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: 29 July 2014
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 $\mathcal{F}$ (with respect to inclusion), i.e. a filter
$\mathcal{F}$ such that there is no filter on $X$
which is strictly finer than $\mathcal{F}$.
Let $X$ be a topological space. The following are equivalent.
(C) 
If $\mathcal{J}$ is a filter on $X$ then there exists $x\in X$ such that
$x$ is a cluster point of $\mathcal{J}\text{.}$

(C') 
If $\mathcal{G}$ is an ultrafilter on $X$ then there exists $x\in X$ such that
$x$ is a limit point of $\mathcal{G}\text{.}$

(C'') 
If $\mathcal{C}$ is a collection of closed sets such that
$\bigcap _{K\in \mathcal{C}}K=\varnothing $ then there
exists $\ell \in {\mathbb{Z}}_{>0}$ and
${K}_{1},\dots ,{K}_{\ell}\in \mathcal{C}$
such that ${K}_{1}\cap {K}_{2}\cap \cdots \cap {K}_{\ell}=\varnothing \text{.}$

(C''') 
If $\mathcal{S}$ is a collection of open sets such that $\bigcup _{U\in \mathcal{S}}U=X$
then there exists $\ell \in {\mathbb{Z}}_{>0}$ and
${U}_{1},\dots ,{U}_{\ell}\in \mathcal{S}$
such that ${U}_{1}\cup {U}_{2}\cup \cdots \cup {U}_{\ell}=X\text{.}$



Sketch of proof. 

(C'') $\iff $ (C''') by taking complements.
(C) $\Rightarrow $ (C'): Assume (C).
To show: If $\mathcal{G}$ is an ultrafilter on $X$ then there exists $x\in X$ such that
$x$ is a limit point of $\mathcal{G}\text{.}$
Assume $\mathcal{G}$ is an ultrafilter on $X\text{.}$
By (C), there exists $x\in X$ such that $x$ is a cluster point of $\mathcal{G}\text{.}$
Since $\mathcal{G}$ is an ultrafilter $x$ is a limit point of $\mathcal{G}\text{.}$
(C') $\Rightarrow $ (C): Assume (C').
To show: If $\mathcal{J}$ is a filter on $X$ then there exists $x\in X$ such that
$x$ is a cluster point of $\mathcal{J}\text{.}$
Assume $\mathcal{J}$ is a filter on $X\text{.}$
Since the collection of filters on $X$ satisfies the hypotheses of Zorn's lemma, there exists an ultrafilter $\mathcal{G}$
such that $\mathcal{G}\supseteq \mathcal{J}\text{.}$
By (C') there exists $x\in X$ such that $x$ is a limit point of $\mathcal{G}\text{.}$
So $x$ is a cluster point of $\mathcal{G}\text{.}$
Since $\mathcal{G}\supseteq \mathcal{J}$ and $x$ is a cluster point of $\mathcal{G}$ then
$x$ is a cluster point of $\mathcal{J}\text{.}$
(not C'') $\Rightarrow $ (not C): Assume that there is a collection $\mathcal{C}$ of closed sets such that
$\bigcap _{K\in \mathcal{C}}K=\varnothing $ and, if
$\ell \in {\mathbb{Z}}_{>0}$ and
${K}_{1},\dots ,{K}_{\ell}\in \mathcal{C}$
then ${K}_{1}\cap \cdots \cap {K}_{\ell}\ne \varnothing \text{.}$
Let $\mathcal{J}$ be the set of subsets of $X$ which contain a set in $\mathcal{C}\text{.}$
Then $\mathcal{J}$ is a filter.
Since $\bigcap _{N\in \mathcal{J}}\stackrel{\u203e}{N}\subseteq \bigcap _{K\in \mathcal{C}}\stackrel{\u203e}{K}=\bigcap _{K\in \mathcal{C}}K=\varnothing ,$
$\mathcal{J}$ does not have a cluster point.
(not C) $\Rightarrow $ (not C''): Assume that there exists a filter $\mathcal{J}$ on $X$ with no cluster point.
Then $\bigcap _{N\in \mathcal{J}}\stackrel{\u203e}{N}=\varnothing \text{.}$
Since $\mathcal{J}$ is a filter, if $\ell \in {\mathbb{Z}}_{>0}$ and
${N}_{1},\dots ,{N}_{\ell}\in \mathcal{J}$
then ${N}_{1}\cap \cdots \cap {N}_{\ell}\ne \varnothing $
and therefore $\stackrel{\u203e}{{N}_{1}}\cap \cdots \cap \stackrel{\u203e}{{N}_{\ell}}\ne \varnothing \text{.}$
Let $\mathcal{C}=\left\{\stackrel{\u203e}{N}\hspace{0.17em}\right\hspace{0.17em}N\in \mathcal{J}\}\text{.}$
Then $\mathcal{C}$ is a collection of closed sets such that $\bigcap _{K\in \mathcal{C}}K=\varnothing $
but there does not exist ${K}_{1},\dots ,{K}_{\ell}\in \mathcal{C}$
such that ${K}_{1}\cap {K}_{2}\cap \cdots \cap {K}_{\ell}=\varnothing \text{.}$
$\square $

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


Proof. 

Let $X$ be a Hausdorff topological space with topology $\mathcal{T}\text{.}$ Let
$K\subseteq X\text{.}$
Assume $K$ is compact.
To show: $K$ is closed.
To show: ${K}^{c}$ is open.
To show: If $x\in {K}^{c}$ then $x$ is an interior point of ${K}^{c}\text{.}$
Assume $x\in {K}^{c}\text{.}$
To show: There exists $U\in \mathcal{T}$ such that $x\in U$ and $U\subseteq {K}^{c}\text{.}$
For each $y\in K$ there exist ${U}_{xy}\in \mathcal{T}$
and ${V}_{xy}\in \mathcal{T}$ such that
$x\in {U}_{xy}$ and
$y\in {V}_{xy}$ and
${U}_{xy}\cap {V}_{xy}=\varnothing \text{.}$
Then
$$\left\{{V}_{xy}\hspace{0.17em}\right\hspace{0.17em}y\in K\}$$
is an open cover of $X\text{.}$
Since $K$ is compact there exists $N\in {\mathbb{Z}}_{>0}$ and
${y}_{1},{y}_{2},\dots ,{y}_{N}\in K$
such that
$$\{{V}_{x{y}_{1}},{V}_{x{y}_{2}},\dots ,{V}_{x{y}_{N}}\}$$
is an open cover of $K\text{.}$
Let
$$U={U}_{x{y}_{1}}\cap \cdots \cap {U}_{x{y}_{N}}\text{.}$$
Since $U$ is a finite intersection of open sets,
$U$ is open.
Also $x\in U$ and
$$\begin{array}{ccc}{\displaystyle U\cap K}& =& {\displaystyle ({U}_{x{y}_{1}}\cap \cdots \cap {U}_{x{y}_{N}})\cap ({V}_{x{y}_{1}}\cup \cdots \cup {V}_{x{y}_{N}})}\\ & =& {\displaystyle \varnothing \text{.}}\end{array}$$
So $x\in U$ and $U\subseteq {K}^{c}\text{.}$
So $x$ is an interior point of ${K}^{c}\text{.}$
So ${K}^{c}$ is open.
So $K$ is closed.

Let $x\in \stackrel{\u203e}{K}$. The neighbourhood filter
$\mathcal{B}\left(x\right)$ of $x$
induces a filter ${\mathcal{B}}_{K}$ WHAT DOES THIS MEAN???
on $K$ which has a cluster point $y\in ????$.
Since
$\mathcal{B}\left(x\right)$ is coarser than
${\mathcal{B}}_{K}$ (considered as a filter base on
$X$) the point $y$ is a cluster point
of $\mathcal{B}\left(x\right)$. So
$y=x$ since $X$ is Hausdorff.
$\square $

ExampleHomework: Let
$X$ be a set with more than one point with topology
$\mathcal{T}=\{\varnothing ,X\}\text{.}$
Show that every subset
$A\subseteq X$ is compact, but the only closed subsets of
$X$ are
$\varnothing $ and
$X\text{.}$ Note that
$X$ is not Hausdorff.
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
This proof that compact implies closed in Hausdorff spaces is taken from notes of J. Hyam Rubinstein for a course Metric and Hilbert spaces at the University of Melbourne. The proof in
[BR, Theorem 2.34] is of the same intent but stated only for metric spaces.
These notes follow Bourbaki [Bou, Ch.I §9 no. 1,2] and [Bou, Ch.I §10 no.1,2].
Theorem 2.1 characterizes compact spaces as spaces where limits exist.
WHAT IS HEINEBOREL??? WHAT IS BORELLEBESGUE???
The characterization of compact spaces by proper mappings is fundamental in algebraic geometry
[REFERENCE?? EGA???].
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, SpringerVerlag, 1989.
MR1726779.
[BR]
W. Rudin, Principles of mathematical analysis, Third edition,
International Series in Pure and Applied Mathematics, McGrawHill 1976.
MR0385023.
[Ru]
W. Rudin,
Real and complex analysis, Third edition, McGrawHill, 1987.
MR0924157.
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