Let $X_1, X_2$ be subspaces of $X$ and $A \subset X_1\cap X_2$. Suppose that $A$ is open in induced toplogy on $X_1$ and it's also open in induced topology on $X_2$. Prove that, $A$ is open in $X_1\cup X_2$.

The definition of induced topology on $X_1, X_2$:

$\tau_{X_1} = \{U \cap X_1: U \in \tau\}$, $\tau_{X_2} = \{U \cap X_2: U \in \tau\}$. Where $\tau$ is a topology on space X.

From conditions above, we get

$A = U_1\cap X_1\in \tau_{X_1}$ and also $A = U_2\cap X_2\in \tau_{X_2}$, now I have to show that $A\in \tau_{X_1\cup X_2}$. So there must exists $U\in \tau$ such as $A = U\cap(X_1\cup X_2)$. That's the moment when I can't go any further.

Is it correct to this moment?


You really should say explicitly that $U_1,U_2\in\tau$:

There are $U_1,U_2\in\tau$ such that $A=U_1\cap X_1\in\tau_{X_1}$ and $A=U_2\cap X_2\in\tau_{X_2}$.

HINT: What is $(U_1\cap U_2)\cap(X_1\cup X_2)$?


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