[Topology]Determining compact sets of R

[GatorPower]

Homework Statement

Is A = {0} union {1/n | n [tex]\in[/tex] {1,2,3,...}} compact in R?
Is B = (0,1] compact in R?

Homework Equations

Definition of compactness, and equivalent definitions for the space R.

The Attempt at a Solution

A is compact, but I can't seem to find a plausible proof of it... It should be homeomorphic to [0,1] and then compactness would follow if I can do that?

B is not compact. A open cover could be C = {(1/n,1] | n [tex]\in[/tex] {1,2,3,...}} which contains no finite subcollection that covers B (atleast my textbook says so, but I don't quite understand that).

How would you prove compactness (or not) on these sets?

 

[micromass]

The set in A is certainly compact, but it is not homeomorphic to [0,1], so that's not a way to prove it.

Have you seen equivalences for compactness?? Maybe you should try those. (Hint: Heine-Borel theorem)

 

[GatorPower]

The set in A is certainly compact, but it is not homeomorphic to [0,1], so that's not a way to prove it.

Have you seen equivalences for compactness?? Maybe you should try those. (Hint: Heine-Borel theorem)

Sorry for the homeomorphic thing, thought about it completely wrong. Heine-Borel theorem (for metric spaces) gives that A is compact if it is complete and bounded. A is complete since any sequence in A cannot have a limit that lies outside A, and is obviously bounded by the inherited metric. Hence A is compact.

More generally A is compact if (Heine-Borel) A is closed and bounded. R - A = (-inf, 0) union (several open intervals between the elts of the sequence 1/n) union (1, inf) which is open so hence A is closed. A is also bounded and we can say that A is compact.

B = (0, 1]. I think one could use sequentially compact criterion here.. If Sn = 1/n it converges to 0 in R, but 0 is not in B. All subsequences converges to 0 as well, but since 0 is not in B we conclude that B is not sequentially compact and hence not compact.

 

[Fredrik]

Heine-Borel theorem (for metric spaces) gives that A is compact if it is complete and bounded.

Actually it's complete and totally bounded. Totally bounded implies bounded, but not the other way round.

The rest of what you said looks good, but you're making it a bit too complicated with B. You can just say that "B isn't closed, so..."

 

[GatorPower]

Actually it's complete and totally bounded. Totally bounded implies bounded, but not the other way round.

The rest of what you said looks good, but you're making it a bit too complicated with B. You can just say that "B isn't closed, so..."

Totally bounded => bounded, but in this particular case (from wikipedia): A subset of the real line, or more generally of (finite-dimensional) Euclidean space, is totally bounded if and only if it is bounded. Archimedean property is made use of.

Saying that B isn't closed is pretty much better. Thanks