[Math] Limit point of an infinite set in a compact space

compactnessgeneral-topology

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.

Best Answer

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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