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Let X be a set and let B = { {x} : x ∈ X }. I'm trying to show that the only topology on X that contains B on a subset is the discrete topology.I have been stuck on this for quite a while, I can't seem to be able to prove it rigorously:

Since $T_{\text discrete}$ = $P(X)$ i.e the set of all open subsets of X, since {x} for x $\in$ X are also open they are contained in $T_{\text discrete}$. Now let T be any topology on X containing B, we show that T can only be the discrete topology, by the definition a Topology generated by basis W is $T_W$ = [$\bigcup$C : C ⊆ W ]∪ {∅}, we can show that B is a basis on X, and then union of all the elements of B is exactly equal to the power set which is the discrete topology on R.

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Just pick any $U\in\mathcal{P}(X)$ and write is as the union of opens $$U=\bigcup_{x\in U}\{x\}$$

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If all the singletons are open, then taking unions any subset of $X$ is open, which means that the topology is exactly the discrete topology.

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