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I have a question regarding an answer given in an older post:

The original question was:

Let $f$ be an entire function. Let $Z(f)$ be the set of all zeros of $f$ which is further uncountable. We have to show that $Z(f)$ has a limit point in $\mathbb{C}$.

And the answer was given as:

Let $K_n:=\{z∈\mathbb{C}:|z|≤n\}$ for $n∈\mathbb{N}$ and suppose that $Z(f)$ has no limit point in $\mathbb{C}$. Since $f$ is continuous and $K_n$ is compact, $Z(f)∩K_n$ must be finite or empty. But then $$Z(f)= \bigcup _{n=1}^{\infty}Z(f) \cap K_n$$ is at most countable.

Why is it that if $f$ is continuous and $K_n$ is compact, then $Z(f)∩K_n$ must be finite or empty?

And why is $Z(f)= \bigcup _{n=1}^{\infty}Z(f) \cap K_n$ at most countable?

Asaf Karagila
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    This is sequential compactness: any sequence in a compact set has a convergent subsequence, and hence every infinite subset of a compact set has a limit point. It follows that any subset of a compact set with no limit point is finite. – Jair Taylor Apr 24 '20 at 05:57
  • Note that the question really doesn't have anything to do with complex analysis. It proves that any uncountable set of points in $\mathbb{R}^n$ must have a limit point. – Jair Taylor Apr 24 '20 at 05:57
  • I don't know why the answer mentions continuity of $f$ as that is not relevant. (note the above argument only works for subsets of a metric space. In general compactness != sequential compactness.) – Jair Taylor Apr 24 '20 at 06:06

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