# 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?

• It's even true for the product of countably infinitely many spaces. You can try to prove that if this seems too easy. – user2345215 Jan 18 '15 at 1:33
• Dear @user0430 I see that, although you have already asked 18 question in this site and received answers in most of them, you have not mark a best answer in any of them. You can do it so by clicking on the checkmark next to the answer that you think is the one that helped you the most. Please read here for more detail. – Leo Sera Jul 2 '15 at 21:36

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

• To the O.P. You can equivalently let $d((x,y),(x',y'))=d_X(x,x')+d_Y(x',y').$ You do not want the product $d_Xd_Y$ as this would make $0=d((x,y_1),(x,y_2))$ when $y_1\ne y_2.$ – DanielWainfleet Jul 17 '17 at 2:21
• @Henno Brandsma Isn't it easier to use that the metric you defined makes the projection continuous to show that the product topology is contained in the topology induced by the metric? – user370967 May 10 '18 at 18:14
• @Math_QED that’s a good alternative yes – Henno Brandsma May 10 '18 at 18:15

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.

• Dividing by $1+d_x+d_y$ is unnecessary, and will make the proof much harder than it needs to be. – Pedro Tamaroff Jan 18 '15 at 1:33
• @PedroTamaroff you are right. I will edit. – spatially Jan 18 '15 at 1:34

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.

• @user0430 Or, if you want something harder, try $\sqrt{d_x^2+d_y^2}$. – user2345215 Jan 18 '15 at 2:00
• haha, or just $(d_X^p+d_Y^p)^{1/p}$ for $1\leq p\leq \infty$, they are all equivalent! – spatially Jan 18 '15 at 2:08