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 and $Y\subset X$ be a non-empty subset. Is the map given by$$f(x)=\inf\lbrace d(x,y)\colon y\in Y\rbrace$$ a Lipschitz map? And does the equivalence $f(x)=0\iff x\in$ closure$(Y)$ hold?

Thank you in advance.

share|cite|improve this question
yes it is Lipschitz and yes the equivalence holds for $\Leftarrow$ use continuity and for $\Rightarrow$ use the definition of closure by sequences. – t.b. Mar 19 '12 at 21:06

Let $x_1,x_2\in X$, and $y\in Y$. We have $d(x_1,y)\leq d(x_1,x_2)+d(x_2,y)$ and taking the infimum over $y\in Y$ we get $f(x_1)\leq d(x_1,x_2)+f(x_2)$ so $f(x_1)-f(x_2)\leq d(x_1,x_2)$ and by symmetry $|f(x_1)-f(x_2)|\leq d(x_1,x_2)$ so $f$ is $1$-Lipschitz.

If $x$ is in the closure of $Y$ then $\{y,d(x,y)<n^{-1}\}$ is a neighborhood of $x$ so $\{y,d(x,y)<n^{-1}\}\cap Y\neq\emptyset$. Let $y_n\in Y$ such that $d(x,y_n)\leq n^{-1}$. We get $f(x)\leq n^{-1}$ for all $n$ hence $f(x)=0$. Conversely, if $f(x)=0$, for each $n$ we can find $y_n\in Y$ such that $d(x,y_n)\leq n^{-1}$, which shows that $\{y,d(x,y)<n^{-1}\}\cap Y\neq\emptyset$ and $x$ is the closure of $Y$.

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.