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

Let $(X,d)$ be a metric space. How to prove that for any closed $A$ a function $d(x,A)$ is continuous - I know that it is even Lipschitz continuous, but I have a problem with the proof: $$ |d(x,a) - d(y,a)| \leq d(x,y) $$ for any $a\in A$ - but we cannot just replace it by $|d(x,A) - d(y,A)|\leq d(x,y)$ since the minimum (or infimum in general) can be attained in different points $a\in A$ for $x$ and $y$, so we only have that $$ |d(x,A)-d(y,A)|\leq d(x,y)+\sup\limits_{a,b\in A}d(a,b) $$ which does not mean continuity.

share|cite|improve this question
I just proved this in this answer a few hours ago, see point 1. in my answer there. – t.b. Jul 1 '11 at 7:31
thanks, voted there also ) – Ilya Jul 1 '11 at 7:36
up vote 19 down vote accepted

If $A = X$ (or if $\overline{A} = X$) then $d(\cdot,A) = 0$ and there's nothing to prove. If $A$ is empty, the usual conventions in the infimum yield $d(\cdot,\emptyset) = \infty$ and Lipschitz continuity doesn't really make sense. However a constant function is certainly continuous.

So, assume that $\emptyset \neq A$ and let $\varepsilon \gt 0$. Choose $a\in A$ such that $d(x,a) \leq d(x,A) + \varepsilon$. Then the triangle inequality yields $d(y,A) - d(x,A) \leq d(y,a) - d(x,a) + \varepsilon \leq d(y,x) + \varepsilon$. By symmetry we get $|d(x,A) - d(y,A)| \leq d(x,y) + \varepsilon$ and the desired result follows because $\varepsilon$ was arbitrary. Note that I didn't use that $A$ is closed. Update: Alternatively, you can choose Zarrax's way and permute two steps in this paragraph and make the argument nicer by getting rid of the explicit mentioning of $\varepsilon$ or appeal to the valuable general result mentioned by Didier.

If $A$ is non-empty and not dense, i.e., $\overline{A} \neq X$ then $1$ is in fact the best Lipschitz constant. Indeed, there are $x$ and $r\gt0$ such that $B_r(x) \cap \overline{A} = \emptyset$, so $d(x,A) \geq r \gt 0$. For every $\varepsilon \gt 0$ we can find $a \in A$ such that $d(x,a) \leq (1+\varepsilon)d(x,A)$. But then $|d(a,A) - d(x,A)| = d(x,A)\geq \frac{1}{1+\varepsilon} d(x,a)$ and the claim follows.

For a closely related thread, see here.

share|cite|improve this answer

If $a$ is in $A$, by the triangle inequality you have $$d(x,a) \leq d(x,y) + d(y,a)$$ The left hand side is at least $d(x,A)$, so we get $$d(x,A) \leq d(x,y) + d(y,a)$$ Take the infinum of this over all $a$ in $A$ and you obtain $$d(x,A) \leq d(x,y) + d(y,A)$$ Which is the same as $$d(x,A) - d(y,A) \leq d(x,y)$$ Reversing the roles of $x$ and $y$ you have $$d(y,A) - d(x,A) \leq d(x,y)$$ Combining the last two equations gives $$|d(x,A) - d(y,A)| \leq d(x,y)$$ Notice $A$ doesn't even have to be closed for this to work.

share|cite|improve this answer

This is a special case of the following general situation.

For any nonempty set $T$, consider a collection $(u_t)$ of real valued functions defined on $X$ and indexed by $t$ in $T$. Assume that each $u_t$ is $1$-Lipschitz, that is $|u_t(x)-u_t(y)|\le d(x,y)$ for every $x$ and $y$ in $X$ and every $t$ in $T$. Let $v=\inf_{t\in T}u_t$ the function defined for every $x$ in $X$ by $$v(x)=\inf\{u_t(x);t\in T\}.$$ Then:

Either $v=-\infty$ everywhere or $v$ is finite everywhere and $1$-Lipschitz.

To prove this, start from the inequalities $v(x)\le u_t(x)\le u_t(y)+d(x,y)$ for every $x$ and $y$ in $X$ and every $t$ in $T$, and follow the lights...

In your setting one can choose $T=A$ and $u_a=d(a,\ )$ for every $a$ in $A$. Then each $u_a$ is nonnegative and $1$-Lipschitz, hence the function $v=\inf_{a\in A}u_a=d(A,\ )$ is finite everywhere and $1$-Lipschitz.

Note that $A$ can be any nonempty subset of $X$.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.