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How can we show that for the topological space $X$ that $X$ is the only dense subset of $X$.

I feel as if this is true, and the definition I have of being dense is as follows: Let $X$ be a space, and a subspace $D \subseteq X$ is dense if and only if $\overline{D} =X$.

If we want to show that $X$ is the only dense subset of $X$ here is what I have thought up:

Since we have that $X = \mathbb{R}$, we can pick $D=\mathbb{R}$. Then clearly the closure of $D$ is still $\mathbb{R}$ which is $X$. So we would have that $X = \overline{X}$, but this doesn't specify that this is the $only$ dense subset of $X$. What am I missing. I don't want any solutions here, I just want more of an understanding of what's going on.

Joey
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    Many topological spaces $X$ have a wealth of proper subspaces that are dense in $X$. For example the set of nonzero numbers is dense in $\mathbb{R}$ with the usual topology (as is the set of all non-integers, or the set of all rational numbers...). So this statement would not be true without more hypotheses on the topology. Just $X$ being a topological space will not guarantee the result. – leslie townes Feb 26 '21 at 22:48
  • Is part of your question "under what conditions is $X$ the only dense subset of $X$"? If so, see https://math.stackexchange.com/questions/1153055/assume-that-the-only-dense-subset-is-x-itself-what-can-you-say-something-abou – Patrick Stevens Feb 26 '21 at 22:50
  • @leslietownes I meant to put the euclidean topology – Joey Feb 26 '21 at 22:53

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If $X$ has a non-isolated point $x$, so that $\{x\}$ is not open, then $X\setminus \{x\}$ is dense and a proper subset of $X$.

So the only spaces where all dense subsets must be $X$ are the discrete ones.

Henno Brandsma
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