Is the compact interval $[0,1]$ in the usual topology compact in this new topology?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\tau_1=\tau_2$.