Limit point of an infinite set in a compact space I'm reading Kolmogorov's book on Real Analysis and I'm having problems understanding the proof of this theorem:
If $T$ is a compact space, then any infinite subset of $T$ has at least one limit point.
The proof goes like this:
Suppose $T$ contains an infinite set with no limit point. Then $T$ contains a countable set
$X=\{x_1,x_2,...\}$ with no limit point. But then the sets 
$ X_n=\{x_n,x_{n+1},... \} $ form a centered system of closed sets in $T$ (every finite intersection is nonempty) with an empty intersection, i.e. $T$ is not compact.
My questions are:


*

*I don't understand why the sets $X_n$ are closed. 

*If we take the interval $[0,1]$ with the usual topology and the sequence $x_n=1/n$, then the intersection of all the $X_n$ should be the empty set since it can't be a positive number because taking $n$ sufficiently large that number won't be in an infinite collection of the $X_n$ and neither can be zero since $0$ is not in any of the sets. Am I understanding the concept of intersection of an infinite collection of sets? 

*Alternatively I have thought of this proof: assume that the theorem is false , then for every point in $T$ we can take a neighborhood containing at most a finite number of points of $X$, in this way we obtain a covering of $T$, extracting a finite subcovering, at least one of the neighborhoods must contain an infinite number of points of $X$, contradiction. 

 A: The sets $X_n$ are closed because (by assumption) the set $X$ has no limit point.
A point $y \in \overline{X_n}\setminus X_n$ would be a limit point of $X_n$, hence a fortiori of $X$.
Your example of $x_n = \frac1n$ in $[0,\,1]$ has a limit point, namely $0$, and therefore the above cannot be applied to it (the $X_n$ are not closed, if you take the intersection of the $\overline{X_n}$, you get a nonempty infinite intersection; the crux is that for any sequence $(x_k)$, the intersection $\bigcap_{k \in \mathbb{N}} \overline{X_k}$ is the set of limit points of $X = \{x_0,\, x_1,\,\ldots\}$).
Your alternative proof is correct, and one of the (many) standard proofs.
A: I ran into this exact same issue in the same book (I assume you mean Kolmogorov and Fomin, Theorem 7, P. 94 in the Dover edition). It's actually a mistake -- a very rare one in an otherwise fine book -- because of how they define "limit point" on p. 79. What they call a "limit point" is nowadays called an omega-accumulation point, and under their definition the proof doesn't go through and the "theorem" is even false in general (unless the space is T1). Using the modern definition of limit point we avoid this sort of confusion, and the proof is then correct, as others have pointed out here. As for your alternative proof, it's fine under the modern def, but it requires X itself to be closed and that might not be the case under the K&F definition.
