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Let $(E, \tau)$ be a topological space. Show that if $U,V ∈ τ$ are both dense in $E$, then $U ∩ V$ is dense in $E$. Is this still true if $U, V$ are not assumed open ?

Since $U$ and $V$ are both dense in $E$, we know $\forall O \in \tau, U \cap O \neq \emptyset $ and similarly $V\cap O \neq \emptyset $ so we can conclude $U\cap V \neq \emptyset $ and $U\cap V \in \tau$

What I really need to prove is $\overline{U\cap V } = E$,

Now, already know the following (by the assumption) : $E = E \cap E = \overline{U} \cap \overline{V} $ and (always true) $\overline{U \cap V} \subseteq \overline{U} \cap \overline{V} $

Maybe from the nonempty intersections: $\forall O \in \tau,\, U\cap(O \cap V) = U\cap (V \cap O) = (U \cap V) \cap O$ not sure if this a is valid move but if the intersection is dense then it will also have non-empty intersection with every open set in the topology.

And as far as it still being true with $U,V$ not open, I suspect at least one needs to be open.

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If one of them, say $U$, is open (and dense) and the other $V$ is dense, then the intersection is dense.

To see this: let $O$ be a non-empty open set in $X$. Then $O \cap U$ is open (as $U$ is too) and non-empty (as $U$ is dense). So $(O \cap U) \cap V$ is non-empty (as $V$ is dense). But this indeed equals $O \cap (U \cap V)$ and so $O$ intersects $U \cap V$ for all open non-empty $U$. So $U \cap V$ is dense.

We do need the openness, as we can take (as N.S. already suggests), the rationals and irrationals in the reals, both of which are dense, and neither is open.

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Hint For $U,V$ non-open, take $U=\mathbb Q$ and $V=\mathbb R \backslash \mathbb Q$.

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