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Let $F \subset R$ be a nonempty closed set and define $g(x) = \inf \{ | x - a | : a \in F \}$. Show that g is continuous on $R$.

I follow a solution, and this is its approach to prove the statement. First, it proves $\forall x \in R$, there exists a number $a_x \in F$ st, $g(x) = | x - a_x |$, and uses the result to state that for a constant $x_0$ and its corresponding $a_0$, $$ | g(x) - g(x_0) | = \inf \{| |x - a| - |x_0 - a_0| | : \forall a \in F \} \space (1) $$

By triangle inequality, it is straightforward to prove, $$|| x - a | - |x_0 - a_0|| \le |x - x_0 | + |a - a_0|$$

The rest of the proof is obvious following definition of continuity.

I got stuck in the equation (1), though I obtained the equation, $$ | g(x) - g(x_0) | = |inf \{ | x - a | : a \in F\} - |x_0 - a_0| |$$

How to prove (1)? I also appreciate if you provide any other solution.

Jesse
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1 Answers1

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The function is actually Lipschitz continuous.

Let $x,y \in \mathbb R$. Since $F \neq \emptyset$ and $d(x,y)=|x-y|$ is a metric on $\mathbb R$, we have that $\inf\{|x-a|: a \in F\}$ exists and is unique. In other words, $g$ is well-defined.

Let $\varepsilon>0$ be given. Then there exists $a_x, a_y \in F$ so that $|x-a_x| < g(x)+\varepsilon$ and $|y-a_y| < g(y)+\varepsilon$. Since $g(x) \leq |x-a|$ and $g(y) \leq |y-a|$ for all $a \in F$, we have $$g(x)-g(y) \leq |x-a_y|-g(y) \leq |x-y|+|y-a_y|-g(y) <|x-y|+\varepsilon$$ and $$g(x)-g(y) \geq g(x)-|y-a_x| \geq g(x)-|y-x|-|x-a_x|>-|y-x|-\varepsilon=-|x-y|-\varepsilon.$$

Thus $|g(x)-g(y)|<|x-y|+\varepsilon$. Letting $\varepsilon \to 0 $, we have $|g(x)-g(y)| \leq |x-y|$.

Now we may observe that we only needed to assume that $F \subset \mathbb R$ is nonempty.

M A Pelto
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