How to show that a disjoint sum of metric spaces is metrizable? I have encountered this question while tring to figure out why every sequential space $X$, is a quotient spaces of a metric space. If I understand correctly, given a sequential space, every sequence (including it's limit) can be identified with the space $Y=\{0\} \cup \{\frac1{n+1}|n \in N\}$. So The space $X$ can viewed as a quotient space of $\bigoplus_{(x_n) \in C} \{(x_n) \} \times Y$ where $C$ is the set of all converges sequences in $X$. My question is, How can I show that this space is metrizable?
Thank you!
Shir
 A: I'll expand a little on Daniel Fischer's comment.
Let $(M_1,d_1)$ and $(M_2,d_2)$ be two metric spaces and define $d_i^*(x,x')=\dfrac{d_i(x,x')}{d_i(x,x')+1}$ for $i=1,2$. Note that $(M_i,d_i)$ is homeomorphic (although not isometric) to the metric space $(M_i,d_i^*)$. Notice that $d_i^*(x,x')\leq 1$.
We now define the metric space $(M,d)$ by $M=M_1\sqcup M_2$ and $$d(x,x')=
\left\{\begin{array}{lcl}
d_1^*(x,x')&\mbox{ if }&x,x'\in M_1\\
d_2^*(x,x')&\mbox{ if }&x,x'\in M_2\\
2&&\mbox{ otherwise }\\
\end{array}\right.$$
which one can readily check is a metric space and is homeomorphic to the $M_1\sqcup M_2$ as topologic spaces. It follows that $M$ is a metrisable space. One can then extend this definition to arbitrary disjoint unions in the usual way.
A: Suppose $(X_i,d_i),\ i\in I$, is a family of disjoint nonempty metric spaces. For each $i\in I$ choose a base point $a_i\in X_i$. We can define a metric on the set $X=\bigcup_{i\in I}X_i$ by defining $d(x,y)=d_i(x,y)$ if $x,y\in X_i$ and, if $x\in X_i$ and $y\in X_j$ where $i\ne j$, then
$$d(x,y)=d(x,a_i)+1+d(a_j,y).$$
