# How is (0,1) not compact?

1. Jun 23, 2009

### 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.

2. Jun 23, 2009

### 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!

Last edited: Jun 23, 2009
3. Jun 23, 2009

### Newtime

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.

4. Jun 23, 2009

### Office_Shredder

Staff Emeritus
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 nk and the union of these intervals would just be the interval (1/nk,1)

5. Jun 23, 2009

### Newtime

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.

6. Jun 23, 2009

### HallsofIvy

More succinctly, any finite subset of {An} with An= (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).

Last edited by a moderator: Jun 23, 2009
7. Jun 24, 2009

### Newtime

Well put, thanks.

8. Jun 26, 2009

### 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.

9. Jul 30, 2009

### 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.

10. Jul 30, 2009

### rasmhop

{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.

11. Jul 30, 2009

### dasdos

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

12. Aug 12, 2009

### 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.

13. Aug 12, 2009

### dasdos

dasdos

14. Aug 12, 2009

### CaffeineJunky

The proof is the Heine-Borel theorem.

15. Aug 12, 2009

### 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).

16. Aug 13, 2009

### Sina

17. May 30, 2011

### mbarby

hi,
the subject seems rather cold. but there are things i still cant comprehend after reading your discussions several times.
for example why cant 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..

18. May 30, 2011

### micromass

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[.

19. May 30, 2011

### 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 ?

20. May 30, 2011

### 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.

21. May 30, 2011

### 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].

22. May 30, 2011

### micromass

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).

23. May 30, 2011

### mbarby

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

24. May 30, 2011

### 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...

25. May 30, 2011

### mbarby

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