Dense and open subset of a topological space Let $U $ be a dense and open subset of a topological space $X $,  and let $A\subseteq  B $  be a chain of open subsets of $X $ such that $A\cap U=B\cap U $.  Why can we verify that $A=B $?
 A: quasi already gave a counterexample to the claim. In almost any space $X$ we can take $U$ any non-trivial ($\neq X$) dense open set (e.g. the complement of a non-isolated point in a $T_1$ space),and then take $A= U, B = X$ : $A \cap U = B \cap U = U$, while $U = A \subsetneq B = X$. 
We can only say that $\overline{A} = \overline{B}$, which is weaker:
To see that $\overline{B} \subseteq \overline{A}$, the other inclusion already following from $A \subseteq B$, we let $x \in \overline{B}$. Let $O$ be any open set containing $x$, then $O \cap B$ is a non-empty open set so it intersects $U$, which is dense. Hence $$\emptyset \neq (O \cap B) \cap U = (U \cap B) \cap O= (U \cap A) \cap O \subseteq O \cap A\text{,}$$ so $O$ intersects $A$ and $x \in \overline{A}$. So $\overline{A} = \overline{B}$.
A: For $A=\phi$ the conclusion is clear (both empty). If $\phi\neq A\subsetneq B,$ then consider $C=B\setminus \bar{A},$ where $\bar{A}$ means the closure of $A.$ Then $C$ is a non-empty open set and hence $C\cap U\neq \phi,$ which implies $A\cap U\subseteq\bar{A}\cap U\subsetneq B\cap U,$ a contradiction.
This is wrong! Please omit.
A: The claim is false.

For example, let
$$X = \mathbb{R},\;\;\;A = \mathbb{R}\setminus\{0\},\;\;\;B =\mathbb{R},\;\;\;U=A$$
