Let $(X_i)_{i\in I}$ be a indexed family of topological spaces.
If each $X_i$ is hausdorff then so is $X=$ $\coprod_{i\in I}X_i$.
Note: I will identify $X_i$ with its image, $X_i^*$, wrt canonical injection from $X_i$ to $X$. This is okay because a homeomorphism between the two exists.
My attempt: Let $(x,i),(y,j)$ be distinct points in $X$. If $i\neq j$ then since each $X_i$ is disjoint from $X_j$ when $i\neq j$, and each $X_i$ is open in $X$, the result follows.
On the other hand, if $i=j$, then since $X_i$ is hausdorff. we can find open disjoint open sets $U$ and $V$ in $X_i$ containing $(x,i)$ and $(y,i)$. Since $U$,$V$ are open in $X_i$, they are open in $X$
Is my attempt correct?
Best Answer
Yes, it's fine. Maybe notation wise, using the canonical injections $\sigma_i$:
If $i \neq j$ then use $\sigma_i[X_i]$ and $\sigma_j[X_j]$ as disjoint open sets.
Otherwise $i=j$ and so $x \neq y$ and we apply Hausdorffness in $X_i$ to get disjoint $O_x, O_y \subseteq X_i$ and as the $\sigma_i$ are open (open embeddings even), $\sigma_i[O_x]$ and $\sigma_i[O_y]$ are as required.
Other separation axioms are equally boring: $T_1,T_3, T_4$ all get preserved: you van just apply the property in each summand space and "transport" them into the sum via the $\sigma_i$. Also, the sum is metrisable iff all $X_i$ are: just truncate all metrics in the $X_i$ and give the metric between points of distinct summands a large enough value. Paracompactness is also preserved.
Of course sums of infinitely many (non-empty) summands will never be compact (or connected), and for countably many summands separability and Lindelöfness are preserved but not for more summands than that.