A question of topological sum of subspaces I have the following question... I believe it is true but I don't know how to prove it...
Let $X$ be a topological space and $A_1,A_2$ disjoint subspaces of $X$.
If we consider the topology of subspace on $A_1\cup A_2$, is it true that $A_1\cup A_2 = A_1\oplus A_2$?
Thanks.
 A: This can fail terribly. E.g. in $\mathbb{R}$ take the disjoint subsets $A_1 = \mathbb{Q}$ and $A_2 = \mathbb{P} = \mathbb{R} \setminus \mathbb{Q}$. Then in $A_1 \cup A_2$ no non-empty subset of $A_1$ nor one of $A_2$ is open, while in the topology on $A_1 \oplus A_2$ all such sets would be open in the total space.
Stefan's example from the comments (also in $\mathbb{R}$): $A_1 = [-1,0), A_2 = [0,1]$ also works: e.g. their union is compact, their topological sum is not (only if both would be compact). 
It is true if $A_1$ and $A_2$ are disjoint closed, or disjoint open subspaces of a space $X$. For the former case, a set $A \subset A_1 \cup A_2$ is closed iff there is some closed subset $A'$ of $X$ such that $A' \cap (A_1 \cup A_2) = A$, and then $A$ can be written as $(A' \cap A_1) \cup (A' \cap A_2)$, which is a union of two closed sets of $X$ which are also closed in $A_1$ and $A_2$ respectively. So there the description of closed subsets of the union and the topological sum do coincide, and the same argument also holds for disjoint open sets. 
I suppose the same would more generally hold for separated subsets $A_1$ and $A_2$ of a space $X$ (so when $\overline{A_1} \cap A_2 = \emptyset = A_1 \cap \overline{A_2}$), as then $A_1$ and $A_2$ are relatively clopen sets in the union, and we could have the same arguments.
