Compact sets and proper mappings

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 20 May 2011

Proper mappings

A morphism is a continuous function $f:X\to Y$.
A closed morphism is a morphism $f:X\to Y$ such that

if $C$ is closed in $X$ then $f\left(C\right)$ is closed in $Y$.

A proper morphism is a continuous function $f:X\to Y$ such that

if $Z$ is a topological space then $f\times {\mathrm{id}}_Z:X\times Z\to Y\times Z$ is closed.

Compact spaces

A topological space $X$ is quasicompact if the mapping $p:X\to \mathrm{pt}$ is proper.
A topological space is compact if it is quasicompact and Hausdorff.

Let $X$ be a set. An ultrafilter on $X$ is a maximal filter $ℱ$ (with respect to inclusion), i.e. a filter $ℱ$ such that there is no filter on $X$ which is strictly finer than $ℱ$.

Let $X$ be a topological space. The following are equivalent.

(a)   $X$ is quasicompact.

(b)   Every filter on $X$ has a cluster point.

(c)   Every ultrafilter on $X$ has a limit point.

(d)   Every family of closed subsets of $X$ with empty intersection contains a finite subfamily with empty intersection.

(e)   Every open cover of $X$ contains a finite subcover.

		Proof.
	(c) $\Rightarrow$ (b): Let $ℱ$ be a filter and let $\overline{ℱ}$ be an ultrafilter containing $ℱ$. Let $x$ be a limit point of $\overline{ℱ}$. Then $x$ is a limit point of $ℱ$. (b) $\Rightarrow$ (c): Let $\overline{ℱ}$ be an ultrafilter. Let $x$ be a clusterpoint of $\overline{ℱ}$. Then $x$ is a limit point of $\overline{ℱ}$. (b) $\Rightarrow$ (d): Let $𝒞$ be a closed family with empty intersection. If every finite subfamily has empty intersection then $𝒞$ generates a filter $ℱ$. Let $x$ be a cluster point of $ℱ$. Then $x\in C$ for all $C$ in $𝒞$. This is a contradiction. So there exists a finite subfamily that does not have empty intersection. (d) $\Rightarrow$ (b): If there exists a filter $ℱ$ without a cluster point then $𝒞=\left\{\bar{F}\right|F\in ℱ\}$ is a family of closed sets contradicting (c). (d) $\iff$ (e): By taking complements. $\square$

Let $X$ be a metric space and let $E$ be a subset of $X$. The set $E$ is compact if and only if $E$ has a limit point in $E$.

		Proof.
	$\Leftarrow$: Let $K$ be a compact set and let $E$ be an infinite subset of $K$. If there is no limit point of $E$ in $K$ then for each $p\in K$ there is a neighborhood $N_p$ of $p$ which contains no other element of $E$. Then the open cover $$𝒩=\left\{N_p\right|p\in K\}\text{,}$$ of $K$ has no finite subcover. $\Rightarrow$: Let $S$ be an infinite subset of $E$. The metric space $E$ has a countable base. So every open cover of $E$ has a countable subcover $𝒞=\{C_1,C_2,\dots \}$. If $𝒞$ does not have a finite subcover then, for each $n$, $${\left(C_{\le n}\right)}^c\ne \varnothing \text{but}\underset{n}{\cap }{C}_{\le n}^c=\varnothing .$$ Let $S$ be a set which contains a poitn from each $C_{\le n}^c$. Then $S$ has a limit point. But this is a contradiction. $\square$

Let $X$ be a Hausdorff topological space and let $K$ be a compact subset of $X$. Then $K$ is closed.

		Proof.
	Let $x\in \bar{K}$. The neighbourhood filter $ℬ\left(x\right)$ of $x$ induces a filter ${ℬ}_K$ on $K$ which has a cluster point $y\in$. Since $ℬ\left(x\right)$ is coarser than ${ℬ}_K$ (considered as a filter base on $X$) the point $y$ is a cluster point of $ℬ\left(x\right)$. So $y=x$ since $X$ is Hausdorff. $\square$ The proof in Baby Rudin [R]: We will show that $K^c$ is open. Let $p\in K^c$. Let $𝒩$ be the open cover of $K$ given by $$𝒩=\left\{N_q\right|q\in K\}\text{where}{N}_q=\left\{B_{\frac{1}{2}d(p,q)}\left(q\right)|q\in K\right\}\text{.}$$ Let $\{N_{q_1}$ be a finite subcover of $K$. Then $$M=M_{q_1}$$, is an open set such that $p\in M\subseteq K^c$. So $p$ is an interior point of $K^c$. So $K$ is open. $\square$

Let $X$ be a metric space and let $E$ be a compact subset of $X$. Then $E$ is closed and bounded.

		Proof.
	Since a metric space is Hausdorff, $E$ is closed. If $E$ is not bounded then there is an infinite sequence in $E$ that does not have a limit point. $\square$

1. A k-cell is compact. [DEFINE K-CELL?]
2. Let $E$ be a subset of ${ℝ}^k$. If $E$ is closed and bounded then $E$ is compact.

		Proof.
	If $E$ is closed and bounded then $E$ is a closed subset of a $k$-cell. Since closed subsets of compact sets are compact $E$ is compact. $\square$

Locally ??? spaces

Let $E$ be a subset of $X$ and let $x$ be an element of ????. The set $E$ is locally closed at $x$ if there is a neighborhood $N_x$ of $x$ such that $N_x\cap E$ is closed in $N_x$.
The space $X$ is locally compact if each point of $x$ has a compact neighborhood.
The space $X$ is locally connected if each point of $X$ has a fundamental system of connected neighborhoods.

Notes and References

These notes follow Bourbaki [Bou???] Chapter II???. The condition that $\Delta$ is closed is the condition used in algebraic geometry for a separated scheme. The treatment of metric spaces and completion follows [BR] Chapter 2 Exercise ??. Separability appears in [BR] Chapter 2 Exercises 22 and 23, and in [Ru] Chapter 4 Exercises 2, 3, 4 and 18. A uniform space is almost a metric space: By [Bou??] the separable Hausdorff uniform spaces are exactly the separable metric spaces.

Graphs of relations and functions are an interesting point. Maybe not. Put this in exercises? This seems to be special to compact and locally compact X and f: X to X/R, where R is an equivalence relation.

References

[Bou] N. Bourbaki, General Topology, Springer-Verlag, 1989. MR??????.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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