Coproduct of totally disconnected sets is totally disconnected. Let $(X_i)_{i \in I}$ be a family non empty topological spaces. Prove that $\coprod_{i \in I} X_i$ is totally disconnected if $X_i$ is totally disconnected for all $i \in I$
My book gives the hint:
Consider the subspace $\{(i,y): i \in I, \forall k \neq i: y_k =x_k\}$ of $I \times Y$ where $I$ is equipped with the discrete topology, $Y:= \prod_{i \in I} X_i$
I can see that there is a natural bijection between the coproduct and the given subspace, but I can't prove it is a homeomorphism. I'm also unable to see how this homeomorphism would imply the result I want.
 A: I don't see the need for showing a homeomorphism with that product subspace. It's true, but it doesn't add much (I assume that $(x_i)$ is some fixed point of $Y$, essentially). It would help if you'd already have shown two basic facts on totally disconnected spaces:


*

*Any product of totally disconnected spaces is totally disconnected.

*Any subspace of a totally disconnected space is totally disconnected.


But you can just show the result directly: Let $C$ be any connected subset of $X:=\coprod_{i \in I} X_i$. Let $e_i: X_i \to X$ be the canonical injections by which the topology is (co-)induced. 
Suppose that $C$ has at least two points. If there are two distinct $i, j \in I$ such that $e[X_i] \cap C \neq \emptyset$ and $e[X_j] \cap C \neq \emptyset$ then $e[X_i]\cap C$ is a non-empty clopen subset of $C$ unequal to $C$ (as all $e[X_k]$ are open and disjoint in $X$ for $k \in I$, so each is the complement of the union of the others, hence the closedness) and this shows that $C$ is not connected. The other case is when there is an $i_0 \in I$ such that $C \subseteq e[X_{i_0}] \simeq X_{i_0}$ and so $C$ is disconnected as $X_{i_0}$ is totally disconnected. 
Hence $X$ is totally disconnected, as required. IMHO this shows much clearer why $X$ is totally disconnected when all $X_i$ are. 
