Eigenvalue of Compact Operators

Question

To prove that the set of eigenvectors of a compact linear operator on a normed space $X$ is countable, I read "it suffices to show that for every real $k > 0$ the set of all eigenvalues whose absolute values are greater than or equal to $k$ is finite". I cannot see why this is true. Even if the quoted statement is true, it may still be possible to have uncountablely many eigenvalues between $-k$ and $k$. Right? What stops this from happening? I must missed something here. Could anyone help me, please? Thank you!

Answer 1

The fact that this is true for every $k$ gives you a lot more information than you're considering. Let's follow Nates comment.

Define $A_n$ to be the set of eigenvalues whose magnitude is greater than $1/n$. By our statement, each $A_n$ is finite, and so certainly qualifies as "at most countable".

Now, consider the set $S$ of all non-zero eigenvalues. Well, if $\lambda > 0$, then $|\lambda| > 1/n$ for some value of $n$, which means that $\lambda \in A_n$. So, we may in fact state that $$ S \subseteq \bigcup_{n=1}^\infty A_n $$ but each $A_n$ is countable, so $\bigcup_{n=1}^\infty A_n$ has to be countable. So, the set of non-zero eigenvalues is certainly countable.

$0$ might be an eigenvalue, but at any rate: the set of all eigenvalues is certainly a subset of $\{0\} \cup S$, which is countable. The conclusion follows.