The property that compact sets in a space are closed lies strictly between T1 and T2
If a topological space is Hausdorff (), then every compact subset of that space is closed. If every compact subset of a space is closed, then (since singletons are always compact) then the space is accessible (). There are spaces that are and have compact sets that are not closed, and there are spaces in which compact sets are always closed but that are not .
Let be an infinite set with the finite complement topology. Singletons are finite, and therefore closed, so is . Let , and let be an open cover of . Let . Then is finite. Choosing a member of for each remaining element of shows that has a finite subcover. Thus, every subset of is compact. An infinite subset of will then be compact, but not closed.
Let be an uncountable set with the countable complement topology. No two open sets are disjoint, so is not Hausdorff. Let be a compact subset of . I shall show that is finite. Suppose is infinite, and let be an infinite sequence in without any repetitions. For any natural number , let be all the elements of except for all the , where . Then is open for each , and covers , but has no finite subset that covers , contradicting the fact that is compact. This contradiction arose by assuming a compact subset of was infinite, all compact subsets of are finite. is (singleton sets are countable), so all compact subsets of are closed.
These examples were suggested by the person known as Polytope on EFNet’s math channel.
|Title||The property that compact sets in a space are closed lies strictly between T1 and T2|
|Date of creation||2013-03-22 17:38:59|
|Last modified on||2013-03-22 17:38:59|
|Last modified by||dfeuer (18434)|