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Let's consider $X$ a topological space. Let's say that $X$ is Hausdorff but not compact.

Can there be in $X$ two disjoint dense subsets $S_1$ and $S_2$? If the answer is yes, then is there a limit to the number of disjoint dense subsets that we can find in $X$? Does the situation change if I drop the Hausdorff condition?

Edit: $X$ is not discrete, but for simplicity let's start with $X=\mathbb{R}$.

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  • $\begingroup$ No $X=\mathbb{N}$. $\endgroup$ – Ali Taghavi Mar 9 '16 at 18:32
  • $\begingroup$ You should say something more about the question, otherwise it won't make that much sense to me, for example if you choose a discrete metric space, then it has only one dense set, on the other hand if you choose $\mathbb R$ , then you might end up with uncountably many distict dense set $\endgroup$ – Anubhav Mukherjee Mar 9 '16 at 18:33
  • $\begingroup$ Take $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ : they are not only distinct but disjoint. Actually in $\mathbb{R}$ you can get uncountably many disjoint dense subsets. $\endgroup$ – Captain Lama Mar 9 '16 at 18:36
  • $\begingroup$ thank you! that's the answer i needed, but which are these subsets? $\endgroup$ – user3419 Mar 9 '16 at 18:38
  • $\begingroup$ You can take the cosets of $\mathbb{R}$ modulo $\mathbb{Q}$. $\endgroup$ – Captain Lama Mar 9 '16 at 18:39
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As Captain Lama noted in the comments, you can get uncountably many pairwise disjoint dense subsets of $\Bbb R$.

Define a relation $\sim$ on $\Bbb R$ by $x\sim y$ if and only if $x-y\in\Bbb Q$. It’s easy to check that $\sim$ is an equivalence relation, so the $\sim$-equivalence classes form a partition of $\Bbb R$. Each of these equivalence classes is countable, and their union is $\Bbb R$, so there must be uncountably many of them (indeed, $|\Bbb R|=\mathfrak{c}$ of them).

Finally, each of them is dense in $\Bbb R$. To see this, let $(a,b)$ be a non-empty open interval in $\Bbb R$, and let $x\in\Bbb R$ be arbitrary; we want to show that there is some $y\in(a,b)$ that is in the same $\sim$-class as $x$, i.e., such that $x\sim y$. $\Bbb Q$ is dense in $\Bbb R$, so there is a $q\in(a-x,b-x)\cap\Bbb Q$. Let $y=x+q$; then $x\sim y\in(a,b)$, as desired.

The answer in general is a bit more complicated. A space is resolvable if it is the union of disjoint dense subsets. It’s clear that a space with an isolated point cannot be resolvable, since an isolated point must belong to every dense subset of the space. Thus, from the start we should limit our attention to spaces without isolated points. The accepted answer to this question shows that if $X$ is a metric space in which every open ball has cardinality $\kappa>\omega$, then $X$ can be partitioned into $\kappa$ dense subsets; this generalizes the result for $\Bbb R$. However, as hot_queen mentions at the end of that answer, there are irresolvable Hausdorff spaces in which every non-empty open set is uncountable. You can find an example (which is even connected) at the end of the paper On connected irresolvable Hausdorff spaces, by Douglas R. Anderson.

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In general could be many dense subset in a topological space $X$. As suggested take $X=\mathbb{R}$, then since $\mathbb{Q}$ is dense in $X$, every subset like $\mathbb{Q}(\sqrt{n})$ is dense, where $n$ is not a square. Clearly there are others.

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