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Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Show that the topology induced by the metric on $X\times Y$ is equal to the product topology.

Note that the product metric is maximum of the metrics.

My attempt: It suffices to show that the basis for the product topology is a basis for the topology induced by the product metric. Let $\mathbb{B}$ the basis which generates the product topology. If $B\in \mathbb{B}$ then $B=U_1\times U_2$ where $U_1$ is open in $X$ and $U_2$ is open in $Y$. Hence $B=\bigcup_jB(x_j,r_j) \times \bigcup_iB(y_i,r_j)=\bigcup_{ij}(B((x_i,y_j),min(r_i,y_j))$. Thus, $B$ is the union of sets open in the metric topology. By a similar reasoning, every open set in the metric topology is the union of some collection of members in $\mathbb{B}$.

Is my attempt correct?

1 Answers1

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The basic observation is that (using $d$ the max metric on $X \times Y$):

$$B_d((x,y),r) = B_{d_X}(x,r) \times B_{d_Y}(y,r)$$

So a $d$-ball is product open, so $\mathcal{T}_d \subseteq \mathcal{T}_{\text{prod}}$

And if $O$ is product open in $X \times Y$, and $(x,y) \in O$ we have $(x,y) \in O_1 \times O_2 \subseteq O$ for some open $O_1 \subseteq X$, $O_2 \subseteq Y$. We have $r_1,r_2 > 0$ such that $B_{d_X}(x,r_1) \subseteq O_1$ and $ B_{d_Y}(y,r_2) \subseteq O_2$. Then with indeed $r=\min(r_1,r_2)$ we have that $B_d((x,y),r) \subseteq O$ and so $O$ is $d$-open and hence $\mathcal{T}_{\text{prod}} \subseteq \mathcal{T}_d$ and we have an equality of topologies.

So your proof contains the right idea, but is less systematic IMHO.

Henno Brandsma
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