Distance between a point and the diagonal set.

Question

Problem:

Given a metric space $(M,d)$ and the diagonal set $\Delta:= \{(x,x)|x\in M\}$. Show that $$z = (x,y)\notin \Delta \implies d'(z, \Delta)>0,$$

where $d'$ is a metric associated with $M\times M$.

My thoughts:

My idea is to look for a term $c$, probably dependent of $d(x,y)$, where $d'(z, \Delta)>c>0$. How to find that $c$? There's another way to solve this problem?

Answer 1

$\Delta$ is a closed set because every metric space is Hausdorff.

Thus if $x_0 \in \Delta^c$ then exists $R>0$ such that $B(x_0,R) \cap \Delta=\emptyset \Longrightarrow B(x_0,R) \subset \Delta^c \Longrightarrow d'(x,y) \geq R ,\forall y \in \Delta \Longrightarrow d'(x,\Delta) \geq R$

Also note that:

$\textbf{Lemma}$ In a general metric space,if $A$ is closed and $B$ is compact then $\exists C>0$ such that $d(x,y) \geq C,\forall x \in A, \forall y \in B$

From this, with $A=\Delta$ and $B=\{x_0\}$ you have again the desired conclusion.

You can prove the lemma i mentioned as an exercise,if you want.

Answer 2

A direct proof, without appeal to general topology (I'm assuming we're using the supremum distance, i.e. $d'((a,b),(c,d))=\max\{d(a,c),d(b,d)\}$):

Let $r=d(x,y)$ and consider $B:=B(z,\frac{r}{3})$ in $M\times M$. We claim that $B\cap \Delta=\emptyset$.

Indeed, if $(w,w)\in B,$ then both $d(w,x)<\frac{r}{3}$ and $d(w,y)<\frac{r}{3}$. In particular, $d(x,y)\leq d'((x,x),z)\leq d'((x,x),(w,w))+d'((w,w),z)< \frac{2r}{3},$ which is a contradiction.