2

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.

Motaka
  • 710
  • 8
    In what context? – Nigel Overmars May 22 '17 at 13:32
  • 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. – user 1987 May 22 '17 at 13:34
  • My guess: Let $X$ be a topological space. Take any open cover $U$ of $X$ ... We obtain a finite subcover $C \subseteq U$ of $X$. Hence $X$ is compact. – Alex Vong May 22 '17 at 13:36
  • 2
    This might refer to an argument that we can do things globally if we can do them locally on a compact space. – MooS May 22 '17 at 13:41
  • 2
    It would help if you explained where you encountered the phrase. – Ben Grossmann May 22 '17 at 13:50
  • here in the last comment of @Pietro Majer, he suggested to me to use "Standard compactness argument" to prove that $g$ is continuous. – Motaka May 22 '17 at 13:57
  • 1
    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? – Mirko May 22 '17 at 15:54
  • related https://mathoverflow.net/questions/143569/first-occurrence-of-by-the-usual-compactness-argument and http://recursed.blogspot.com/2013/09/by-usual-compactness-argument.html – Mirko May 24 '17 at 02:46

1 Answers1

1

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.

Henno Brandsma
  • 250,824