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Math Help - Covering compact

  1. #1
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    Covering compact

    Can anyone help me show directly that [a,b] is covering compact? By directly, I meant that I am not allowed to show that [a,b] is sequentially compact which implies covering compact, by a theorem we learned.

    Hint:
    If we let U be an open covering of [a,b] and let C={ x in [a,b]: finitely many members of U cover [a,x]}. Try to show that b is in C by the least upper bound property.

    I really have trouble with the covering compact concept. We have the definition of covering compact as : M is a metric space, A in M is covering compact if every covering U of A has a finite subcovering.
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  2. #2
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    Here is an attempt. It seems you are not trying to use Heine-Borel theorem. Which makes is harder. Let C be defined as in the hint. This set is non-empty because a\in C. And it has an upper bound, ii.e. b. Thus, by completeness there is a least upper bound c. Note, c\leq b, we will show c<b is impossible. Let \mathcal{C} be an open covering of [a,b]. By construction of C it means there exists finitely many open sets \mathcal{U}_1,...,\mathcal{U}_k which cover [a,c]. So it means c\in \mathcal{U}_1 without lose of generality. But since it is an open set it means there is \epsilon > 0 such that (c-\epsilon,c+\epsilon)\subset \mathcal{U}_1. Also since a<c<b it means there is \delta > 0 such that a<c-\delta <c+\delta < b. Let \eta = \min\{ \delta, \epsilon \}. Then by construction (c-\eta,c+\eta)\in \mathcal{U}_1 and [a,c+\eta] can be covered too. So c+\eta \in C. Which contradicts that c is the least upper bound. Thus, it means c=b. And so [a,b] is covering compact.
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