Are all open sets compact in the discrete topology?

[Useful nucleus]

A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.
Does not this imply that every open set is compact. Because let F is open, then
F= F $\bigcup$ ∅. Since F and ∅ are open , we obtained a finite subcover of F.
Am I missing something here?

 

[lavinia]

A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.
Does not this imply that every open set is compact. Because let F is open, then
F= F $\bigcup$ ∅. Since F and ∅ are open , we obtained a finite subcover of F.
Am I missing something here?

Every cover must have a finite subcover not just one or some of them.

 

[Useful nucleus]

But F$\bigcup$∅ is contained in every open cover for F?

 

[SteveL27]

But F$\bigcup$∅ is contained in every open cover?

As a subset, yes. But F and ∅ might not be elements of the open cover.

Compactness says that if you have an open cover -- a collection of open sets whose union covers F -- then some finite subcollection will also cover F. In other words you have to be able to pick out a finite collection of sets from your open cover, and show that the finite collection is also an open cover.

As an example, take F = (0,1), the open unit interval. Let A_n = (1/n, 1). Then the collection {A_n} for n = 1, 2, 3, ... is an open cover of F; but no finite subcollection of the A_n's covers (0,1). Working through this example will be helpful.

 

[Useful nucleus]

Then, I think the definition of the open cover of a set F has to emphasize that F has to be a proper subset of the cover. Otherwise the argument in my first post will be correct.

 

[lavinia]

But F$\bigcup$∅ is contained in every open cover for F?

As SteveL7 said, F and ∅ may not be elements of the open cover. An open cover is a collection of open sets. F may not be in the collection.

 

[SteveL27]

Then, I think the definition of the open cover of a set F has to emphasize that F has to be a proper subset of the cover. Otherwise the argument in my first post will be correct.

Well if F is an open set, then {F} is indeed an open cover of F consisting of one set. So that's an example of one open cover of F that happens to have a finite subcover.

But compactness requires that EVERY open cover has a finite subcover.

I think you would find it helpful to work through the example of F = (0,1) that I gave earlier.

FWIW pretty much everyone has trouble learning about compactness. It's very unintuitive. Not like connectedness, say, whose technical definition matches our intuition about what a connected set should be. With compactness, it's like, "How did they ever think of that?"

 

[Useful nucleus]

SteveL27 and lavinia thank you very much for your help! I have a better understanding now of the definition. The open unit interval example was particularly helpful.

 

[Deveno]

the usual example "in the other direction" (unbounded sets, rather than open sets) is R:

certainly {R} is an open cover of R. yet it hardly seems "right" to call R "compact", given how large it is.

and, of course, it is not:

{(n,n+2): n in Z} is an open cover of R, but we need "all |Z| of them" (and |Z| is not finite) to cover R, no finite subcover will do.

 

[lavinia]

SteveL27 and lavinia thank you very much for your help! I have a better understanding now of the definition. The open unit interval example was particularly helpful.

Here is a stark example.

Take the integers with the discrete topology. Here each singleton,{n}, is an open set.
Cover the integers with the set of singletons. Not onl;y is there no finite subcover, there is no subcover.