Disjoint union of Hausdorff spaces Recently I have read the post: Hausdorff Disjoint Union which proves that disjoint union of Hausdorff spaces are Hausdorff.
My understanding of the proof is that:

*

*If two elements are in the different indexed sets (suppose we have $X,Y$ only) then we can define two open sets $X\coprod \emptyset$ and $\emptyset\coprod Y$ such that they are disjoint.

*This is the part I do not understand. I know that we can have disjoint $U,V\subseteq X$ such that they contain two elements and disjoint but I do not see why they can form two disjoint sets in $X\coprod Y$. Can I say that one of them is in $U\coprod \emptyset$ and the other is in $V\coprod \emptyset$?

Thanks!
 A: To set a proper notation:
$\coprod_{i \in I} X_i$ is the set $\{(x,i): i \in I \land x \in X_i\}$, with standard embeddings $e_i: X_i \to \coprod_{i \in I} X_i$ defined by $e_i(x)=(x,i)$ and topology $\{O \subseteq \coprod_{i \in I} X_i \mid \forall i \in I: e_i^{-1}[O] \text{ open in } X_i\}$. In this topology all $e_i$ are open continuous injections from the $X_i$ into $\coprod_{i \in I} X_i$ and in many texts $X_i$ and its image $X_i \times \{i\}$ under $e_i$ are identified (they are homeomorphic after all) but I find that in proofs like this it pays to be exact.
If we have two points $(x,i)$ and $(y,i)$ in $\coprod_{i \in I} X_i$ coming from the same $X_i$ we find $U$ and $V$ disjoint in $X_i$ such that $x \in U$ and $y \in V$. Then $e_i[U] = U \times \{i\}$ and $e_i[V] = V \times \{i\}$ are also disjoint and open in $\coprod_{i \in I} X_i$. The former contains $(x,i)$, the latter $(y,i)$.
Indeed, if $(x,i) \neq (x', j) \in \coprod_{i \in I} X_i$ where $i \neq j$ we can just take $e_i[X_i]$ and $e_j[X_j]$ (which are open too) and are trivially disjoint because they differ on their index coordinates.
So $\coprod_{i \in I} X_i$ is Hausdorff when all $X_i$ are.
A: Yes, intuitively you are working with already separated sets. This means that the ways of arranging are separated.
