Does infinite neighborhood of point $p$ necessarily imply $p$ is a limit point

general-topologyreal-analysis

I'm reading Principles of Mathematical Analysis by Walter Rudin, and a question popped into my head when reading the following theorem:

If p is a limit point of a set E, the every neighborhood of p contains infinitely many points of E.

I know that this questions has been discussed several times here before, so I'm not gonna repeat it again. However, I'm wondering if the converse (sort of) statement is true? That if a neighborhood of p contains infinitely many points of E, then must p be a limit point of E?

In other words, if a point $p \in E$ is an isolated point, must the neighborhood $N_p$ be a finite set?

Tried to find the answer to this question but couldn't find it exactly.

EDIT: I realize this is a really stupid question, since if the neighborhood of p is large enough, it may contain an infinite amount of points given that $E$ is infinite? So let's put the questions into context.

I'm trying to answer the question regarding if $E \subset X$, where $X$ is a metric room, $E$ only consists of isolated points and $E$ is compact, then can $E$ contain an infinite amount of points?

My idea was to show that the if $\{N_p\}$ is the collection of neighborhoods to all points $p \in E$, then its union is an open cover. Then by the definition of compactness, there must exist a finite subcover containing $E$. Thus, there must exist a sequence of points $\{p_1, …, p_n\} \in E$ such that $E \subset N_{p_1} \cup \ …\ \cup N_{p_n}$. Finally I'm trying to make an argument that if all points in p are isolated, these neighborhood can only contain a finite amount of points, and thus can $E$ not contain an infinite amount of points.

I don't even know if this is the case, and if my reasoning is somewhat reasonable? Sorry for long text guys..

Best Answer

That if a neighborhood of p contains infinitely many points of E, then must p be a limit point of E?

Clearly entire space is a neighbourhood of $p$ containing infinitely many points of $E$. So the answer is "no".

I'm trying to answer the question regarding if $E \subset X$, where $X$ is a metric room, $E$ only consists of isolated points and $E$ is compact, then can $E$ contain an infinite amount of points?

What's an isolated point? It's a point $x\in E$ such that there is an open subset $U\subseteq X$ such that $U\cap E=\{x\}$. So if all points of $E$ are isolated then clearly $E$ is discrete.

And a compact discrete space $E$ has to be finite. Indeed, assume that $E$ is discrete and infinite. Since it is discrete then any convergent sequence is eventually constant. But since it is infinite then there is a sequence $(x_n)\subseteq E$ which is injective (as a function $\mathbb{N}\to E$). This sequence cannot have a convergent subsequence. Contradiction.

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