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Let $X$ be a discrete topology and $A$ is a subspace of $X$. Then,

$A$ is compact $\iff$ A is finite.

My attempt is general:

($\implies$) Let $X$ be compact. Suppose that X is infinite.

Consider a open-cover $C=\{\{x\} : x \in X\} $.

By the hypothesis, and by the definition of compactness of a space,

$\exists H\in C : \bigcup_{H\in C}H=X$ where H is finite.

Since $X$ is infinite but the subcover $H$ is finite, this is a contradiction.

Now, my question is: How can I solve this through a subspace $A$ of $X$?

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    $\begingroup$ Replace $X$ with $A$ everywhere in your argument. What is the problem? $\endgroup$ Commented Feb 27, 2018 at 20:53

2 Answers 2

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You're proof is the right idea but you should be more formal about things when doing topology. Assume that $A$ is compact and for the sake of contradiction, infinite. Then, consider the following open cover of $A$, $C=\{\{x\}:x\in A\}$. Let $H$ be any finite open subcover, then we can write $H=\{\{x_i\}:1\leq i \leq N\}$ for some $N\in\mathbb{N}$. Since $A$ is infinite, there is an element $y\in A$ such that $\{y\}\not\in H$. This contradicts that $H$ is a cover, so such a finite sub cover cannot exist.

For the other direction, assume that $A$ is finite. Then we have that $A = \bigcup\limits_{i=1}^N \{x_i\}$. Let $C$ be an open cover of $A$, $C=\{U_j: j\in J\}$ for some open sets $U_j$ with index set $J$ (possibly infinite). Since $C$ is a cover, each $x_i\in A$ must belong to some open set in $C$, which we denote $U_i$. Taking $K = \bigcup\limits_{i=1}^N U_i$ gives a finite open subcover of $A$, thus $A$ is compact (since $C$ was arbitrary, all open covers have a finite open subcover).

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    $\begingroup$ This is an excellent solution. Thank you $\endgroup$
    – AYARcom
    Commented Feb 27, 2018 at 21:13
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You don't have to. Compactness and discreteness are properties of spaces rather than of a subspaces. So if $A$ is a compact subset of a discrete space $X$, then $A$ is a compact discrete space on its own.

You should also note, that clearly a finite space is compact.

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  • $\begingroup$ Okay, so how do we prove the other part? $\endgroup$
    – AYARcom
    Commented Feb 27, 2018 at 20:57
  • $\begingroup$ You mean that finite spaces are compact? You have to cover all points and if there is only finitely many… $\endgroup$
    – user87690
    Commented Feb 27, 2018 at 20:59

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