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:XY.
A closed morphism is a morphism f:XY such that
if C is closed in X then f(C) is closed in Y.
A proper morphism is a continuous function f:XY such that
if Z is a topological space then f×idZ : X×Z Y×Z is closed.

Compact spaces

A topological space X is quasicompact if the mapping p:Xpt 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 (with respect to inclusion), i.e. a filter such that there is no filter on X which is strictly finer than .

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

(C) If 𝒥 is a filter on X then there exists xX such that x is a cluster point of 𝒥.
(C') If 𝒢 is an ultrafilter on X then there exists xX such that x is a limit point of 𝒢.
(C'') If 𝒞 is a collection of closed sets such that K𝒞K= then there exists >0 and K1,,K𝒞 such that K1K2K=.
(C''') If 𝒮 is a collection of open sets such that U𝒮U=X then there exists >0 and U1,,U𝒮 such that U1U2U=X.

Sketch of proof.

(C'') (C''') by taking complements.

(C) (C'): Assume (C).
To show: If 𝒢 is an ultrafilter on X then there exists xX such that x is a limit point of 𝒢.
Assume 𝒢 is an ultrafilter on X.
By (C), there exists xX such that x is a cluster point of 𝒢.
Since 𝒢 is an ultrafilter x is a limit point of 𝒢.

(C') (C): Assume (C').
To show: If 𝒥 is a filter on X then there exists xX such that x is a cluster point of 𝒥.
Assume 𝒥 is a filter on X.
Since the collection of filters on X satisfies the hypotheses of Zorn's lemma, there exists an ultrafilter 𝒢 such that 𝒢𝒥.
By (C') there exists xX such that x is a limit point of 𝒢.
So x is a cluster point of 𝒢.
Since 𝒢𝒥 and x is a cluster point of 𝒢 then x is a cluster point of 𝒥.

(not C'') (not C): Assume that there is a collection 𝒞 of closed sets such that K𝒞K= and, if >0 and K1,,K𝒞 then K1K. Let 𝒥 be the set of subsets of X which contain a set in 𝒞.
Then 𝒥 is a filter.
Since N𝒥NK𝒞K=K𝒞K=, 𝒥 does not have a cluster point.

(not C) (not C''): Assume that there exists a filter 𝒥 on X with no cluster point.
Then N𝒥N=.
Since 𝒥 is a filter, if >0 and N1,,N𝒥 then N1N and therefore N1N.
Let 𝒞={N|N𝒥}.
Then 𝒞 is a collection of closed sets such that K𝒞K= but there does not exist K1,,K𝒞 such that K1K2K=.

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 𝒯. Let KX.
Assume K is compact.
To show: K is closed.
To show: Kc is open.
To show: If xKc then x is an interior point of Kc.
Assume xKc.
To show: There exists U𝒯 such that xU and UKc.
For each yK there exist Uxy𝒯 and Vxy𝒯 such that xUxy and yVxy and UxyVxy=. Then {Vxy|yK} is an open cover of X.
Since K is compact there exists N>0 and y1,y2,,yNK such that { Vxy1, Vxy2, , VxyN } is an open cover of K.
Let U=Uxy1 UxyN. Since U is a finite intersection of open sets, U is open. Also xU and UK = ( Uxy1 UxyN ) ( Vxy1 VxyN ) = . So xU and UKc.
So x is an interior point of Kc.
So Kc is open.
So K is closed.

  1. Let x K. The neighbourhood filter (x) of x induces a filter K WHAT DOES THIS MEAN??? on K which has a cluster point y????. Since (x) is coarser than K (considered as a filter base on X) the point y is a cluster point of (x). So y=x since X is Hausdorff.

Example-Homework: Let X be a set with more than one point with topology 𝒯={,X}. Show that every subset AX is compact, but the only closed subsets of X are and X. 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 Nx of x such that NxE is closed in Nx.
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 HEINE-BOREL??? WHAT IS BOREL-LEBESGUE??? 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, Springer-Verlag, 1989. MR1726779.

[BR] W. Rudin, Principles of mathematical analysis, Third edition, International Series in Pure and Applied Mathematics, McGraw-Hill 1976. MR0385023.

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

page history