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Help,about subcover!

  1. Jan 28, 2009 #1
    I am studying Heine-Borel theorem.
    I finally get the sense of this theorem after i read a lots posts here.
    But I found a question posted by Rach3
    why for S={1,1/2,1/3……} there is no subcollection of it is a cover of S
    but every open cover of T={0,1,1/2,1/3……} has a finite subcover?
     
  2. jcsd
  3. Jan 28, 2009 #2
    I know that I can find a cover e.g Ia=(1/n,1+1/n),union of Ia from n=1 to inifinte is a cover for S but not finite
     
  4. Jan 28, 2009 #3
    but how to prove that every open cover of T has a finite subcover
     
  5. Jan 28, 2009 #4

    quasar987

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    There is no universally recognized statement of the Heine-Borel theorem so you'd better tell people what you call the "Heine-Borel" theorem when you speak about it.

    I will assume that you take it to mean "A subset of R is compact if and only if it is closed and bounded."

    So the answer to your question is that every open cover of T admits a finite subcover because it is compact. And why is it compact? Because it is closed and bounded. On the other hand, the set S above is not compact because it is not closed.
     
  6. Jan 28, 2009 #5
    I need to prove every every open cover of T has a finite subcover.
    My version of Heine-Borel theorem is every open cover of a closed interval has a finite subcover.
    It is easy to prove if it is a closed interval. I mean like [a,b]
    not a interval like T.
     
  7. Jan 28, 2009 #6

    quasar987

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    Don't forget the word "bounded": your version of the Heine-Borel theorem is that every open cover of a closed and bounded interval has a finite subcover.

    It is important not to forget "bounded" because for instance, the whole real line R is a closed interval but clearly there is no finite subcover of the open cover [itex]\{(-n,n)\}_{n\in\mathbb{N}}[/itex].


    So, why does your version of the H-B theorem implies that every open cover of T has a finite subcover?

    Well, notice that T is a closed subset of the closed and bounded interval [0,1]. Now consider [itex]\{U_i\}_{i\in I}[/itex] an open cover of T. Then add to that cover the open set V:=R\T so that [itex]\{V\}\cup \{U_i\}_{i\in I}[/itex] is an open cover of [0,1]. By the H-B theorem then, there exists a finite subcover of that open cover. Of that subcover, keeping only the U_i's (that is to say, removing V if it in a member of the subcover), we get a finite subcover of T.
     
  8. Jan 28, 2009 #7
    Thank you.
    I think I need some times to understand the last step" Of that subcover, keeping only the U_i's (that is to say, removing V if it in a member of the subcover)"
     
  9. Jan 28, 2009 #8
    I think you are using the result to prove the question .
    you mentioned that t is closed and bounded,then processing to prove in the way T is bounded and closed...
    If you using S instead of T , your prove still hold.
    "Now consider an open cover of T. Then add to that cover the open set V:=R\T so that is an open cover of [0,1]. By the H-B theorem then, there exists a finite subcover of that open cover. Of that subcover, keeping only the U_i's (that is to say, removing V if it in a member of the subcover), we get a finite subcover of T.
    "
     
  10. Jan 28, 2009 #9

    quasar987

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    If I understood correctly, the question was "Use the Heine-Borel theorem to prove that every open cover of T has a finite subcover.", and it was not "Show that T is closed and bounded."

    However, you're right that in solving the problem, I did use the fact that T closed. But that is easy to show.
     
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