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I was reading "Introductory Real Analysis" by Kolmogorov and Fomin and came across this theorem:

Every infinite subset $S$ of a compact space $T$ has a limit point

Note: In this book a point $x$ is called a limit point of $E$ if every neighborhood of $x$ contains an infinite number of points of $E$

Proof: Suppose $S$ has no limit point. Then there is a countable subset $R \subset S$ that has no limit point. Let $R = \{x_1, x_2, ...\}$. But then the sets $R_n = \{x_n, x_{n+1}, ...\}$ are closed sets with the finite intersection property with empty intersection. Hence $T$ is not compact.

What I don't understand is, why are these sets closed? If we assume that $T$ is a $T_1$-space, then I was able to show that they are in fact closed. Is this true if we don't assume that $T$ is $T_1$? If yes, how does one prove this?

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    $R$ is not closed in general. Consider the trivial topology. – 2017-02-06
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    Ok, somehow I didn't consider this! Then this proof is wrong. Can you write your comment as an answer so I can accept it? – 2017-02-06
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    @Arthur If a set has no limit points, then it is closed. – 2017-02-06
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    @Arthur But then every point of $T$ is a limit point of $R$ – 2017-02-06

2 Answers 2

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This proof is indeed incorrect (with your definition of limit point) if the space is not $T_1$, since the sets $R_n$ need not be closed. Here is a correct argument you can give.

Let $C_n$ be the closure of the set $R_n$, so that the sets $C_n$ are clearly closed and have the finite intersection property. We only need to check that $\bigcap C_n$ is empty. But if $x\in\bigcap C_n$, that means that for all $n$, every neighborhood $U$ of $x$ intersects $R_n$. If $U$ contained only finitely many points of $R$, then $U$ would be disjoint from $R_n$ for some $n$ (just pick $n$ larger than the index of any of the points of $U\cap R$). So this means $U\cap R$ is infinite for every neighborhood $U$ of $x$. That is, $x$ is a limit point of $R$ and hence of $S$, which is a contradiction.

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A set without limit points is trivially closed, since it contains all of its limit points

To elaborate on Arthur's comment, while it's true that this sets will not be closed in a space with the trivial topology that's not relevant here, since they contain plenty of converging subsequences, in fact every sequence in a trivial space converges to every point of the space.

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    How do you prove that if a set $S$ contains all of its limit points, then it is closed? – 2017-02-06
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    A point $x$ is called a limit point of $S$ if every for every open $U$ with $x \in U$ we have $|U \cap S| = \infty$ (We may have different definitions of this) – 2017-02-06
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    suppose $S$ is not closed, then $S^C$ is not open. A set $A$ is open if for every $x\in A$ there is a nbhd $U$ of $x$ with $U\subseteq A$, so $S^C$ not open means that there is an $x\in S^C$ such that for all nbhds $U$ of $x$ we have $U\cap S\neq\varnothing$, which is the definition of a limit point – 2017-02-06
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    Alright, we have different definitions. A point with the property you have mentioned is called "contact point" in this book. So I guess this proof is wrong – 2017-02-06
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    I should have said $(U\setminus\{x\})\cap S\neq\varnothing$ rather than $U\cap S$ to make clear that $x$ is not an adherent point. Your definition [isn't a standard one](https://en.wikipedia.org/wiki/Limit_point) and doesn't work for finite spaces – 2017-02-06
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    Yes, I didn't know it until now. But if $U\setminus\{x\}\cap S\neq\emptyset$ for any $U$, then it still does not mean that every neighborhood of $x$ contains an infinite number of points, does it? – 2017-02-06
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    it doesn't, you need some stronger assumption on your topological space, the space being $T_1$ is enough, as Eric's answer shows – 2017-02-06