Is the function distance continuous?

Question

And if not, what are the most common hypotheses (on open sets it induces, for example) under which distance is a continuous function?

Answer 1

As Qiaochu points out $d(x,y)$ is continuous for fixed $x$. You may like to see this as well, as this is a familiar result in Topology.

• If $A$ is a non empty closed subset of a metric space $(X,d)$ then show that the function f on X given by $f(x)=d(x,A)$ is continuous.

Because $$| f(x) - f(y) | = | d(x,A) - d(y,A) | \leq d(x,y)$$.

This means that $f$ is uniformly continuous (use $\delta = \epsilon$ in any point)

Let $x$ and $y$ be points in $X$, and $p$ any point of $A$.

Then $d(x,p) \leq d(x,y) + d(y,p)$ (triangle inequality) and so $d(x,A) \leq d(x,y) + d(y,p)$ (as $d(x,A)$ is the infimum). But then $d(y,p) \geq d(x,A) - d(x,y)$ (for all $p$, obtained by subtracting from the previous ןinequality) so that $d(y,A) \geq d(x,A) - d(x,y)$ (as $d(y,A)$ is the infimum). So : $d(x,A) - d(y,A) \leq d(x,y)$.

Now reverse the roles of $x$ and $y$ to get $d(y,A) - d(x,A) \leq d(x,y)$.

This is taken from http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2004;task=show_msg;msg=1323.0001

Answer 2

Yes. The standard definition of the topology induced by a metric ensures this; in fact it's not hard to see that it's the coarsest topology such that $d(x, y)$ is continuous for fixed $x$.