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Possible number of dense subset of metric space $X$

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

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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 ?

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    $\begingroup$ Perhaps give a link to the original question? It is not clear what your assumptions on $X$ are. $\endgroup$ – Keen-ameteur Oct 7 at 9:43
  • $\begingroup$ What do you mean $p=\mathbb Q$? $\endgroup$ – Calvin Khor Oct 7 at 11:59
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Your question makes no sense to me.

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.

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