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

[Newtime]

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.

 

[matticus]

consider the family of open sets of the form (1,1/n) where n = 1,2,...

note the definition of compact is that EVERY open cover contains a finite subcover.

edit: oops that should be (1/n,1) of course!

 

[Newtime]

consider the family of open sets of the form (1,1/n) where n = 1,2,...

note the definition of compact is that EVERY open cover contains a finite subcover.

I'm not the greatest when it comes to series and sequences, but if the right term, 1/n, converges to 0 can we regard it as simply 0? If so then wouldn't the cover and finite subcover of (1/n,1) be the set itself? If we can't regard it as zero then it seems that clearly there is not a finite subcover since there will always be some n such that 1/n is greater than 0 but then again that implies that the set does not cover (0,1) to begin with. Thank you for the quick reply though, this has been bugging me for a while.

 

[Office_Shredder]

I'm not the greatest when it comes to series and sequences, but if the right term, 1/n, converges to 0 can we regard it as simply 0? If so then wouldn't the cover and finite subcover of (1/n,1) be the set itself? If we can't regard it as zero then it seems that clearly there is not a finite subcover since there will always be some n such that 1/n is greater than 0 but then again that implies that the set does not cover (0,1) to begin with. Thank you for the quick reply though, this has been bugging me for a while.

The set of all intervals (1/n,1) DOES cover the interval (0,1). If x is in (0,1) then necessarily there exists N such that 1/N < x and hence x is in (1/N,1). So taking the union of all the intervals (1/n,1) for each n gives us the interval (0,1) as required. And trivially there is no finite subcover, since if there was there would be a list $(1/n_1, 1), (1/n_2, 1),...$ and we can pick the largest n_k and the union of these intervals would just be the interval (1/n_k,1)

 

[Newtime]

The set of all intervals (1/n,1) DOES cover the interval (0,1). If x is in (0,1) then necessarily there exists N such that 1/N < x and hence x is in (1/N,1). So taking the union of all the intervals (1/n,1) for each n gives us the interval (0,1) as required. And trivially there is no finite subcover, since if there was there would be a list $(1/n_1, 1), (1/n_2, 1),...$ and we can pick the largest n_k and the union of these intervals would just be the interval (1/n_k,1)

Very helpful, thanks. Just to make absolutely sure I've got this though: the reason there cannot exist a finite subcover is because the finite subcover would end up being the cover itself which is not finite.

 

[HallsofIvy]

More succinctly, any finite subset of {A_n} with A_n= (1/n, 1), must have a largest n. Call that largest n, "N", so the the subset contains no member of (0, 1) less than 1/N and so can't be a cover for (0, 1).

 

[Newtime]

More succinctly, any finite subset of {A_n} with A_n= (1/n, 1), must have a largest n. Call that largest n, "N", so the the subset contains no member of (0, 1) less than 1/N and so can't be a cover for (0, 1).

Well put, thanks.

 

[zhentil]

A lot of people confuse "for all" with "there exists." A way to keep it straight is that there is always an open cover consisting of precisely one set.

 

[dasdos]

We all know that [0,1] is compact but if we look at the set {{0},{A_n},{1}} where A_n = (1/n,1) where n is an integer value which covers [0,1], but there is no finite subset of this set which cover [0,1]. I understand there is a flaw in my argument I just want to know what it is.

 

[rasmhop]

We all know that [0,1] is compact but if we look at the set {{0},{A_n},{1}} where A_n = (1/n,1) where n is an integer value which covers [0,1], but there is no finite subset of this set which cover [0,1]. I understand there is a flaw in my argument I just want to know what it is.

{0} and {1} are not open sets so you only provided a cover with no finite subcover, but compactness only guarantees that an OPEN cover has a finite subcover. Otherwise no infinite set could be compact because we could just cover it by all it's one-element subsets.

 

[dasdos]

Wow why didn't I think of that. This stuff escapes me some times. Thanks for the quick response and the detailed answer.

 

[Sina]

If you want to use machinery, in Reals (or any complete space) a set is compact iff it is closed and bounded and you can look at proof of the theorem to get an insight on how so.

But of course at the initial stages of your learning it is best to try to do this via constructing counter examples etc.

 

[dasdos]

Could you please post a link showing me the proof?

dasdos

 

[CaffeineJunky]

The proof is the Heine-Borel theorem.

 

[dasdos]

Doesn't Heine-Borel only prove in one direction? I was taught that if its closed and bounded then its compact. How do we know that every compact set has to be closed and bounded (in a Hausdorff space).

 

[Sina]

http://en.wikipedia.org/wiki/Heine–Borel_theorem

in R^n and its subsets both ways are true but in Haussdorf spaces one way works.

 

[mbarby]

hi,
the subject seems rather cold. but there are things i still can't comprehend after reading your discussions several times.
for example why can't we take (0 , 1/n(k) ) U ( 1/n(k+1), 1) for simplicitys sake. this is a finite subcover is it not ? (k are indexes)
ps: i am not a mathematician so be generous with the explanations..

 

[micromass]

hi,
the subject seems rather cold. but there are things i still can't comprehend after reading your discussions several times.
for example why can't we take (0 , 1/n(k) ) U ( 1/n(k+1), 1) for simplicitys sake. this is a finite subcover is it not ? (k are indexes)
ps: i am not a mathematician so be generous with the explanations..

A finite subcover of what??
If you take $(]1/n,1[)_n$, then this forms an open cover of ]0,1[ but not a finite subcover. You can't take ]0/1n[ as an element of a finite subcover, because it was not an element of the original cover all along, i.e. ]0,1/n[ is not of the form ]1/n,1[.

 

[mbarby]

let me rephrased this so that i understand it right: when we say open we mean open in the topological sense, i.e. if (0,1] is given as open then all the open subsets should be in the form of (x,y], is that right ?

 

[HallsofIvy]

For example, in the set of rational numbers with the topology inherited from the real numbers (the metric topology with d(p,q)= |p- q|) all compact sets are closed and bounded but there exist sets that are both closed and bounded that are NOT compact.]

For example, the set $\{x\in Q| x^2\le 2\}$ is both closed and bounded, in that topology, but not compact.

 

[micromass]

No, not all open subsets are of the form (x,y] there. The (x,y] certainly are open and form a "subbasis for the topology", but there are more open sets than just the (x,y].

For example, $(0,1/2)\cup (1/2,1]$ is also open in (0,1].

 

[micromass]

For example, in the set of rational numbers with the topology inherited from the real numbers (the metric topology with d(p,q)= |p- q|) all compact sets are closed and bounded but there exist sets that are both closed and bounded that are NOT compact.]

For example, the set $\{x\in Q| x^2\le 2\}$ is both closed and bounded, in that topology, but not compact.

Also $[0,1]\cap \mathbb{Q}$ is closed and bounded, but not compact in $\mathbb{Q}$. There are very few compact sets in Q (in technical terms: the rationals are not locally compact).

 

[mbarby]

No, not all open subsets are of the form (x,y] there. The (x,y] certainly are open and form a "subbasis for the topology", but there are more open sets than just the (x,y].

For example, $(0,1/2)\cup (1/2,1]$ is also open in (0,1].

but the result of the unions itself is of the form (x,y] . so this is considered open i guess...

 

[micromass]

Not really, the result of the union is simply (0,1] with the point 1/2 removed. So it's not from the form (x,y]. It is open however...

 

[mbarby]

thx for the help, i will have to go through all this once more , i think :). ...

 

[lavinia]

A compact metric space must be complete.

 

[mathwonk]

f(x) = 1/x is continuous but unbounded on (0,1), so (0,1) is not compact.

 

[zhentil]

I'm not sure the original poster will understand using the big guns. Not understanding why "but (0,1/2),(1/4,1) IS an open cover with a finite subcover" doesn't contradict noncompactness is a misunderstanding of logic, not topology.

 

[colstat]

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].

 

[Newtime]

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].

Note that $$\bigcup_{i=1}^{\infty} (1/n,1)$$ is not a cover of [0,1] because neither 0 nor 1 are in this set.

 

[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.