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If $\mathbb{R}$ has the topology consisting of sets $A$ such that $\mathbb{R}-A$ is either countable or all of $\mathbb{R}$, is $[0,1]$ a compact subspace?

I've found the example $A_n = [0,1]-\{\frac{1}{n},\frac{1}{n+1},\cdots\}$ but I don't know how it answers the question. Why this set cannot be covered by finitely many open subspaces?

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Clearly, $\{A_n\}$ is an open cover of $[0,1]$. If $[0,1]$ is compact, then there exists a finite set $F \subset Z^+$ such that $\{A_n: n\in F\}$ covers $[0,1]$. Let $n_0 \in Z^+$ such that $n\le n_0$ for all the $n\in F$ and hence $\frac{1}{n_0} \in [0,1] \setminus \bigcup_{n\in F}A_n$ which is a contradiction.

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  • $\begingroup$ thanks a lots @ paul $\endgroup$
    – jasmine
    Apr 12, 2018 at 14:34
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The sequence $\{A_n\}$ is a sequence of open sets that covers $[0,1]$. It admits no finite subcover.

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  • $\begingroup$ thanks a lots @ Martin $\endgroup$
    – jasmine
    Apr 12, 2018 at 14:34

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