# Compact Sets

Homework Statement

Show that every compact set must be closed.

I am looking for a simple proof.

This is supposed to be Intro Analysis proof.

Relevant equations

Any compact set must be bounded.

## The Attempt at a Solution

Suppose A is not closed, so let a be an accumulation point of A such that a is in A etc. (I think I can finish this proof, but I do not like it.)

Does anyone know of a simpler proof.

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I don't understand: if you are working in R^n (which I imagin you may be doing, since you mention an analysis book), then Heine-Borel is the resul. If not, the result is true only if you have a compact subset of a Hausdorff space, where you show that the complement in the space of the compact set is open, by finding a 'hood (neighborhood) of a point x in the complement that is disjoint from the compact space, by using Hausdorffness (so you can choose disjoint pairs of open sets containing x, disjoint from finitely-many open subsets of the cover. This is the main idea ). So this last one should take care of the case of R^n, or for any Hausdorff space.

Yes, what is the context you are working in?? Are you working in $\mathbb{R}^n$?? In a metric space? A topological space? A Hausdorff space??

And how did you define compact??

The section in the book is called Topology of the Real Number System. We have not covered metric spaces, topological spaces, Hausdorff space, etc.

A set A subset R is compact if every open cover C of A contains a finite subcover F of A.

And by the way can anyone explain what does a cover, a subcover, etc. really represent? I understand the definitions but what is really meant by something like C covers A?

C covers A if the union of the open sets contains A; a subcover of C is a subcollection of the subsets of C, so you want a finite subcollection that contains C.

An example, maybe somewhat-trivial, is that of a convergent sequence {xn} in R, with limit x; take a collection of open intervals (an,bn)--take it, please! (sorry, prepping for my 2nd job)-- with an:= xn-0.1, (any small number will do ), and bn:=xn+0.1, and a 'hood (neighborhood) (x-0.1,x+0.1) of the limit point x . Then the union of the intervals (a,n,bn) and (x-0.1,x+0.1)covers {xn}, together with a 'hood , trivially, since each member of {xn} is in the n-th interval. Now, you can find a subcover ( you, and you alone!), by using (x-0.1,x+0.1) , which , by convergence of {xn}, will contain all-but-finitely many of the elements of the sequence, so that you can use finitely-many intervals (an,bn) to cover all the remaining elements.

It may be good too, for contrast, to consider a cover of the integers in the real line, by
open intervals (n-1/4,n+1/4), and see if you can find a finite subcollection of these open sets that covers the integers. For a more advanced problem, show that a half-open interval [a,b) in the real line is not compact. I'm pretty sure you will eventually see Heine-Borel theorem which tells you that, for Rn, the compact sets are always those that are close and bounded.

HallsofIvy