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We know that $[1,\infty)$ is not compact in $\Bbb{R}$. A simple proof: the cover $\mathfrak{C}=\bigcup B(n,\frac{3}{4})$ for all $n\in\Bbb{N}$ of $[1,\infty)$ does not have a finite subcover.

What is wrong with the following argument (which closely resembles the proof of Heine-Borel's Theorem):

Let $B=\{x\in [1,\infty): [1,x]\text{ has a finite subcover of }\mathfrak{C}\}$. If $B$ is not bounded, then the whole of $[1,\infty)$ has a finite subcover. So let us assume $B$ is bounded, and let $m=\sup B$. All the open sets $m$ is contained in are of the form $(a,b)$, where $b>m$ and the point $\frac{b+m}{2}$, which is larger than $m$, is also contained in $(a,b)$. We can select one such open set. So there are points larger than $m$ in $B$, which is a contradiction. Hence, $B$ is unbounded, and the whole of $[1,\infty)$ is compact.

Thanks in advance!

2 Answers2

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The mistake is the implication $B$ unbounded $\Rightarrow [1, \infty)$ has a finite subcover. But really $B$ being unbounded just means that for any $x$ (no matter how large), $[1,x]$ has a finite subcover of $\mathfrak{C}$, which is not surprising since for any such $x, [1,x]$ is compact.

walcher
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  • If for any point $x$ we choose in $[1,\infty)$, we can choose a bigger point $y$ such that $[1,y]$ is compact, does that not prove all values above $1$ satisfy this criteria, hence $[1,\infty)$ is compact? –  Aug 17 '13 at 11:42
  • For some reason, I am convinced both by your argument and by mine. –  Aug 17 '13 at 11:51
  • There is a difference between saying 'for all $x\ge 1$ there is $y \gt x$ s.t. $[1,y]$ is compact' and saying '$[1,\infty )$ is compact'. For example: Let $\mathfrak{C}={ (n-1,n+1):n\in \Bbb N}$. For each $x\ge 1$ there is a finite subset $C_x$ of $\mathfrak{C}$ such that $[1,x]\subseteq \bigcup C_x$, but as $x$ increases, $|C_x|$ increases too, so if we let $x\to \infty$, then $|C_x|$ will not be finite anymore. – walcher Aug 17 '13 at 18:15
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    Note also the similarity with the following argument: for each $n\in \Bbb N$ the set ${ 0,1,...,n}$ is finite, hence $\Bbb N$ is finite. The mistake is that just because something can be arbitrarily approximated by finite objects, does not make it finite itself. – walcher Aug 17 '13 at 18:19
  • After a lot of hair pulling, I think I got the basic gist. I'll present my understanding: let us take the finite set $(0,n)$. If we say the subset $(0,x)$ is compact for any point $x\in (0,n)$, then $x$ can also be $n$, and hence the whole of $(0,n)$ is compact. However, let us now take $(0,\infty)$. For the whole set to be compact, $x$ has to equal the largest element in $(0,\infty)$. However, such an element does not exist. Hence, although we can say $(0,x)$ is compact $\forall x\in (0,\infty)$, we can't comment on whether the whole of $(0,\infty)$ is compact. –  Aug 18 '13 at 04:39
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$B$ is actually unbounded.

What the proof shows is that for each $x\in [1,\infty)$, $[1,x]$ has a finite subcover of $\mathfrak C$, which is not surprinsing as these intervals are compact. But it does not imply that $[1,\infty)$ has a finite subcover.

Davide Giraudo
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  • If for any point $x$ we choose in $[1,\infty)$ we can choose a bigger point $y$ such that $[1,y]$ is compact, does that not prove all vaues above $1$ satisfy this criterion, and hence $[1,\infty)$ is compact? –  Aug 17 '13 at 17:33