Let $X$ be a topological space and let $X_1, X_2 \subseteq X$ two subsets with
$X = X_1 \cup X_2$. We view $X_1$ and $X_2$ as topological spaces equipped with the subspace topology inherited from $X$. Let $U \subseteq X_1 \cap X_2$ be a subset with the property that $U$ is open as a subset of $X_1$ and as a subset of $X_2$.
Since $U$ is open w.r.t. $X_1$ we can find an open $V_1 \subseteq X$ such that $U = V_1 \cap X_1$. Analogously, since $U$ is open w.r.t. $X_2$ we can find an open $V_2 \subseteq X$ such that $U = V_2 \cap X_2$.
I have to show that $U$ is open w.r.t $X$, but I do not really see how to continue. I thought about considering $U = (V_1 \cap X_1) \cup (V_2 \cap X_2)$ and somehow find a way to write this as $A \cap (X_1 \cup X_2)$ for some $A$ expressed in $V_1$ and $V_2$, but it didn't really work out well. Could someone help me with this?
Best Answer
Let $V=V_1\cap V_2$. Then
$$\begin{align*} V\cap X_1&=(V_1\cap V_2)\cap X_1\\ &=V_2\cap(V_1\cap X_1)\\ &=V_2\cap U\,. \end{align*}$$
$U=V_2\cap X_2$, so $U\subseteq V_2$, and therefore
$$V\cap X_1=V_2\cap U=U\,.$$
A similar argument shows that $V\cap X_2=U$. Can you finish it from here?