Metric defined on the union of two disjoint metric spaces

Question

I am working on this question:

Given $2$ metric spaces $(X, d_{X})$ and $(Y, d_{Y})$ with some bound $M$ of $d_{X}+d_{Y}$ and $X\cap Y=\varnothing.$ Then, we can define a metric on $X\cup Y$, by $$d(u,v)=d_{X}(u,v)\ \text{if}\ u,v\in X,$$ $$d(u,v)=d_{Y}(u,v)\ \text{if}\ u,v\in Y,$$ $$d(u,v)=M\ \text{if}\ u\in X,\ v\in Y.$$ Check that this satisfies the axioms to show that $d(u,v)$ induces a metric on $X\cup Y$.

A similar question was asked here: Disjoint Union of Metric Spaces is a Metric Space, but this post was about the proof of triangle inequality and this post specified the upper bound $M=1$.

I have two other questions:

(1) How to show $d(u,v)=d(v,u)$? This equality holds immediately for $u,v\in X$ or $u,v\in Y$, but the case of $u\in X$ and $v\in Y$ is a little bit confusing. If we have chosen $u\in X$ and $v\in Y$, $d(u,v)=M$ is true by definition, but is it true that the $d(v,u)$ is not defined? (since the definition specifies that the first coordinate is in $X$ and the second is in $Y$).

(2) Can $M=0$? At first I thought $M$ cannot be $0$, but I could not prove it.

If $M=0$, then for all $x_{1}, x_{2}\in X$ and $y_{1}, y_{2}\in Y$, we have $$d_{X}(x_{1}, x_{2})+d_{Y}(y_{1}, y_{2})\leq 0,$$ implies $$d_{X}(x_{1}, x_{2})\leq -d_{Y}(y_{1}, y_{2}),$$ but both of them are nonnegative, and thus for all $x_{1}, x_{2}\in X$ and $y_{1}, y_{2}\in Y$, $$d_{X}(x_{1}, x_{2})=d_{Y}(y_{1}, y_{2})=0,$$ which happens if and only if $x_{1}=x_{2}$ and $y_{1}=y_{2}.$

This implies that $X$ and $Y$ are just two distinguished points. Then if you choose $x\in X$, and $y\in Y$, then by definition $d(x,y)=M=0$.

Then I want to use $d(x,y)=0$ as a contradiction since $X\cap Y=\varnothing$. However, given the post here:

Find two closed subsets or real numbers such that $d(A,B)=0$ but $A\cap B=\varnothing$

and here:

Prove that if $A$ is compact and $B$ is closed and $A\cap B = \emptyset$ then $\text{dist}(A,B) > 0$

I don't think it gives me any contradiction.

Answer 1

For (1), if $d(v, u) $ is left undefined, then they assumed you'd figure out that it's equal to $M$, because there's nothing else it can be.

For (2), if $M=0$ then this violates the requirement that $d(x, y) =0$ implies that $x=y$. So $M$ can't be zero for this to work. The linked questions are about distances between sets, not distances between points. The infimum of the distances between points in two sets can be $0$ because it can be arbitrarily small. For a single pair of distinct points (or disjoint finite sets of points) the distance can't be $0$.

Answer 2

1. The definition of $d$ is clearly incomplete, since $d(v,u)$ is undefined when $v\in Y$ and $u\in X$. Of course, the only way of defining it at those points is to put $d(v,u)=M$.
2. No, assuming that $X,Y\neq\emptyset$ . Because then you take $x\in X$ and $y\in Y$ and then $0\neq d(x,y)=M$ .

Answer 3

No, $M$ should be a bound on $d_X + d_Y$ and if $X$ or $Y$ has at least two points, the distance between those alone is $>0$ by the definition of a metric for $d_X$ resp. $d_Y$. And if $|X|=|Y|=1$ we must take $M>0$ for the distance between the unique point of $X$ and that of $Y$.

Having such a bound is not really restrictive because $d'_X=\min(d_X,1)$ is bounded by $1$ and induces the same topology as $d_X$ on $X$. So do the same for $Y$ and take $M=2$, e.g.

The final condition could better be formulated as $d(u,v)=M$ when $\{u,v\} \cap X \neq \emptyset$ and $\{u,v\} \cap Y \neq \emptyset$ and $d(u,v)=d_X(u,v)$ when $\{u,v\} \subseteq X$ and likewise for $Y$. Or alternatively:

$$d(u,v)=\begin{cases} d_X(u,v) &\text{ if } \{u,v\} \subseteq X\\ d_Y(u,v) &\text{ if } \{u,v\} \subseteq Y\\ M & \text{ otherwise}\\ \end{cases}$$