Prove or disprove: If $A \subseteq B$ and $B$ is denumerable, then $A$ is denumerable Claim: If $A \subseteq B$ and $B$ is denumerable, then $A$ is denumerable Proof: Assume $A \subseteq B$ and $B$ is denumerable, then it follows that $B$ is countable. Every subset of a countable set is countable, so $A$ is countable (denumerable). Is this proof valid? If $A$ is countable, then it could be either finite or denumerable, right? How can I prove that it's denumerable and not just finite? Similarly, if $A \subseteq B$ and $A$ is denumerable, then $B$ is denumerable. I know that $A$ is countable, so $B$ is countable as well since $A$ is a subset of $B$. However, how can I show that $B$ is denumerable and not just finite?

The first claim is indeed true. The key to the proof is to use the following (equivalent) definition of denumerability ( = **countably infinite** ) Def: a set $X$ is denumerable if all of its elements can be enumerated in a sequence, i.e.all of its elements can be listed as a sequence $x(1), x(2), ...x(n)$, ... So using this definition we can write a proof for this claim. Suppose $A$ is a subset of $B$ and $B$ is denumerable. So we can enumerate the elements of $B$ as a sequence $x(1), x(2), ..., x(n),.$. The subsequence consisting of those elements that belong to $A$ is then an enumeration of $A$. We are done.

The second claim is not true. Take $A = \mathbb{N}$ and $B = \mathbb{R}$, then $A$ is a subset of $B$ and $A$ is denumerable while $B$ is not.