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Q: What are the compact subsets of $(\mathbb{R}, T_{fin})$? Where $T_{fin}$ is de cofinite topology.

I'm sorry if this question is too basic or easy. I just feel that I don't have a well enough understanding of compactness when it concerns subsets of topologies. I've tried to work with the definition of the cofinite topology but I can't figure it out!

Thanks in advance!

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I think that every set is compact.

Let $S \subseteq \mathbb{R}$, and let {$U_\lambda$}$_{\lambda \in \Lambda}$ be an open cover of $S$, i.e: $S \subseteq \cup_{\lambda \in \Lambda}U_\lambda$ while $U_\lambda \in \tau$ for every $\lambda \in \Lambda$.

There exists $\lambda_0 \in \Lambda$ , such that $|\mathbb{R} \backslash U_{\lambda_0}| < \infty$. Then we can write $(\mathbb{R} \backslash U_{\lambda_0}) \cap S=$ {$y_k$}$_{k=1}^n$.

From the fact that {$U_\lambda$}$_{\lambda \in \Lambda}$ is a cover we know that there exists {$\lambda_k$}$_{k=1}^n \subset \Lambda$, such that $y_k \in U_{\lambda_k}$ for every $k=1,...,n$.

Then $S\subseteq \cup_{k=0}^n U_{\lambda_k}$ ,and {$U_{\lambda_k}$}$_{k=0}^n$ is a finite open subcover of $S$.

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  • $\begingroup$ You are correct. They're all compact. $\endgroup$ – DanielWainfleet Mar 19 '17 at 3:54
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First we need to understand what this topology describes : an open set for what you call $T_{fin}$ will either be a set $A \subseteq \mathbb{R}$ such that $\mathbb{R} \backslash A$ is finite (note that $\mathbb{R}$ is such a set), or the empty set $\emptyset$.

Now take any subset $B \subseteq \mathbb{R}$. Let $\left\{U_{\alpha}|\alpha \in I\right\}$ be a collection of open sets (for the $T_{fin}$ topology) that cover $B$ i.e. : $$ B \subseteq \bigcup_{\alpha \in I}U_{\alpha} $$

You can verify that the number of $U_{\alpha}$ needed for such a cover is actually finite (remember what the open sets for $T_{fin}$ look like). This will be true for any open cover of $B$. The following characterization of compactness then allows you to conclude that $B$ is a compact subset of $\mathbb{R}$ for $T_{fin}$.

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  • $\begingroup$ Your wording could easily misunderstood as saying that $B$ is compact because there is some finite open cover of $B$. The point is that $B$ is compact because every open cover of $B$ can be refined to a finite subcover. $\endgroup$ – Alex Kruckman Mar 18 '17 at 21:28

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