Product of Two Metrizable Spaces I am having trouble with a practice exam question:
$$\text{Show that if $X$ and $Y$ are metrizable, then so is $X\times Y$}$$
What I have so far:
Given metric spaces $(X,d_x)$ and $(Y,d_y)$, I know that since $d_x$ and $d_y$ are continuous, then $d_x \times d_y$ is also continuous, so all that I need to prove is that $d_x \times d_y$ defines a metric on $X \times Y$.  The first two properties are trivial, I am just a little confused as to the last property (Triangle Inequality)
So let $(x_1,y_1),(x_2,y_2),(x_3,y_3) \in X \times Y$.  Then
$$
\begin{align}
& d_{X \times Y}((x_1,y_1),(x_3,y_3)) = d(x_1,x_3) \times d(y_1,y_3) \\[4pt]
\le {} & (d(x_1,x_2) + d(x_2,x_3)\times(d(y_1,y_2)+d(y_2,y_3))
\end{align}
$$
By just expanding out the right side, I get 4 terms. Is this what I am suppose to get? Or is my approach wrong?
 A: There are two things to be done: 
a) Define a metric on $X \times Y$ (based on $d_X$ and $d_Y$). The product of the metrics won't work (as then the distance is already $0$ for points like $(x,y)$ and $(x,y')$, to name an easier reason), but the sum or max both work and are convenient choices. So e.g. (my favourite choice for 2 spaces) define 
$$d((x_1, y_1), (x_2, y_2)) = \max(d_X(x_1, x_2), d_Y(y_1, y_2))$$
and check that this indeed gives a metric on $X \times Y$.
b) Secondly, and quite importantly, one has to check that the topology of $(X \times Y, d)$ is actually the same as the product topology on $X \times Y$. That will show that the product of $X \times Y$, which by definition/convention will have the product topology, can also be induced by a metric, i.e. is metrisable.
E.g. check that $$B_d((x,y), r) = B_{d_X}(x,r) \times B_{d_Y}(y, r)$$
which shows that open balls under $d$ are indeed open in the product topology of $(X,d_X)$ and $(Y,d_Y)$, so $\mathcal{T}_d \subseteq \mathcal{T}_{\text{prod}}$. The reverse also needs showing: suppose $O$ is product open, and let $(x,y) \in O$. Then there is an open set $U$ in $\mathcal{T}_X$ (topology induced by $d_X$) and an open set $V$ in $\mathcal{T}_Y$, such that 
$(x,y) \in U \times V \subseteq O$. This means that there is some $r_1>0 $ such that $x \in B_{d_X}(x,r_1) \subseteq U$ and some $r_2>0$ such that $y \in B_{d_Y}(y,r_2) \subseteq V$. But then, setting $R  =\min(r_1, r_2)>0$:
$$(x,y) \in B_d((x,y), R) = B_{d_X}(x,R) \times B_{d_Y}(y,R) \subseteq B_{d_X}(x,r_1) \times B_{d_Y}(y,r_2) \subseteq U \times V \subseteq O$$
showing $O \in \mathcal{T}_d$ as required. So the product topology coincides with the topology generated by $d$.
As a final remark: one can show that this extends to countable products of metric spaces as well. This is somewhat more work, and then a sum metric is the most suitable: 
$$d((x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}}) = \sum_{n \in \mathbb{N}} \frac{1}{2^n}\min(d_{X_n}(x_n, y_n), 1)$$
will then do as a metric on $\prod_n X_n$ that induces the product topology. See this answer for more details.
Also s
A: Usually multiplication will not preserve the triangle inequality. I suggest you try
$$ d_{x\times y}:=\frac{d_x+d_y}{1+d_x+d_y} \tag 1$$
Yes, as comment suggest $d_X+d_Y$ is good enough. But I still recommend you to try to prove $(1)$, which I am sure you will deal with later in your class.
A: You can use $d_X+d_Y$, and you may also use $\max(d_X,d_Y)$. In both cases, showing they are metrics is not hard. 
