Let $X$ be a metric space ; then which of the following is possible ?

1) $X$ has exactly $3$ dense subsets

2) $X$ has exactly $4$ dense subsets

3) $X$ has exactly $5$ dense subsets

4) $X$ has exactly $6$ dense subsets

I know that if $X$ has a proper dense subset then for some $a \in X$ , we should have $X \setminus \{a\}$ is dense in $X$ and then $\{a\}$ is not open in $X$ ; but I can't relate this to no. of dense subsets except that if $X$ has only finitely many dense subsets then the topology of $X$ cannot be discrete . Please help . Thanks in advance

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    $\begingroup$ “If $X$ has only finitely many dense subsets then the topology of $X$ cannot be discrete.” On the contrary, $X$ is discrete if and only if it has just one dense subset. $\endgroup$ – user87690 Dec 13 '15 at 17:18
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    $\begingroup$ Wasn't this asked in yesterday's Tifr exam? $\endgroup$ – Sameer Kulkarni Dec 14 '15 at 7:37

It must be that $X$ is almost discrete. Because if $p$ is not an isolated point of $X$, $C(p) = X \setminus \{p\}$ is open and dense. The intersection of finitely many open dense subsets is open and dense as well.

If $X$ has one non-isolated point $p$, then $X$ and $C(p)$ are the only dense subsets (as every dense subset must contains all isolated points, and those are $C(p)$). So this does not qualify.

So if $X$ has two non-isolated points $p \neq q$, then $X$, $C(p)$, $C(q)$ and $C(p) \cap C(q)$ are the only dense sets. So 4 of them.

If $X$ has 3 non-isolated points $p,q,r$, then every dense set contains $X\setminus \{p,q,r\}$ (all isolated points) and we can add any subset of $\{p,q,r\}$ to get different dense subsets, so we have 8 of them.

So 4 is the only one that can occur, among your list. E.g. for the metric space $X = \{0\} \cup \{\frac{1}{n}: n = 1,2,3,\ldots\} \cup \{2\} \cup \{2 + \frac{1}{n}: n =1,2,3.\ldots\}$ as a subspace of the reals.

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    $\begingroup$ In other words: If the number of dense subsets is finite, it is a power of $2$. $\endgroup$ – Hagen von Eitzen Dec 14 '15 at 16:49
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    $\begingroup$ @HagenvonEitzen Indeed. For $T_1$ spaces at least. $\endgroup$ – Henno Brandsma Dec 14 '15 at 17:00
  • $\begingroup$ @HennoBrandsma Can you elaborate this point a bit- "if $p$ is not an isolated point of $X$, $C(p) = X \setminus \{p\}$ is open and dense". I didn't understand it properly. $\endgroup$ – Kushal Bhuyan Apr 13 '16 at 6:49
  • $\begingroup$ @Quintic $\{p\}$ is closed (in metric spaces surely) so $X \setminus \{p\}$ is open. $X \setminus \{p\}$ is dense iff it is not closed (so then its closure is $X$) iff $\{p\}$ is not open, i.e. iff $p$ is not an isolated point. $\endgroup$ – Henno Brandsma Apr 13 '16 at 19:46
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    $\begingroup$ @Sushil It's explained in the posting: there can only be finitely many dense sets if there is a finite set $F$ of non-isolated points (Say $n$ points). Then all dense subsets are of the form $(X \setminus F) \cup G$ where $G \subseteq F$, and there are $2^n$ choices for $G$, so that many dense sets. $\endgroup$ – Henno Brandsma Nov 13 '16 at 13:05

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