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?
 A: 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).
A: 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.
