If $(X,d)$ is a compact metric space, then we know that $X \times X$ is also a compact metric space. But how to get it's metric?

Question

I know that product of finite compact topological spaces is compact.

But when we take product $X \times X$ of a compact metric space $(X,d)$, how will the metric of this space look like? We can define a metric as follows :

$$d'((x,y),(z,w))=\max\{d(x,z),d(y,w)\}.$$ Then using sequential compactness criterion, it is easy to show that $(X \times X,d')$ is a compact metric space. But how can I be sure that $d'$ is the metric for $X \times X$?

I know how to show that $d'$ is a metric using the definition of a Metric space. But are there any mathematical steps that lead us to showing that $X \times X$ is a compact metric space without defining a metric like $d'$ as above? Or if possible showing that $d'$ is the metric?

Answer 1

This question is context dependent. There are many metrics which can be put on $X \times Y$, many of which give homeomorphic, or uniformly homeomorphic, or Bi-Lipschitz metric structures. For example, you could write $d''((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2)$. This spaces $(X \times Y, d')$ and $(X \times Y, d'')$ are homeomorphic, uniformly homeomorphic, and bi-Lipschitz. That is, $d' \leq d'' \leq 2 d'$. So they are both, in some way, perfectly good metrics to put on $X \times Y$.

Let me give a criterion which forces us to cook up the $d'$ you give. If $(U,d_U),(V,d_V)$ are metric spaces, say that $f:U \to V$ is short if $d_V(f(u_1),f(u_2)) \leq d(u_1,u_2)$. Note that if $f$ is short, bijective, and if $f^{-1}$ is short, then $f$ is an isometry; i.e. $d_V(f(u_1),f(u_2)) = d(u_1,u_2)$ for all $u_1,u_2 \in U$. Say that a metric space $(V,d_V)$ is the product of metric spaces $(X,d_X)$ and $(Y,d_Y)$ if there exist short maps $p_X : V \to X$ and $p_Y : V \to Y$, called projection maps, such that, if $(U,d_U)$ is a metric space, and $\phi_X : U \to X$ and $\phi_Y : U \to Y$ are short, there exists a unique short map $\psi : U \to V$ such that $\phi_X = p_X \circ \psi$ and $\phi_Y = p_Y \circ \psi$. If $(U,d_U)$ and $(U', d_{U'})$ are products of $X$ and $Y$, with projection maps $p_X,p_Y,p'_X,p'_Y$ respectively, then there exists an isometry between $U$ and $U'$. Indeed by definition there exists a unique map $\psi : U \to U'$ such that $p_X = p'_X \circ \psi$ and $p_Y = p'_Y \circ \psi$, and a unique map $\psi' : U' \to U$ such that $p'_X = p_X \circ \psi'$ and $p'_Y = p_Y \circ \psi'$. We then have $p_X = p'_X \circ \psi = p_X \circ \psi' \circ \psi$, and similarly $p_Y = p_Y \circ \psi' \circ \psi$. But there exists a unique $\eta : U \to U$ such that $p_X = p_X \circ \eta$ and $p_Y = p_Y \circ \eta$, and one candidate for $\eta$ is $\text{Id}_U$, so $\psi' \circ \psi = \text{Id}_U$, and similarly $\psi \circ \psi' = \text{Id}_V$, giving an isometry between $U$ and $U'$. This justifies calling $U$ the product of $X$ and $Y$.

I claim that $X \times Y$ with the metric you gave, i.e. $$d_{X \times Y}((x_1,y_1),(x_2,y_2)) = \max(d_X(x_1,x_2),(y_1,y_2)),$$ is the product of $X$ and $Y$ in the above sense, with $p_X$ and $p_Y$ the standard projection maps. Indeed, let $(U,d_U)$ is a metric space, and let $\psi_X : U \to X$ and $\phi_Y : U \to Y$ be short. Then if $p_X \circ \psi = \phi_X$ and $p_Y \circ \psi = \phi_Y$, we must have $\psi(u) = (\phi_X(u),\phi_Y(y))$. Note that $\psi$ is short: $$d_{X \times Y}(\psi(u_1),\psi(u_2)) = \max(d_X(\phi_X(u_1),\phi_X(u_2)),d_Y(\phi_Y(u_1),\phi_Y(y_2)))$$ $$ \leq d_U(u_1,u_2),$$ since $\phi_X$ and $\phi_Y$ are both short.

Answer 2

Okay, this is my understanding and it may be naive and if it is wrong I will take this answer down. And, for simplicity, I'll limit dimensions to 2.

You can't define the cross product of metric spaces $X\times Y$ as a metric space without defining a metric to go on it. Otherwise $X \times Y$ is simply a set; a collection of ordered pairs". We can define any metric, $D$, we want but we want one that will somehow "naturally extend" the metrics $d_X$ of $X$ and and $d_Y$ of $Y$.

By "naturally extend" I think we mean that if $U \subset X$ is open with metric $d_X$ and $V \subset Y$ is open with $d_Y$ then $U \times V$ will be open $D$ and if $A \subset X \times Y$ is open $D$ and every "cross-section" of $A$ restricted to $X$ or $Y$ is open with $d_X$ or $d_Y$ respectively.

(A "cross-section" (for now) can be described: We can define $\phi_X:X\times Y\rightarrow X$ via $\phi_X(x,y) = x$ for any $(x,y) \in X \times Y$ and $\phi_Y(x,y) = y$ similarly. (These are the projections from $X\times Y$ into $X$ and $Y$). The two strictly $X$ and $Y$ cross-sections of $A$ at any point $(w,z) \in X\times Y$ would be $\phi_X(A \cap X \times\{z\})$ and $\phi_Y(A \cap \{w\} \times Y)$.)

There might/will be many/infinite such compatible metrics. However if two different metrics yield the same class of open sets, they yield the same topology. And thus are considered equivalent.

We can chose any metric that satisfies the condition that the projection of open sets in $X\times Y$ yield open sets when projected into $X$ and $Y$. Your maximum metric ($D(a=(x_1,y_1), b=(x_2,y_2)) = \max (d_X(x_1,x_2),d_Y(y_1,y_2)$) is one such. As would be the Euclidean metric ($D_e(a,b) = \sqrt{d_X^2(x_1,x_2) + d_Y^2(y_1,y_2)}$). These would be considered equivalent metrics and will yield the same topology with the same class of open sets.

I hope I got this right, and I hope it was simple to follow.