Our professor gave us the following theorem. Let $T$ be a compact topological space then any infinite subset of T has at least a limit point.

Proof $T$-compact space. Suppose $B$ subset of $T$ with infinitely points and no limit points. Take a sequence $\{x_n\} ,\, n=1,2... $, $x_n \in B$. Consider sets $Xn=\{x_n,x_{n+1},...\}$. These sets form a centred system, Moreover $X_n$ are closed. Moreover infinite intersection of $X_n=\emptyset$ set (which shouldn't be by a previous theorem) so contradiction.

But what I don't understand from the proof why are all $X_n$ closed? I think that a point that is not itself a limit point may indeed have limit points and hence closure of $X_n$ may differ for each $X_n$.


Suppose B subset of T with infinitely points and no limit points

The $X_n$ are closed because $B$ has no limit points. If $X_n$ had a limit point $x$, every neighbourhood of $x$ would contain points of $X_n\setminus \{x\}$, and then it would also contain points of $B\setminus\{x\}$, hence $x$ would be a limit point of $B$.

  • $\begingroup$ thanks a lot...now it makes sense $\endgroup$ – loukanikos Feb 15 '14 at 14:34

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