# How can we prove that there does not exist any metric space $4,5,6$ dense subsets?

Let $(X,d)$ be a metric space. Which of the following is possible?

(A) $X$ has exactly $3$ dense subsets.

(B) $X$ has exactly $4$ dense subsets.

(C) $X$ has exactly $5$ dense subsets.

(D) $X$ has exactly $6$ dense subsets.

I think it's answer is (A) as if we consider $\mathbb R$ with usual metric then $\mathbb R$ contains three dense subsets $\mathbb Q$ ,$\mathbb Q^c$ and $\mathbb R$ itself.

Am I correct?Are there other metric spaces which makes (B),(C),(D) true? If not How can we prove that there does not exist any metric space $4,5,6$ dense subsets?

• $\mathbb{R}$ has a lot more dense subset than that. Any subset $X$ with $\mathbb{Q}\subset X\subset \mathbb{R}$ is dense ! Commented Jun 5, 2017 at 9:45
• hint: if $D$ is dense in $X$ then every set that contains $D$ as a subset is dense too. Commented Jun 5, 2017 at 9:46
• means all options are true? @ArnaudD.@drhab Commented Jun 5, 2017 at 9:47
• @PKStyles No, each option says "exactly $n$ dense subsets" so none of these are true for $\mathbb{R}$. Commented Jun 5, 2017 at 9:49
• @ArnaudD.:then which metric space should we consider? Commented Jun 5, 2017 at 9:51

Hint: Consider the set $X = A\cup D$ with $A=\{0\}$ and $D=\left\{\frac1n: n\geq 1\right\}$ equipped with the metric from ${\Bbb R}$. Show that the only dense subsets are $A\cup D$ and $D$ (so there are precisely two dense subsets).

Now, this example is constructed using a discrete set $D$ with precisely one accumulation point $0$. Try now to construct sets using a discrete set and having 2, 3,... accumulation points and draw conclusions from this.

• :Clearly any non-empty open set in X must contain a point of X.So,X is dense in X.Now,we'll show that A is dense in X.Taking any point $p \in X$.If $p=0$,then $B(o,r) \cap X \neq \phi$ for any $r>0$.If $p\neq 0$ then $p \in D$.Hence,$B(p,r) \cap X \neq \phi$.Thus,Thus only dense subsets are $X$ and $D$. Commented Jun 5, 2017 at 13:57
• :$B=${$1+ \frac{1}{n} : n\geq 1$} has eaxactly one accumulation point namely 1.Similarly,$B=${$2+ \frac{1}{n} : n\geq 1$} has exactly one accumulation point namely 2.Then $B \cup C$ is a set with exactly two limit points.On extending this construction we get, $\cup_{j=0}^k${$j+\frac{1}{n}$} has exactly $k$ accumulation points. Commented Jun 5, 2017 at 14:11
• :Sorry but i'm not getting any conclusion.Will you please give me some hint? Commented Jun 5, 2017 at 14:15
• First: Take $X=B\cup C$. Show that if $D$ is dense in $X$ then it must contain all points except possibly two. Why and which two points? (making a drawing might help) Commented Jun 5, 2017 at 14:48
• :How can $D$ be dense in $X=B \cup C$ as $D$ consists of numbers on LHS of 1 while $X$ consists of numbers on RHS of 1?I tried it by making picture of it i'm getting if $D$ is dense in $X$ then $D$ must be {1}. Commented Jun 5, 2017 at 15:09

In this answer I showed that only exactly 4 dense sets is possible: this can happen if almost all points of $X$ are isolated. Say there is a set $N$ of $n$ non-isolated points . Then $D\subset X$ is dense iff $D= (X\setminus N) \cup A$ where $A\subseteq N$. So there are as many dense subsets of$X$ as there are subsets of $N$, i.e. $2^n$.

• I understand your answer, but what do you mean by 'almost' in "almost all points of X are isolated" ?
– user422112
Commented Dec 9, 2017 at 15:02
• @ThatIs All but finitely many points. A finite metric space has only one dense subset (itself, as all of its poinst are then isolated), so $X$ is infinite. If $X$ has 2 non-isolated points, the rest being isolated, it has 4 dense subsets, if 3, then 8 etc. Commented Dec 9, 2017 at 15:20