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I cant seem to find what's wrong in my proof regarding the compactness of Bolzano-Weierstrass Hausdorff topological space.

So consider an infinite open cover $\mathcal C$ of a Bolzano-Weierstrass Hausdorff space X, construct a largest open sub-cover $\mathcal C\mathcal S$ for $C$ such that $C_1 \bigcup C_2 \notin \mathcal C\mathcal S, \forall C_1,C_2 \in \mathcal C\mathcal S$ and $C_1 \nsubseteq \bigcup_{\mathcal C \mathcal S \setminus\{C_1\}} C, \forall C_1 \in \mathcal C\mathcal S$. Therefore, each $C\in \mathcal C\mathcal S$ has a point that's not in other set in $\mathcal C\mathcal S$

Now, for each $C$ in $CS$, pick a point that's not contained in other $C$ of $\mathcal C\mathcal S$, and call it $F$ the set of such points

Assuming $\mathcal C\mathcal S$ is infinite which implies $F$ is also infinite, and because $X$ has Bolzano-Weierstrass property $F$ has a limit point, because of Hausdorff property any open set that contains a limit point of $F$ must contains infinitely many points of $F$ which implies at least 1 $C \in \mathcal C\mathcal S$ contains more than 1 point of $F$ which leads to contradiction

So I conclude that $\mathcal C\mathcal S$ cant be infinite which means $\mathcal C\mathcal S$ is a finite subcover for $X$ hence $X$ is compact

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  • $\begingroup$ Is a Bolzano-Weierstrass space a sequentially compact space? $\endgroup$ – Jonathan Davidson Jul 9 '17 at 5:18
  • $\begingroup$ Bolzano - Weierstrass space is a space that every inifinite subset has a limit point $\endgroup$ – JokingBear Jul 9 '17 at 5:19
  • $\begingroup$ Ok, I wanted to make sure. Sequentially compact is another term for that property. $\endgroup$ – Jonathan Davidson Jul 9 '17 at 5:20
  • $\begingroup$ @JonathanDavidson no it's limit point compact, not sequentially compact. $\endgroup$ – Henno Brandsma Jul 9 '17 at 12:38
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There is no reason that such a subcover $\mathcal{CS}$ should exist. For a simple example to illustrate why you shouldn't expect to be able to find such a subcover, let $X=\mathbb{R}$ and consider the cover consisting of the sets $C_n=(-n,n)$ for each positive integer $n$. Any subcover must contain $C_n$ for infinitely many different $n$. In particular, every element of $\mathbb{R}$ must be contained in infinitely many of the sets in any subcover, since $x\in\mathbb{R}$ is in $C_n$ for all $n>|x|$.

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  • $\begingroup$ nice, tks, I was totally not considering that possibility $\endgroup$ – JokingBear Jul 9 '17 at 15:35
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The ordinal space $\omega_1$ has the Bolzano-Weierstrass property but is not compact. Try to find your $\mathcal{CS}$ for the cover of initial segments.

You don't actually give any argument why such a $\mathcal{CS}$ should exist. You just state it.

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