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I'm studying real analysis, and I am wanting to clarify something in the proof of Bolzano-Weierstrass.

The theorem states, as I'm sure you know, that in any metric space, an infinite subset $E$ of a compact set $K$ has a limit point in $K$.

The first step in the proof is to assume no $K$ is a limit point of $E$. Then (this is the part I need help on), every $q \in K$ would have a neighborhood that contains at most one point of $E$.

From my understanding, a point is a limit point of a set if any neighborhood around that point intersects that set. So, if no $K$ is a limit point of $E$, then shouldn't any neighborhood of $q \in K$ not hit $E$ at all, rather than at at most one point?

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The definition of a limit point (at least in this context) is instead that $x$ is a limit point of $E$ iff every neighborhood of $x$ contains a point of $E$ other than $x$. If you didn't include the "other than $x$" condition, then trivially any point which is in $E$ would be a limit point of $E$, so it would be trivial that $E$ has limit points.

So if $x$ is not a limit point of $E$, there exists a neighborhood of $x$ which contains no point of $E$ other than possibly $x$ itself. Such a neighborhood then contains at most one point of $E$.

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What if $q \in E$? Any neighborhood of $q$ contains $q$, so that's one point that belongs to $E$.

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