I saw a proof on proofwiki but couldn't understand it. I only know the bare basics of topology (closed, open, limit points, neighbourhoods). Is there a simple proof for this?
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In the discrete topology every subset is its own closure (this property characterizes the discrete topology). – Qiaochu Yuan Jan 13 '21 at 01:06
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1Why not give a link to the proof you don't understand? – Rob Arthan Jan 13 '21 at 01:21
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Assume we have an uncountable discrete space. If you have a discrete topological space, every subset is closed and open, so the closure of a subset is itself. So if you have a dense subset, then it must be the whole space, but the space is uncountable. (I am going off of the definition: a space is separable iff it has a countable dense subset).
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