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Let $X$ be the real numbers. Is the following topology stronger than the cofinite topology? And is $X$ then compact?

$T = \{ U \subset\mathbb{R}\:|\:0\notin U\text{ or }\mathbb{R}\setminus U\text{ finite}\} $

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What is $X$? How does finiteness of $X\setminus \mathbb R$ affect $U$? What is a compact topology? It is homework? – Ilya Jan 11 at 14:28
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And does "stronger" mean having more or less open sets? – Chris Eagle Jan 11 at 14:31
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More open sets.. Because now you have finite open sets right? Every set without $0$ belongs to the topology (along with all cofinite sets). – omar Jan 11 at 14:34
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@Chris: Where have you seen it used to mean fewer open sets? – Brian M. Scott Jan 11 at 14:36
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@ChrisEagle: I like this "especially analysts"! I always knew one shall be careful with these people – Ilya Jan 11 at 14:42
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You can write this as the union of the cofinite topology and the union of all sets not containing $0$, so it is stronger. It is also compact. Every open covering must contain an element containing $0$ and this element will have a finite complement. Supply the details yourself.

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So the proof that the above given space X is compact would be: Let $\mathcal{U}$ be an opencover of $X$, then there exists a $U_0 \in \mathcal{U}$ which contains $0$ such that $X\setminus{U_0}$ is finite. Then, for every $x \in X\setminus{U_0}$ there is a $U_x \in \mathcal{U}$ containing x. The union $U_0 \cup \{U_x : x \in X\setminus{U_0} \}$ is a finite open cover of X. So X is compact. – omar Jan 11 at 18:02
@omar Exactly.${}$ – Michael Greinecker Jan 11 at 18:37

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