my proof on equinumerous sets

Question

Hello i am hoping that you can tell me how my proof is and the improvements i can do. If i am complexity wrong please do tell.

Prove that if $W$ is denumerable, then $W$ is equinumerous with a proper subset of itself.

$S$ is denumerable. therefore the is a bijection between S and the natural number: $f : S \rightarrow \mathbb{N}$. Say that we have a infinite Set $ A \subset \mathbb{N}$. $A$ would also be denumerable as a infinite subset of a denumerable set is also denumerable. Then we do $f^{-1}(A) = B$. Then $ T \subset S $ is a proper subset. Lets take another function $g: \mathbb{N} \rightarrow A$. g wouuld also be a bijection. Now let $h = g\circ f, h$ would be a function that went $S \rightarrow B$ and would also be a bijection .

Therefor you can say that S and T are equinumermous. Hence $S$ is equinumermous with a proper subset of itself.

How did i do?

Answer 1

The idea is good. A few small issues:

1. You didn't assume that $A$ was proper, so $B$ doesn't have to be a proper subset of $S$, which I think you want.
2. Also where did $T$ come from? You just introduced a new variable $T$ without explanation.
3. You took a function $g$ from $\Bbb{N}\to A$. $g$ doesn't have to be a bijection. However since $A$ is denumerable, there is a bijection, which we can call $g$.
4. I think $B$ and $T$ are supposed to be the same thing, but you keep switching back and forth between the two variables.
5. Not so much a mathematical issue, but you keep typing "equinumermous." (At least three times, though I fixed one earlier when I noticed it and before I noticed the others.) It's actually spelled "equinumerous."