# Standard compactness argument

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.

• 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
• 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
• It would help if you explained where you encountered the phrase. – Omnomnomnom May 22 '17 at 13:50
• 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

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.