0

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?

Error 404
  • 6,006
  • See here for more choices. – Dietrich Burde Mar 24 '17 at 13:59
  • @DietrichBurde I haven't came across metrizable spaces. But I just got acquainted with their concept and understood it. So what I should do is : Since $X$ is metrizable, $X \times X$ is also metrisable and hence it is a compact metric space since $X$ is compact. Right? – Error 404 Mar 24 '17 at 14:12
  • Your question was "But how can I be sure that $d'$ is the metric for $X \times X$? " So the other question also gives other choices, like $ d_{x\times y}:=\frac{d_x+d_y}{1+d_x+d_y} \tag 1$. – Dietrich Burde Mar 24 '17 at 14:19
  • @DietrichBurde Exactly! There are different choices for $d'$, but instead of just defining $d'$, aren't there any mathematical steps that intrinsically give us the correct $d'$? – Error 404 Mar 24 '17 at 14:42
  • But giving these metrics is really a mathematical step. What else do you expect? Any metric inducing the product topology is fine. – Dietrich Burde Mar 24 '17 at 15:06
  • @DietrichBurde So is there an exhaustive list of metrics for this problem? – Error 404 Mar 24 '17 at 15:15
  • I imagine there are an infinite number of such metrics and no, no such exhaustive list exists. All such metrics will yield the exact same open sets and only the same open sets. Thus all such metrics will yield the exact same topology (which is a definition of which sets are open and which are not) and thus can be considered equivalent. Choose any one you want. – fleablood Mar 24 '17 at 17:51
  • @fleablood So you say that all metrics yield the same topology for $X \times X$? That's cool then. This might be perhaps another question to ask on this site. – Error 404 Mar 24 '17 at 18:01
  • NO!!!! I am saying there are metrics that do and metrics that do not. But all the ones that do will yield the same open sets. And a topology is nothing more than a definition of which sets are open and which are not. SO the ones that do yield the same class of open sets are all equivalent and for all practical purposes can be considered the "same". But metrics that do NOT yield the same open sets are not equivalent. – fleablood Mar 24 '17 at 18:10
  • KEY WORD "such". – fleablood Mar 24 '17 at 18:10
  • @fleablood okay I got what you want to say. – Error 404 Mar 25 '17 at 09:01

2 Answers2

0

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.

Justthisguy
  • 1,551
  • Okay. I don't know these concepts : uniformly homeomorphic, bi-lipschitz. Also does your essence of argument is that all metrics are equivalent for $X \times X$? – Error 404 Mar 24 '17 at 17:58
  • Obviously not ALL metrics. The discrete metric, d(a,b) = 1 if $a \ne b$ but d(a,a) = 0, makes all sets open is obviously not eqivalent to the euclidean metric. However, if you have two different metrics that both yield the same open sets. (the max metric and the euclidean metric yield the same open sets as each other; the discrete metric and d(a,b) = 57.3 if a $\ne$ b yield the same open sets) then those to metrics are equivalent to each other. – fleablood Mar 24 '17 at 18:16
  • @fleablood Yeah! I got your point. – Error 404 Mar 25 '17 at 09:03
0

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.

fleablood
  • 124,253
  • Your word "naturally extend" sums up my question very well. Indeed, I was thinking of getting a single metric for $X \times X$ using some deductions. But now I see that I was wrong. We can define as many metrics as we please. One caution we need to take is that their induced topology matches with the product topology of $X \times X$. – Error 404 Mar 24 '17 at 18:22
  • Well, i think your max metric is the most "natural" interpretation. But it isn't the only interpretation. And there's really no authoritative reason to claim it is "the". We may, for whatever reasons, have very good reasons for another. E.g. In RxR then $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ describes actual geometric distance. But $\max(|x_1 - x_2|, |y_1-y_2|)$ yields an equivalent topology (with often much easier math). – fleablood Mar 24 '17 at 19:05
  • okay.. Thanks for elaborating. – Error 404 Mar 25 '17 at 09:02