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If we take the standard topology on $\mathbb{R}$ we can easily find two disjoint sets that are dense, namely $\mathbb{R}\setminus\mathbb{Q}$ and $\mathbb{Q}$. Similarily, if we take the same topology and restrict it to $\mathbb{Q}$ we can again find two disjoint dense sets. Define for every prime $p$: $$A_p:=\left\{ \frac{i}{p^j} : i,j\in\mathbb{Z}\right\} \setminus \mathbb{Z}$$ For any pair of distinct primes $p, q$ one can easily see that $A_p \cap A_q=\emptyset$, and both $A_p$ and $A_q$ are dense.

This made me curious about general topologies, and I attempted to prove or disprove the following proposition:

Given any topological space $(X,\mathcal{T})$, such that for all non-empty $S\in \mathcal{T}$ we have $|S| \ge2$, there exist two disjoint dense subsets of $X$.

My Initial Attempt:

Use the axiom of choice to define a function $f$ that selects an item from each non-empty open set in $X$. Now, use the axiom again of define a function $g'$ that selects an item from each set in: $$ \mathcal{T}':=\left\{ S\setminus\left\{f(S)\right\}\,:\,S\in\mathcal{T}\setminus\{\emptyset\} \right\} $$ and then define $g(S):=g'\left(S\setminus\left\{f(S)\right\}\right)$ for all non-empty $S\in \mathcal{T}$. This means that both $f(\mathcal{T})$ and $g(\mathcal{T})$ are dense, and moreover $f(S) \neq g(S)$ for all non-empty $S\in \mathcal{T}$. My attempt was to show that $f(\mathcal{T})$ and $g(\mathcal{T})$ are disjoint, but I later realized that this is incorrect. Any help will be very appriciated.

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2 Answers

up vote 6 down vote accepted

In technical terms, your question can be stated as "Is every perfect space resolvable?"

The answer to that is no.

A standard counterexample is a space $(X, T)$ where $T \setminus \{ \emptyset \}$ is a free ultrafilter on $X$. Since the ultrafilter is free, all nonempty open sets are infinite. Because it is an ultrafilter every nonempty proper subset is either open or closed, and therefore any dense proper subset must be open. On the other hand, no two sets in a filter are disjoint, and it follows that no two dense subsets can be disjoint. Note that the last part works more generally for any "door space".

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I'll have to read up on ultrafilters. I trust your answer is correct, and I'll confirm it as soon as I understand it. –  SomeStrangeUser Feb 12 at 16:39
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@SomeStrangeUser: For a quick introduction into filters, I can recommend section 5 of math.uga.edu/~pete/convergence.pdf –  Niels Diepeveen Feb 12 at 16:46
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I'm going to assume the space has a countable basis consisting of open sets with uncountable cardinality.

Choose a point from every basis set to get a dense set $A$. Now if there is no open set such that all of its points have been chosen, we are done, just choose the complement of $A$. But because we have just countably many points, not every point in each chosen set has been chosen.

Edit: Ok, this case was simpler than I thought. And not even very interesting.

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That is a HUGE assumption. It would be similar to asking whether or not every number is the sum of two primes, and you'd say "Well, I'm going to assume that the number is of the form $p+2$ for a prime number $p$". –  Asaf Karagila Feb 12 at 16:13
    
@Asaf Karagila: I agree. I was hoping that my argument would have something in it that would let me relax the assumptions eventually. But I'm not so sure anymore. –  J. J. Feb 12 at 16:14
    
My original method would work also for finite sets. I.e. select a point from each open subset and then deselect them one-by-one from "full" open sets to obtain a dense set with dense complement. –  J. J. Feb 12 at 16:30
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