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I want to prove that given a metric space $(M,d)$ and $F \subset M$, then the function $f_F: M \to \Bbb R$ given by $f_F(x) = d(x,F) = \inf\{d(x,y) \ : \ y \in M\}$ is continuous.

Take $x \in M$. If $x \in F$ it is obvious. I am convinced that it is continuous, since if you take a sequence $(x_n)_{n \in \Bbb N}$ which converges to $x \in M$, then the distances from the $x_n$ to $F$ go close to the distance from $x$ to $F$, since the $x_n$ go close to $x$.

But I'm having much trouble writing it. My teacher said to "just forget about the $\inf$ and work with sequences", but that doesn't help me at all.

Given $\epsilon > 0$ and a sequence $(x_n)_{x \in \Bbb N}$ such that $x_n \to x$, I must prove that $f_F(x_n) \to f_F(x)$, that is, find $n_\epsilon \in \Bbb N$ such that: $$n > n_\epsilon \implies |d(x_n,F) - d(x,F)| < \epsilon $$

Surely if $y \in F$, we have $d(x,y) \geq d(x,F)$. Other than this, I don't know how to deal with these infs, and this absolute value. I suppose I'll have to use the triangle inequality. Can someone help me do it?

Ivo Terek
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  • Hi I was bit confuse here , to prove the distance function is continuous,we don't need to condition that set is closed correct? – yi li Oct 17 '20 at 03:40
  • You're right, the condition is not used. Cases of interest end up being when $F$ is compact, so that given $x\in X$ there is a point $y = y(x)\in F$ with $d(x,F)= d(x,y)$ (i.e., the distance is realized by a point in $F$) – Ivo Terek Oct 17 '20 at 03:57

2 Answers2

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From $d(x',y)\leq d(x',x)+d(x,y)$, by taking infimum we get $d(x',F)\leq d(x',x)+d(x,F)$. Similarly, $d(x,F)\leq d(x',x)+d(x',F)$. Consequently, $|d(x',F)-d(x,F)|<d(x',x)$.

Fabio Lucchini
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4

Here is a more worked example without "taking infimum" using the inverse triangle inequality:

We wish to show for all $x, y \in M$, $\lvert f_F(x) - f_X(y)\rvert \le d(x, y)$.

Case: $d(x, z) \ge d(y, z)$.

Fix the $z$ such that both bounds are true. By the definitions of infimum, we have two bounds (let $\epsilon > 0$):

\begin{align} \forall z \in F \quad &f_F(x) \le d(x, z) \\ \exists z \in F \quad &f_F(x) + \epsilon > d(x, z) \end{align}

By the inverse triangle inequality,

\begin{align} & d(x, z) - d(y,z) \le d(x,y) \\ \implies & f_F(x) - d(y, z) \le d(x,y) \\ \implies & f_F(x) - f_F(y) - \epsilon \le d(x,y) \\ \implies & f_F(x) - f_F(y) \le d(x,y) + \epsilon \end{align}

Since this holds for all $\epsilon > 0$, $f_F(x) - f_F(y) \le d(x,y)$. The case of $d(x,z) < d(y,z)$ is similar, so we have proved $\lvert f_F(x) - f_F(y)\rvert \le d(x, y)$. Now simply apply our definition of continuity using $\delta = \epsilon$ (actually $f_F$ is Lipschitz continuous with constant $1$).

qwr
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