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Let $(X, d)$ be a metric space. Assume that the only dense subset is $X$ itself. What can you say something about the topology, that is, the family of open sets?

My Try: Since the only dense subset of $X$ is $X$ itself, so every subset of $X$ is a closed set.

User8976
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    I'm not really sure how in your try, you made the jump from $X$ being the only dense subset of $X$ to suddenly every subset of $X$ being a closed set. It seems like too big of a leap for me. When writing proofs, you have to detail your logical steps so that someone looking over your proof understands how one step follows logically from another. – layman Feb 17 '15 at 15:05

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What does it mean for a subset $A$ of $X$ to be dense? It means for each $x \in X$ and every open $U \subseteq X$ with $x \in U$, $U \cap A \neq \emptyset$.

So if $X$ is the only dense subset of itself, that means for each $x \in X$, $X - \{ x \}$ is not dense in $X$. That means, in particular, that there is some $y \in X$ and some open $U \subseteq X$ such that $y \in U$ and $U \cap (X - \{ x \}) = \emptyset$. But if this intersection is empty, then it must be that $y = x$ (why?). It also must be that $U = \{ x \}$ (why?), and so $\{x \}$ is open. But then that means that singletons are open.

But if $\{ x \}$ is open for each $x$, then it is easy to see that the topology on $X$ is actually the discrete topology (why?), i.e., the set of all subsets of $X$, which is the finest topology possible on $X$.

layman
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One thought: Every point is isolated. For if $x$ is a limit point of $X$, then the closure of $X-x$ is $X$. We assume this cannot happen, so $x$ must be isolated; hence the singleton of the arbitrary point $x$ is open.

Cass
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  • This is equivalent to the OP's original conjecture. If every set is closed, then also every set is open. In particular, single points are open, so they constitute neighborhoods of themselves -- so single points are isolated. This is just the discrete topology. – MPW Feb 17 '15 at 15:04
  • Yeah, I mean every answer is going to be equivalent. But he asked the question, so what shall I say? – Cass Feb 17 '15 at 15:05
  • I accept your answer, was just pointing that out. Not meant as criticism, sorry. Will give you +1 as compensation ;) – MPW Feb 17 '15 at 15:07
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The original comment by user46944 is very much on point: the gap between ‘$X$ is the only dense subset of $X$’ to ‘every subset of $X$ is closed’ is far too large to be leaped in a single bound. There actually is a way to arrive at that result without using the approach of user46944 and Cass, but it takes a bit more work, and I definitely recommend their approach. In case you’re interested in trying it, I’ll sketch the main steps.

Suppose that $A\subseteq X$ is not closed. Let $D=A\cup(X\setminus\operatorname{cl}A)$.

  • Prove that $D\ne X$.
  • Prove that $\operatorname{cl}D=X$ and hence that $D$ is dense in $X$.
Brian M. Scott
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  • So your approach is by contradiction in a sense: suppose there is a subset of $X$ that is not closed. Then we find a strict subset of $X$ that is dense, which contradicts that $X$ is the only dense subset of $X$. – layman Feb 18 '15 at 01:29
  • @user46944: One can look at it that way, but one needn’t: I’m simply proving the contrapositive of tone’s assertion. – Brian M. Scott Feb 18 '15 at 01:37