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:

  1. I don't understand why the sets $X_n$ are closed.
  2. 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?
  3. 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.
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    $\begingroup$ A closed subset $X$ of $T$ is a set which contains all of its limit points in $T$. If you assume that $X$ has no limit points, then it is trivially closed. $\endgroup$ – Dan Rust Jul 3 '13 at 11:24
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    $\begingroup$ Note that your alternative proof works whenever the space is assumed the be compact (and, by the way, it is correct). On the other hand, the cited proof can be applied if the space is just countably compact (every countable open cover has a finite subcover). So it proves that in a countably compact space every infinite subset has a limit point. $\endgroup$ – Stefan Hamcke Jul 3 '13 at 11:50

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.


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