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 March 20th, 2009, 11:00 AM #1 Newbie   Joined: Mar 2009 Posts: 9 Thanks: 0 open covers and compactness I'm a hobby mathematician (i.e., I've done some college analysis and measure theory but not too much), and I'm trying to understand the notion of open covers / open subcovers / finite cover in relation to compactness: The set consisting of zero and 1/n, n=1,2,..., is compact - how do I prove this using the notion of covers? (i.e., not prove it by the Heine-Borel theorem). I can't see how I can find a finite subcover, which I need to prove compactness? Any explanations and suggestions on how to construct these covers? Thanks
 March 20th, 2009, 12:19 PM #2 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: open covers and compactness Take any open cover and pick the open set containing 0, now there are only finitely many points of the form $\frac{1}{n}$ that are not in this open set. (This follows immediately from the definition of open set and the fact that $\frac{1}{n}$ converges to 0.) Now pick an open set containing each of the finitely many points that remain and you have your finite subcover. Hope this helps.
 March 20th, 2009, 05:45 PM #3 Newbie   Joined: Mar 2009 Posts: 9 Thanks: 0 Re: open covers and compactness Yes, that is what I thought as well. But in the definition of cover in wikipedia, it looks like the cover should be contained in the set - so if I take an open set around zero, this won't be contained in the set of 0 and 1/n (lets call this set K). Another thing I was wondering - if we're looking at (0,1), why isn't thisset compact from this definition? the cover (0,1) is finite or...? And as this covers (0,1), it should be compact... I think I have somehow misunderstood the definitions, but looking at them again and again doesn't make me realise where I'm wrong..
 March 20th, 2009, 07:00 PM #4 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: open covers and compactness If were are really restricting ourselves to elements in the set (we do not need to) the open set around zero (from our cover) will only contain elements from our set. That is it will be a Ball in the normal sense intersect our set, which will still be open in the topology induced from R, so we are ok. As for your second question, the definition says that every open cover has a finite subcover, (0,1) is just 1 open cover and I agree it is certainly finite, still this is not enough. Can you think of a cover that doesn't have a finite subcover? An interesting note is that for any finite dimensional normed linear space (e.g. R) the only compact sets are those that are closed and bounded.
 March 27th, 2009, 08:54 PM #5 Newbie   Joined: Mar 2009 Posts: 9 Thanks: 0 Re: open covers and compactness I gave it some thought (had some stat exams so had to lay of the hobby maths) - I think I have understood the concept now. The theorem states that ANY open cover of the set should have a finite subcover in order for the set to be compact. Thus, the set (0,1) is covered by the union of the sets (0,1/2),(1/2,3/4),(3/4,7/....., but there exists no finite subset of this cover that covers (0,1) Thanks for the help!
 March 28th, 2009, 10:16 PM #6 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: open covers and compactness What you found is not actually a cover, you should think about that for a while.

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