I need to know what does "standard compactness argument" means, I google it but But I did not find any satisfactory results.

Any help: Thanks.

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    $\begingroup$ In what context? $\endgroup$ – Nigel Overmars May 22 '17 at 13:32
  • $\begingroup$ It means the following: a space $X$ is compact if from every open cover of $X$ we can find a finite subcover. So basically, the definition. $\endgroup$ – user 1987 May 22 '17 at 13:34
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    $\begingroup$ This might refer to an argument that we can do things globally if we can do them locally on a compact space. $\endgroup$ – MooS May 22 '17 at 13:41
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    $\begingroup$ It would help if you explained where you encountered the phrase. $\endgroup$ – Omnomnomnom May 22 '17 at 13:50
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    $\begingroup$ you may pick a convergent subsequence of a given sequence, thereby proving the existence of a certain object which happens to be the limit of the convergence subsequence. As the first context suggests, you need to know the context: There are many standard arguments, each standard in a respective context. If you encounter this expression in a particular proof or construction, which is it? $\endgroup$ – Mirko May 22 '17 at 15:54

For me the canonical compactness argument is the argument that shows that a compact Hausdorff space is regular, in fact showing a common maxim: "compact sets behave like points (often)"

Suppose $x\in X$, $C \subseteq X$ compact and $x \notin C$, all in a Hausdorff space $X$. Then $x$ and $C$ have disjoint open neighbourhoods (just as two distinct points already have).

For every $p \in C$ pick $U_p$ and $V_p$ open in $X$ such that $x \in U_p, p \in V_p , U_p \cap V_p = \emptyset$, which can be done as $x \neq p$ and $X$ is Hausdorff. The $V_p$ cover the compact set $C$, so finitely many cover them, say $C \subseteq V:=V_{p_1} \cup \ldots \cup V_{p_n}$, for finitely many $p_1,\ldots, p_n \in C$. But then $U = \cap_{i=1}^n U_{p_i}$ is also open (a finite intersection!) and contains $x$ and

$$U \cap V = \cup_{i=1}^n (V_{p_i} \cap U) \subseteq \cup_{i=1}^n (V_{p_i} \cap U_{p_i}) = \emptyset$$ showing that $U$ and $V$ are the required open neighbourhoods of $x$ and $C$.

A totally similar argument, using the above as a lemma, shows that $C$ and $D$ compact disjoint subsets of a Hausdorff space $X$ have disjoint open neighbourhoods as well.

The compactness allows one to fix things for points, and then using a finite subcover, to fix them for the whole compact set. Many compactness arguments use this idea.


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