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Show that a space $(X,T)$ is discrete if and only if each set consisting of only one point is open.

I believe I achieved the proof going from left to right: The discrete space is a set with the discrete topology and by that, in the discrete topology every subset of X is open by definition. (please critique if necessary)

But I am unsure how to prove in the other direction.

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First prove that if $(X,T)$ is discrete then every set of one element is open.

This is easy because if $X$ is discrete then every set is open.


Then prove that if every set of one element is open the $(X,T)$ is discrete.

This is the hard part. To prove $(X,T)$ is discrete we prove every set is open. So let $U\subseteq X$ be an arbitrary set.

We prove it is open by noticing $U=\bigcup\limits_{u\in U} \{u\}$. The set on the right is clearly open since it is a union of open sets.

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  • $\begingroup$ Thank you!!! I couldn't get the structure right $\endgroup$ – Jess Savoie Nov 1 '17 at 2:51
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If subsets consisting of a single point are open, then so are arbitrary unions of them, i.e. all subsets, because all subsets are unions of the single-element subsets formed by the elements they contain.

Conversely, if the topology is discrete, then all subsets are open. In particular all single element subsets.

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