show that a distance function is continuous

Question

Let $X$ be a metric space with metric $d$. Define $d: X \times X \to \mathbb{R}$, show that $d$ is continuous.

I would like to show that the function is continuous the topology way (since it is a course on topology).

Let $(a,b)$ be a basic open set in $\mathbb{R}$, then $d^{-1}(a,b) = \{(x,y): a<d(x,y)<b\}$. Define $A= \bigcup_{x \in X} B_{b} (x)$, where $B_b (x) = \{y \in X| d(x,y)<b$}. Clearly $A$ is open. Now define $C = \{(x,y):d(x,y)>a \}.$ Essentially I hope that $A \cap C$ is open. But I am stuck in showing that $C$ is open.

Any help please? :)

Perhaps you can show $C^c$, the complement of $C$ is closed? — Rudy the Reindeer, Feb 26 '13 at 22:13
@MattN. I have not thought about that, but I think that should work too. — Daniel, Feb 26 '13 at 23:09

score 7 · Accepted Answer · answered Feb 26 '13 at 22:17

7

HINT: Suppose that $d(x,y)>a$. Let $r=\frac12\big(d(x,y)-a\big)$. If $\langle u,v\rangle\in B_r(x)\times B_r(y)$, then

$$d(x,y)\le d(x,u)+d(u,v)+d(v,y)\;,$$

so $$d(u,v)\ge d(x,y)-d(x,u)-d(v,y)\;.$$

answered Feb 26 '13 at 22:17

Brian M. Scott

616,228

Thanks to your inequality I think I just solved this :) I am not sure if my solution will be correct, but I think it won't be far away – Daniel Feb 26 '13 at 23:06
@jsk: I’ll keep my fingers crossed. :-) – Brian M. Scott Feb 26 '13 at 23:15
The $r$ of this answer doesn't address that $d(u,v)$ must be less than $b$. – JustAskin Nov 12 '14 at 21:09
@Justaskin_: That's intentional: it's a hint. – Brian M. Scott Nov 12 '14 at 21:12
I realized that just now but couldn't edit my comment (strict time limit!). Good hinty answer in that case :) – JustAskin Nov 12 '14 at 21:18

score 4 · Answer 2 · answered Feb 26 '13 at 22:17

To show that $d^{-1}(a,b)$ is open, you need to specify for each point $(x,y)$ in it a basic open set of the product topology (yes, we should use properties of the product topology somewhere), i.e. a set of the form $U\times V$ with $u,V$ open, $x\in U$, $y\in V$. As $X$ is a metric space, we may try open balls $U=B_r(x)$, $V=B_r(y)$ for suitable $r$. How can we choose $r>0$ to enforce $B_r(x)\times B_r(y)\subseteq d^{-1}(a,b)$? (Simply translate what this means) You will need (alas!) the defining properties of metric for this.

Thanks for the reminding of the product topology! – Daniel Feb 26 '13 at 23:04 — Daniel, Feb 26 '13 at 23:04

show that a distance function is continuous

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