# How there are $4$ possible dense subset?

Possible number of dense subset of metric space $$X$$

I found the answer but i have some confusion in my mind

My attempt : If i take two non-isolated point that is $$p= \mathbb{Q}$$ , $$q=\mathbb{R}\setminus \mathbb{Q}$$ where $$p$$ and $$q$$ are two non -isolated point , Then according henno Brandsma sir answer
Number of possible dense subset are

$$1. c(q)$$

$$2.c(p)$$

$$3.c(p) \cap c(q)$$

$$4.\mathbb{R}= X$$

But Here option 3 is not possible because $$c(p) \cap c(q) = \emptyset= \mathbb{Q} \cap( \mathbb{R} \setminus \mathbb{Q}) = \emptyset$$

we know that empty set is not dense.

Then How there are $$4$$ possible dense subset ?

• Perhaps give a link to the original question? It is not clear what your assumptions on $X$ are. – Keen-ameteur Oct 7 at 9:43
• What do you mean $p=\mathbb Q$? – Calvin Khor Oct 7 at 11:59

I am talking about points $$p$$ and $$q$$ from the space, you take subsets ? $$p=\Bbb Q$$?
If we're working in the reals: the reals have uncountably many different dense subsets, e.g. already all sets of the form $$\Bbb R\setminus F$$ where $$F$$ is finite (we can even take countable $$F$$), plus sets like the irrationals.
So we are given that we are in a metric space $$X$$ with only finitely many dense subsets. So for sure $$X=\Bbb R$$ is impossible. I give an example of a subspace $$X$$ of the reals that indeed has finitely many (4) dense subsets. So it is possible to have such $$X$$.
In the remainder of the answer I try to explain why if $$X$$ has finitely many dense subsets this number is a power of $$2$$ (so $$1$$, $$2$$, $$4$$, $$8$$ etc.) and that is why $$4$$ was the right answer to the multiple choice in the original question.
$$C(p)$$ is the complement of a point! Your question is based on a misunderstanding.