# Thread: Compact Topology Definition

1. ## Compact Topology Definition

I am having trouble understanding the definition of a compact topological subset..

This is defined as a subset of a topological space where every open cover has a finite subcover...

however as a subcover is a subset of a cover, isnt every open cover a subcover of itself..

therefore isnt the definition eqivalent to saying that the set has a finite open cover, which is obviously not the correct definition....

can anyone see where i am going wrong??

2. Originally Posted by johnbarkwith
trouble understanding the definition of a compact topological subset.. This is defined as a subset of a topological space where every open cover has a finite subcover.
However as a subcover is a subset of a cover, isnt every open cover a subcover of itself.. therefore isnt the definition eqivalent to saying that the set has a finite open cover, which is obviously not the correct definition....
The definition means that if we can cover a compact set with any particular collection of open sets then there is a finite subcollection of those very sets which also covers the compact set.
It does not say that if a set has a finite open covering then the set is compact.
Rather, for a compact set, from any open covering of that set we can find a finite subcovering.

I have tried to put that two different ways. I hope that helps.

3. This is a common misunderstanding by people who are just learning the definition of "compact set" Every set has a finite open cover: the entire set itself is open and covers all subsets. But just having a finite open cover is not enough: not every open cover contains a finite subcover.

For example, let A= (0,1), a subset of the real numbers. That is itself an open set so it "covers" itself. But- let U be the collection of all open intervals, {(0, (m-1)/m)} when m is every positive integer. That is an open cover for A: if x is any number in A, it is larger than 0 and less than 1. But (m-1)/m converges to 1: If $\displaystyle \epsilon$> 0 there exist M so that $\displaystyle (M-1)/M> 1- \epsilon$. If we take $\displaystyle \epsilon= 1-x$, x is contained in that set. But any finite subcollection has a largest m: the largest set in it is (0, (m-1)/m), for that largest m, which does not contain any number larger than (m-1)/m. But (m-1)/m< 1 so there exist numbers in A that are not "covered" by that subcollection. (0, 1) is not compact.