Algebraic (Kuratowski axiomatic) proof of a simple topological statement I was trying to prove the following basic result using the Kuratowski closure axioms for topological spaces.
Let $X$ be a space, $A$ and $U$ dense and open subsets respectively. Then $\overline{U} = \overline{U \cap A}$.
 A: Lemma: Let $X$ be a topological space, $U ⊆ X$ open, $A ⊆ X$. Then $\overline{A} ∩ U ⊆ \overline{A ∩ U}$.
Proof: $\overline{A} = \overline{A ∩ U} ∪ \overline{A \setminus U}$ by additivity. We have that $A \setminus U ⊆ X \setminus U$, which is closed, so $\overline {A \setminus U} ⊆ X \setminus U$, so $\overline{A \setminus U} ∩ U = ∅$. In conclusion $\overline{A} ∩ U ⊆ \overline{A ∩ U}$.
Proof of the original statement: $\overline{U ∩ A} ⊆ \overline{U}$ by monotonicity. $U = U ∩ \overline{A} ⊆ \overline{U ∩ A}$ by density and previous lemma. $\overline{U} ⊆ \overline{\overline{U ∩ A}} = \overline{U ∩ A}$ by monotonicity and transitivity/idempotence.
A: Prove: Obviously, $\overline {U\cap A} \subset \overline{U} $. Now we prove that $ \overline{U}  \subset \overline {U\cap A}$. Suppose that $x \in \overline{U}$. If $x \notin  \overline {U\cap A}$, then there is an open set $V$ containing $x$, satisfying that $V \cap (U\cap A)=\emptyset$. It implies that $(V\cap A) \cap U=\emptyset$. Since $A$ is dense, we have $V\cap A$ is not empty, and since $ A$ is open, $V\cap A$ is also open. Since $U$ is dense in $X$, $(V\cap A) \cap U \not=\emptyset$. It is a contradiction.
