Why is (0,1) not compact in topology?

[mathwonk]

learn the definition.

 

[Bacle2]

There is also a result that for compact metric spaces, every infinite sequence has a convergent subsequence, and, e.g., {1/n} has no convergent subsequence. This result has to see with the fact that you will end up with an infinite collection of elements contained in a finite collection of sets, so that some set will contain infinitely-many elements, so that you can then conclude, by Bolzano-Weirstrass (every bounded infinite subset of R^n has a limit point) that there should be a limit point. The covers described in other posts without finite subcovers are not counterexamples.

Now, for more fun, consider all other types of compactness: sequential, limit-point, countable...(can't remember all others) and when they are equivalent to each other.

 

[Sina]

I am confused.

Seems like you can use (1/n, 1) to argue for (0,1) or [0,1] and say it is not compact. Because the subcover is infinite in (0,1) and [0,1].

Nope this is not a cover as explained. Even if you tried other sets instead of these it wouldn't work. Let suppose you tried.

Let sets be (-ε,2/n-ε/2) where ε is as small as you like but non zero and n starts from 1. This will cover your set including 1 and 0 (try building one your self so you will see why I had to build it this way). However almost all of the elements (another way of saying all but finitely many) of this will cluster in in the interval (-ε,0). So you can find a finite subcover.

Caution, this is not a proof that this interval is compact :) This is just an example on how to proceed for understanding things, by building your own examples. It helps you build intiution.

 

[lavinia]

I've only just started getting into Topology and a few examples of compactedness have me a little confused. For instace, the one in the title: how is the open interval (0,1) not compact but [0,1] is? Obivously I'm making some sort of logical mistake but the way I think about it is that there are any number of open covers for (0,1). I'll use {(-2,2),(-1,3),(0,4)}. By definition, there needs to exist a finite subcover which still contains (0,1) right? So couldn't that finite subcover simply be {(-1,3)}? I have a feeling the reason I'm not grasping this example is because the definition states that a finite subcover must exist for all open covers of a given set, but I still can't think of an open cover which does not have a finite subcover which contains (0,1). Thanks.

For the real numbers a set is compact if it is bounded and complete. The open interval is not complete since it contains Cauchy sequences that converge outside of it. The closed interval catches these Cauchy sequences and so is compact.

 

[Bacle2]

Yes, in the most general sense, a metric space being compact is the same as being complete and totally bounded. A complete space must be closed (to avoid convergence to limit points not included in the set), and in R^n, totally bounded and bounded are equivalent.

 

[Jamma]

Here is an easier way to see why (0,1) is not compact (more intuitively, anyway)- topologically, we cannot tell it apart from the real numbers. Sure, to us looking from outside embedded in the real line it "looks small" and "non-infinite" but we could relabel the points so that it wasn't. We could take a homeomorphism from (0,1) (actually, (-1,1)) into the real line by sending x to, say tan(\pi.x).

I'm sure that you see intuitively why the real line is not compact? Think of any argument for this that you want- you will be able to directly translate it to the space (0,1) (we think of (0,1) as being without boundary on both ends).

Try thinking about the sphere with the north pole removed. You should have seen stereographic projection which defines a homeomorphism from the sphere with a point removed to the plane. So the sphere with a point removed is non-compact.

Effectively, what I'm trying to tell you is that when looking purely from a topological point of view, there are good reasons why the spaces you are looking at should be considered as non compact.