Prove that $f(x)=d(x,A)=\inf_{y\in A}d(x,y)$ is continuous on $M$

Question

Let $(M,d)$ be a metric space, $A\subset M$. Prove that $$f(x)=d(x,A)=\inf_{y\in A}d(x,y)$$ is continuous on $M$.

I have tried the following

Let $\epsilon>0$, $x_0\in M$. I have to find $\delta>0$ such that if $d(x,x_0)<\delta$, then $|f(x)-f(x_0)|<\epsilon$.

For that,

$$|f(x)-f(x_0)|<\epsilon\iff \big|\inf_{y\in A}d(x,y)-\inf_{y\in A}d(y,x_0)\big|<\epsilon \iff \big|\inf_{y\in A}\{d(x,y)-d(y,x_0)\}\big|<\epsilon$$ and I do not know how to continue.

Note: My teacher has not defined Lipschitz continuity.

@T'x no, but sup{A+B}=sup{A}+sup{B} and I suposse that inf{A-B}=inf{A}-inf{B} — Marco García, Mar 13 '16 at 17:22
no, i think you will easily find a counterexample for the inf's — T'x, Mar 13 '16 at 17:31
See also Continuity of the function $x\mapsto d(x,A)$ on a metric space — Martin Sleziak, Feb 26 '17 at 08:30

Henricus V. · Answer 1 · 2016-03-13T17:59:48.457

2

Let $y \in A$. By the triangular inequality, $$ d(x,A) \leq d(x,y) \leq d(x,x_0) + d(x_0,y) $$ Therefore, $d(x,A) - d(x,x_0) \leq d(x_0,y)$. From this, you can conclude $$ d(x,A) - d(x,x_0) \leq d(x_0,A) $$ $$ d(x,A) - d(x_0,A) \leq d(x,x_0) $$ Similarly $d(x_0,A) - d(x,A) \leq d(x,x_0)$. Taking absolute value, $$ |d(x,A) - d(x_0,A)| \leq d(x,x_0) $$ It is sometimes quite difficult to do algebra with $\inf$ and $\sup$. Understanding the fact that they are the lower and upper bounds will help.

edited Mar 13 '16 at 17:59

answered Mar 13 '16 at 17:18

Henricus V.

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It is not clear that $d(x,A)-d(x,x_0)\leq d(x_0,A)$ – Marco García Mar 13 '16 at 23:24
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The equation $d(x,A) - d(x,x_0) \leq d(x_0,y)$ holds for all $y \in A$. Thus it holds for the infimum. – Henricus V. Mar 13 '16 at 23:33

Prove that $f(x)=d(x,A)=\inf_{y\in A}d(x,y)$ is continuous on $M$

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