Prove that $C$ is open with respect to $A\cup B$ if $C\subset A\cap B$, $C$ is open with respect to $A$ and $C$ is open with respect to $B$.

Question

Let $A$, $B$ and $C$ be subspaces of a topological space $X$ with $C\subset A\cap B$. Prove that $C$ is open with respect to $A\cup B$ if $C$ is open with respect to $A$ and $C$ is open with respect to $B$.

My attempt: Let $x$ be a point of $C$. Then there is an open set $U$ in $A$ such that $x\in U=A\cap M\subset C$, for some set $M$ open in $X$. But there is also an open set $V$ in $B$ such that $x\in V=B\cap N\subset C$, for some set $N$ open in $X$. Since $x\in M\cap N$ and $x\in A\cup B$ too (because $C\subset A\cap B \subset A\cup B$), then $$x\in \left (M\cap N\right ) \cap \left (A\cup B\right).$$

Since $$\left (M\cap N\right ) \cap \left (A\cup B\right)=\left (M\cap N\cap A\right) \cup \left (M\cap N\cap B\right) \subset \left (M\cap C \right) \cup \left (C\cap N\right) = C\cap \left (M\cup N\right) \subset C, $$ therefore $C=\left (M\cap N\right ) \cap \left (A\cup B\right)$, i.e. $C$ is open in $\left (A\cup B\right)$.

If I'm wrong give me a hint to be right, please. Thanks in advance.

Answer 1

If $C\subset A\cap B$ and $C$ is open with respect to $A$ and $B$, then $C=A\cap U$ where $U$ open in $X$ and $C=B\cap V$ where $V$ open in $X$. Then $$C=(A\cap U) \cup (B\cap V) = (A\cup B)\cap(U\cup V),$$ so that $C$ is open in $A\cup B$.