Hausdorff implies KC: Difference between revisions

From Topospaces
(New page: {{topospace property implication}} ==Statement== ===Property-theoretic statement=== Here are two equivalent formulations: * The property of being a Hausdorff space is stronger than...)
 
No edit summary
Line 29: Line 29:


Note that the crucial thing is to pass to a ''finite'' subcover of <math>A</math>, so that we need to intersect only ''finitely'' many open subsets containing <math>x</math>, because the axioms for open subsets only guarantee that ''finite'' intersections of open subsets are open.
Note that the crucial thing is to pass to a ''finite'' subcover of <math>A</math>, so that we need to intersect only ''finitely'' many open subsets containing <math>x</math>, because the axioms for open subsets only guarantee that ''finite'' intersections of open subsets are open.
==References==
===Textbook references===
* {{booklink|Munkres}}, Page 165 (Theorem 26.3) Also see Lemma 26.4, that provides a slightly more general formulation useful in other proofs, for instance, [[compact Hausdorff implies normal]]
* {{booklink|SingerThorpe}}, Page 27

Revision as of 20:31, 21 April 2008

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property must also satisfy the second topological space property
View all topological space property implications | View all topological space property non-implications
|

Property "Page" (as page type) with input value "{{{stronger}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.Property "Page" (as page type) with input value "{{{weaker}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.

Statement

Property-theoretic statement

Here are two equivalent formulations:

Verbal statement

Any compact subset of a Hausdorff space is closed.

Proof

Suppose X is a Hausdorff space and A is a closed subset of X that is compact in the subspace topology. We need to show that A is closed in X. In order to do this, it suffices to show that the complement of A in X is an open subset. In fact, it suffices to show that for any xAX, there is an open subset U of X containing x, and disjoint from A.

Let's do this:

  • For every yA, use the Hausdorffness of X to find disjoint open subsets Uyx and Vyy
  • Now the Vys, as yA, cover A, hence VyA form an open cover of A in the subspace topology. By compactness of A, we can find a finite collection y1,y2,,ynA such that the union of Vy1,Vy2,,Vyn contains A
  • Now consider the intersection:

U=i=1nUyi

This is an open set containing x, because it is a finite intersection of open sets, and it is disjoint from each of the Vyis, hence it is disjoint from A. This completes the proof.

Note that the crucial thing is to pass to a finite subcover of A, so that we need to intersect only finitely many open subsets containing x, because the axioms for open subsets only guarantee that finite intersections of open subsets are open.

References

Textbook references

  • Topology (2nd edition) by James R. MunkresMore info, Page 165 (Theorem 26.3) Also see Lemma 26.4, that provides a slightly more general formulation useful in other proofs, for instance, compact Hausdorff implies normal
  • Lecture Notes on Elementary Topology and Geometry (Undergraduate Texts in Mathematics) by I. M. Singer and J. A. ThorpeMore info, Page 27