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Let $\mathbb{R}$ be a topological space with topology consisting of the sets $A \cup B$, where $A$ is open in the usual topology, and $B \subseteq \mathbb{R} \setminus \mathbb{Q}$. Is the interval $[0,1]$ compact in this topology?

I think that it might be, as if we have an open cover of $[0,1]$ in this topology, the cover consists of open sets of the form $A \cup B$. $B$ is a set of irrational numbers in the interval $[0,1]$. We can always find an open interval in $[0,1]$ to cover each irrational number, since for any irrational $p \in [0,1]$ there are $a,b \in \mathbb{Q} \cap [0,1]$ such that $a < p < b$, and then $p \in (a,b)$. Therefore now we have an open cover entirely composed of sets from $A$. But we know closed intervals in the usual topology are compact, so then $[0,1]$ is compact in the $A \cup B$ topology.

Could you please tell me whether my argument is correct or not?

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  • $\begingroup$ You need to show that any open cover of $[0,1]$ contains a finite cover. $\endgroup$ – AlexR Nov 21 '13 at 11:34
  • $\begingroup$ It looks strange because there is a subset that is not compact. In fact with this topology any irrational point is open (take $A=\emptyset$). So the set of irrational points has a cover from which we cannot extract a finite cover. $\endgroup$ – amorvincomni Nov 21 '13 at 11:53
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    $\begingroup$ The fact that you can always find an open interval containing any given irrational is irrelevant: it may be that no open interval containing that irrational belongs to your cover. $\endgroup$ – Brian M. Scott Nov 21 '13 at 13:41
  • $\begingroup$ @BrianM.Scott's comment is the only answer to the actual question. $\endgroup$ – Carsten S Nov 22 '13 at 12:27
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It is not a compact space. Let $\pi_n$ be a positive increasing succession converging to an irrational number $p<1$. You had that $\{p\}=\emptyset \cup \{p\}$ is open in $[0,1]$. So in particular $X=\cup_{n>0} \{(0,\pi_{n})\} \cup \{[0,\pi_1),\{p\},(p,1]\}$ is an open cover, from where you cannot extract a finite one.

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HINT: This is not a compact space anymore. In fact not even Lindelof. Take an $\varepsilon$ cover of the rational numbers, then generate from it a cover without a finite subcover. Indeed without a countable subcover.

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I am sorry for not looking at your argument, but if $\tau_1$, $\tau_2$ are topologies on $X$, $\tau_2\supset\tau_1$, $(X,\tau_1)$ compact Hausdorff, then $(X,\tau_2)$ is compact if and only if $\tau_1=\tau_2$.

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