Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
@Renato: Are you talking about the distance function? – anonymous Oct 27 '10 at 16:27
@Renato: Please make your question clear! – anonymous Oct 27 '10 at 16:27
If you are asking whether the distance function $d:X\times X\to\mathbb R$ is continuous when $(X,d)$ is a metric space, then aswer is yes. You should try to prove it yourself; first do it for $\mathbb R$ with its usual metric, and then generalize. – Mariano Suárez-Alvarez Oct 27 '10 at 16:32
@Chandru1 yes, I'm talking about the distance function, sorry for the mistake. – Renato Oct 27 '10 at 16:33
@MarianoSuárez-Alvarez I was also trying the same question. But how to prove it using $\epsilon - \delta$ notation.Let $(x,y) \in X$ and $(x',y') \in X$ then whenever $||(x,y)-(x',y')|| \lt \delta$ then we have $|d(x,y)-d(x',y')| \lt \epsilon$.. right?? How to go ahead with this proof – Mathy Jun 25 '13 at 11:33
up vote 14 down vote accepted

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;task=show_msg;msg=1323.0001

share|cite|improve this answer
where did you use the fact that A is closed? – GAJO Jan 28 '14 at 19:11
Why are we assuming that the metric on $\mathbb{R}$ is the standard one? – The Substitute Aug 31 '14 at 0:19
@GAJO I think $d(x,E)$ is continuous in $x$ for any nonempty subset $E$. – Fang Jing Oct 14 '14 at 20:29

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 $x$.

share|cite|improve this answer
Isn't it also the coarsest one for which $d$ itself is continuous? – Mariano Suárez-Alvarez Oct 27 '10 at 16:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.