# Is there anything wrong in this proof (idea)?

I just want to know what fails when using the same argument of Heine-Borel for trying to prove that an open interval $(a, b)$ is compact.

For a given cover $\{U_i\}$, let $A = \{x \in (a, b) ;(a, x)$ covered by a finite number of $U_i \}$, then, using the same argument as in Heine-Borel (for $[a, b]$) and assuming that $supA> a$, I get that $supA = b$. So, is there anything wrong in this proof (idea)? Is the possibility of $supA = a$ the only reason for $(a, b)$ not being compact?

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 How do you know that $A$ is non-empty? What is $\sup A$ if $A$ is empty? – Thomas Andrews Oct 26 '12 at 15:53 @ThomasAndrews $\sup A = -\infty$ – user40276 Oct 26 '12 at 16:32

You can’t even start the argument for $(a,b)$, because $a\notin(a,b)$: there may be no member of the cover that contains $a$ in the first place. If $\mathscr{U}$ is an open cover of $(a,b)$, it’s entirely possible that $U\subseteq(a,b)$ for every $U\in\mathscr{U}$.

Added: Even if every set $(a,x]$ with $a<x<b$ has a finite subcover, $(a,b)$ need not: let $(a,b)=(0,1)$ and $$\mathscr{U}=\left\{\left(0,1-\frac1n\right):n\ge 2\right\}\;.$$

Every subset $(0,x]$ of $(0,1)$ can be covered by a single member of $\mathscr{U}$, but no finite subset of $\mathscr{U}$ covers $(0,1)$.

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So assuming that $supA>a$, everything else will hold. Is that right? – user40276 Oct 26 '12 at 15:41
@user40276: No. Consider the cover $\left\{\left(0,1-\frac1n\right):n\in\Bbb Z^+\right\}$ of $(0,1)$. – Brian M. Scott Oct 26 '12 at 15:45
So assuming that $\sup A > a$ and b is covered by some open set in the cover, then everything else will hold. Again, is that right? – user40276 Oct 26 '12 at 16:39
@user40276, Whoops, I was wrong about it being equivalent, but the basic statement is true: If that is true then you have just a restriction of an open cover of [a,b], so then you have a finite sub-cover because there is a corresponding finite sub-cover of $[a,b]$ – Thomas Andrews Oct 26 '12 at 16:51

The problem is that it is possible to have $A= \emptyset$, and then the rest of the argument doesn't work.

For $[a,b]$ the key is that $a \in U_i$ for some $i$, thus $[a,x] \subset U_i$ for some $x$, which shows that $A$ is non-empty. But for covers of $(a,b)$ it is possible to get $A=\emptyset$.

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Let $\mathcal U$ be an open cover of the open interval $(a,b)$. If you have the property: $\exists x,y\in(a,b)$ and $U,V\in\mathcal U$ such that $(a,x)\subseteq U$ and $(y,b)\subseteq V$, then $\mathcal U$ has a finite sub-cover.

But that follows since $[x,y]$ has a finite sub-cover, and you can just add $U$ and $V$ to that finite subcover. Alternatively, if this is true, $\mathcal U$ corresponds to the restriction of a cover of $[a,b]$, which then has a finite sub-cover, which then corresponds to an finite sub-cover of $(a,b)$.

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