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Let X be any set, and let d be the discrete metric on X. Show that d generates the discrete topology.

I just want know if my following proof is valid or not:

The discrete topology is the power set of X, which is the set of all subsets of X. This means that in the Discrete Topology all sets are open. With the Discrete Metric all points can have a ball with radius between 0 and 1, that will contain only that individual point. This leads to the fact that all set are both open and closed under the Discrete Metric. Therefore the discrete metric generates the discrete topology.

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It’s correct, but it’s a bit wordy and clumsy. Here’s a neater way to say exactly the same thing:

Let $x\in X$ be arbitrary, and let $0<r\le 1$; then $B_d(x,r)=\{x\}$, so $\{x\}$ is an open set. If $A\subseteq x$, then $A=\bigcup_{x\in A}\{x\}$ is a union of open sets and therefore open, so every subset of $X$ is open, and $X$ has the discrete topology.

It would actually be perfectly acceptable to stop once you’ve shown that each singleton is open: that’s often taken as the definition of the discrete topology anyway.

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    $\begingroup$ $A\subseteq x$ or $A\subseteq X$? why can we say that any $A$ is a union of singletons? $\endgroup$ – newhere Aug 23 '18 at 15:42
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This reasoning looks quite solid. You can say this. For all $x\in X$, $$B_1(x) = \{x\},$$ so all singletons are open subsets of $X$. Since any union of open sets is open, all subsets of $X$ are open.

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