Denseness of a set, whose complement is known to be dense.

Do you think in general that if say $U\subset X$ was dense in $X$, then if we let $V=X−U$, but we know $V$ is of higher cardinality than $U$, does that imply that $V$ must be dense?

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Take the set $X=[0,1] \cup \{2\}\subset \mathbb R$ with the usual topology induced by $\mathbb R$. The set $\mathbb Q \cap X$ is dense in $X$ and it is countable. The complementary set is not countable but also non dense because $2$ is far from every point in $X\setminus \mathbb Q$.

addendum to answer the question in the comments. You can also take $X=[0,1] \cup \mathbb Q \subset \mathbb R$ and you notice that $U=\mathbb Q$ is dense in $X$, the complementary set, however, is the same as before. It is not countable, but not dense (even more so).

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Yes that makes sense, thank you! –  ellya Mar 29 '14 at 8:42
But how does this change if we assert that the topology must be defined on the set $X$, of which $U$ is dense in. i.e. the standard topology on $\mathbb{R}$ with $U=\mathbb{Q}$. –  ellya Mar 29 '14 at 9:05
Here the closure of $\mathbb { Q }$ is $\mathbb { R }$ which is not $X$ –  ellya Mar 29 '14 at 18:02
@ellya: the closure of a set cannot be larger than the ambient space. If you consider $X$ as a topological space, then the closure of $U$ is $X$. –  Emanuele Paolini Mar 29 '14 at 18:39
Good to know, is that because in terms of the topology, the other points cease to exist? –  ellya Mar 29 '14 at 18:57

At an extreme end you can consider any nonempty set $X$ with the particular point topology: pick $x_* \in X$, and declare the nonempty open sets to be exactly those subsets of $X$ which contain $x_*$.

It is easy to see that $\{ x_* \}$ is a dense subset, however $X \setminus \{ x_* \}$ is not (since $\{ x_* \}$ is a nonempty open set disjoint from it). And this is true regardless of the cardinality of $X$.

You'll see that a lot of these examples follow the same line. Not only is there a small dense subset, but there is a small dense subset with nonempty interior. And this is a necessary condition. The basic definition of density is that $\overline{D} = X$, however with the relationship $\overline{A} = X \setminus \mathrm{Int} ( X \setminus A )$ it follows easily that $D$ is dense iff $\mathrm{Int} ( X \setminus D ) = \varnothing$. By choosing dense sets with nonempty interiors, we ensure that their complements cannot be dense. And topological spaces can have small (in terms of cardinality) open sets.

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That was my example for the OP in the comments where this question first came up. (I used the particular $\Bbb R$ and $0$.) I suppose that now it's the time to say something like set theoretical minds think alike? :-P –  Asaf Karagila Mar 29 '14 at 14:51
@Asaf: That's probably more favorable than the German Zwei Dumme, ein Gedanke. (I'll leave their comparative accuracies for others to determine.) ;-) –  Arthur Fischer Mar 29 '14 at 15:04