1
$\begingroup$

Proposition. A subset of a countable set is countable. Proof. Let $X$ be a countable set and let $A \subseteq X$. Then there is a bijection $f:X \to \mathbb{N}$. Let $\varphi: A \to X$ be the inclusion mapping defined by $\varphi(x)=x$. Since the composition of injections is also an injection, it follows that $f \circ \varphi: A \to \mathbb{N}$ is also an injection. Thus $A$ is countable.

If $X$ were to be countably infinite, could we expand the above proof to show that every subset of $X$ is either finite or countably infinite? It seems like we would just have to show that $\varphi$ is also a surjection, but I am not sure how.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Your proof should describe $f$ as an injection, rather than a bijection. Otherwise, it looks fine.

Now, if $X$ is countably infinite, then we can alter the proof as you say. However, it would be easier to prove it as a corollary. To start with, prove that every infinite subset of $\Bbb N$ is in bijection with $\Bbb N.$ Consequently, every countable set is either finite or countably infinite. Can you prove these claims and proceed from there?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .