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Say we have a separable space $X$, and $E$ is its countable dense subset. Let $E$ contain points $\{e_{1},e_{2},\dots e_{n}\}$. $E$ can be dense only if it contains at least one point from each base set containing any point from $X\setminus E$.

Does this imply that if $E$ is a countable dense set, then $X\setminus E$ has to have a countable base? The base sets of $X\setminus E$ can be constructed by taking $(B_{i}\cap X\setminus E)$, where $B_{i}$ is a base set of $X$.

I searched for this property but couldn't find it anywhere. So I thought there might be something wrong with the argument.

Thanks in advance!

2 Answers2

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No, $X\setminus E$ does not have to have a countable base; it can be an uncountable, closed, discrete subset of $X$. The Mrówka space $\Psi$ constructed in this post in Dan Ma’s Topology Blog is such a space. The Čech-Stone compactification, $\beta\Bbb N$, of the natural numbers has $\Bbb N$ as a dense subset, but $\beta\Bbb N\setminus\Bbb N$ is very far from having a countable base.

Brian M. Scott
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  • But why can't an uncountable, closed, discrete subset of $X$ have a countable basis? Especially if the base sets are disjoint, and at least one of them contains an uncountable number of points (so that a countable number of disjoint base sets may contain all but a countable number of points, and the rest of the countable points can be easiy covered by a countabe number of disjoint base sets). –  Jun 17 '13 at 06:26
  • @Ayush: Let $D$ be a discrete space, and let $\mathscr{B}$ be a base of the topology. Then ${x}$ is open for each $x\in D$, so for each $x\in D$ there must be a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq{x}$. Clearly, then, $B_x={x}$. But then $|\mathscr{B}|\ge|D|$, so if $D$ is uncountable, so is $\mathscr{B}$. – Brian M. Scott Jun 17 '13 at 06:29
  • But if $X\setminus E$ is a discrete subset of $X$, then $E$ can't be a dense subset of $X$! I mean, if $x\in X\setminus E$, then ${x}$ will be an open set. This can't be a limit point of $E$. Hence, $E$ is not dense in $X$. My fundamental assumption is $E$ is dense in $X$. Am I missing some important point? –  Jun 17 '13 at 06:33
  • @Ayush: It certainly can; read the construction of $\Psi$ at the link that I gave. For an even simpler example of that phenomenon, though not of what you asked about in the question, note that $\bigcup_{n\in\Bbb Z}(n,n+1)$ is a dense subset of $\Bbb R$ whose complement is the closed, discrete set $\Bbb Z$. – Brian M. Scott Jun 17 '13 at 06:35
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A simple example is this:

Let $X$ be a finite completment topology on an uncountable set. Then it is separable. Any countable subset is dense in $X$. Pick any countable subset $E$ of $X$. Consider $X\setminus E$. It hasn't a countable basis.

Paul
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