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Let $X=[0,1]$. Prove $X$ is compact.

Let $\{U_i\}_{i\in I}$ be an open cover of $X$, or equivalently $$X=\bigcup_{i\in I} U_i~\text{and each}~ U_i~\text{is a open subset of}~X.$$

By definition of compactness we need to show that every open cover of $X$ admits (or has) a finite subcover.

Let $$B=\{x\in X:(\exists J\subset I)\text{ s. t. }|J|<\infty\wedge[0,x]\subset\bigcup_{i\in J} U_i\}.$$

I have showed that $B$ is a nonempty, open and closed subset of $X$. Since intervals in $\mathbb{R}$ are connected, it follows that $X$ is connected.

By definition of connectedness it follows that $B=[0,1]$.

But, I do not understand why $B=[0,1]$ implies that there exists a finite subcover for each open cover of $X$.

I would appreciate any help.

Thank you.

johnny09
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    $B$ is the set of all $x \in [0,1]$ such that $[0,x]$ is covered by a finite subcover of the $U_i$. Our goal is to show that $1 \in B$. – David Wheeler Apr 23 '16 at 18:12
  • We can prove something a little stronger:

    The Heine Borel Theorem states that $A \subseteq \mathbb{R}$ is closed and bounded iff $A$ is compact, but we only need one direction of the theorem for this proof. See here for a similar proof: http://www.math.utah.edu/~bobby/3210/heine-borel.pdf

     – Andres Mejia Apr 23 '16 at 18:12

1 Answers1

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$B$ is the set of $x$ such that the given cover of $X$ (which is also a cover of $[0,x]$) admits a finite sub-cover of $[0,x]$. The connectedness argument shows that $B=X$ and in particular $1\in B$. Hence $[0,1]$, i.e., $X$ itself, admits a finite sub-cover of the cover we started with. As we started with an arbitrary open cover, this shows that every open cover of $X$ admits a finite sub-cover. (That each different cover may in principle evoke a different $B$ doesn’t matter).

Hagen von Eitzen
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  • Thank you very much for your answer. May I ask how do we know that $X=[0,1]$ admits a finite subcover of the cover we started with? – johnny09 Apr 24 '16 at 13:12
  • Because $1\in B$ – Hagen von Eitzen Apr 24 '16 at 14:02
  • It is not obvious to me why it follows from $1\in B$. Why not $0\in B$ implies that $X$ admits a finite subcover? – johnny09 Apr 24 '16 at 17:26
  • $B$ is the set of $x$ such that $[0,x]$ has a finite subcover of $U_i$. 1 is in $B$, so $[0,1]$ admits a finite subcover of $U_i$.

    This argument works for all open covers, so all open covers admit finite subcovers.

     – Chessanator Sep 30 '16 at 09:32