# Compact Set Question

Consider the topology $\tau$ defined on $\mathbb{R}$ by $U\in\tau$ iff $\forall s\in U$, $\exists t>s$ such that$[s,t)\subseteq U$. Show that $[0,1]$ is not compact.

My attempt: We only need to find one open cover {$V_i$} of $[0,1]$ that does not have a finite sub-cover. First, note that any interval of the form $[a,b)\in\tau$. If we let {$V_i$} ={$[1/(n+1),1/n): n = 2,3,4,...$} $\bigcup [1/2,3/2)$, then there is no finite sub-cover of {$V_i$}, so $[0,1]$ is not compact.

Could anyone tell me if they see a problem with my argument? I think there is something wrong because we wouldn't have $0\in\bigcup_iV_i$, right?

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As you define it, $U=[0,1]$ does not belong to $\tau$ since for $s=1$ there does not exist $t>s$ so that $[1,t)\subset[0,1]$. Or do you mean than for opens sets $U$ only or something like that? Can you explain? –  Stef Feb 23 at 6:56
Yes, I mean that $U$ is an open set, so, as you said, $[0,1]$ is not open here. Sorry for any ambiguity –  Adam Feb 23 at 6:58
Perhaps the sets $[0, 1-\frac{1}{n})\cup[1, 3/2)$. I am not sure though. The problem has to occur in $1$ and not in $0$. For the $V_i$'s you suggested indeed $0$ is not in their union, so their not a cover. –  Stef Feb 23 at 7:16

No, $0$ will not be covered. But you made it the wrong way. Consider $[0,1/2)\cup[1/2,2/3)\cup[2/3,3/4)\cup\ldots\cup[1,2)$.

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Awesome! thank you –  Adam Feb 23 at 7:17

Hint: note that the set $[1,2)$ is open

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Finding a cover with no finite subcover is the right way to go. But I think that you are distracting yourself by trying to make the elements of your cover pairwise disjoint: this is not necessary. (It's not wrong either - you're just adding a complication.)

Anyway, as @Omnomnomnom notes, $[1,2)$ is open in $\langle\mathbb{R},\tau\rangle$, so if you can find a collection of $\tau$-open sets covering $[0,1)$ with no finite subcover, then you are done.

But this is easy, since $\tau$ is a finer topology than the usual topology on $\mathbb{R}$! Any such counterexample to the compactness of $[0,1)$ in the usual topology will suffice.

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