$\coprod_{i \in I} X_i$ metrizable iff $\forall i \in I: X_i$ metrizable

Prove that $\coprod_{i \in I} X_i$ metrizable iff $\forall i \in I: X_i$ metrizable

My attempt:

"$\implies$" If $\coprod X_i$ is metrizable, then also the subspace $X_i \times \{i\}$ for each $i \in I$, and because this space is homeomorph with $X_i$, it follows that $X_i$ is metrizable as well.

$"\impliedby"$ Suppose $\mathcal{T}_i$ is induced by a metric $d_i$ for every $i \in I$, which we can assume to be less than $1$. (Indeed, if $d$ is a metric, then the function $d \land 1$ is a metric too that induces the same topology).

Define $d: \coprod X_i \times \coprod X_i \to \mathbb{R}^+: ((x,i),(y,j)) \mapsto \begin{cases} d_i(x,y) \quad i = j \\ 1 \quad i \neq j\end{cases}$

We claim that $\coprod \mathcal{T}_i = \mathcal{T}_d$

"$\subseteq$" Sufficient: $\forall i \in I: B_i \in \mathcal{T}_i: B_i \times \{i\} \in \mathcal{T}_d$

Hence let $(x,i) \in B_i \times \{i\}$. Then $x \in B_i \in \mathcal{T}_{d_i}$, so we can find $\epsilon \in (0,1): B_{d_i}(x, \epsilon) \subseteq B_i$. Then $B_d((x,i),\epsilon) = B_d(x,\epsilon) \times \{i\} \subseteq B_i \times \{i\}$, hence the inclusion follows.

"$\supseteq$" Let $\epsilon \in (0,1)$. Let $(x,i) \in \coprod X_i$.

Then $B_d((x,i), \epsilon) = B_{d_i}(x,\epsilon) \times \{i\} \in \coprod \mathcal{T}_i$. Because the balls with radius smaller than $1$ form a basis, this is sufficient.

This ends the proof.

Does this look correct?

• I could not understand this until I realized that what you are calling $\prod_{i\in I}X_i$ is not the product of the sets $X_i$ but the disjoint union, which should be written $\cup_{i\in I}(X_i\times \{i\})$. In other words the toplogical sum of metrizable spaces is metrizable. – DanielWainfleet May 11 '18 at 1:55

To me it looks quite correct. Not much to add, really. Maybe a short proof that $d$ is actually a metric? (some cases for the triangle inequality, mostly).