Show that $a\to f(a)$ from $A$ to $S$ is continuous.

Question

I am reading "Continuity" in Metric Spaces

Suppose $S\subset \mathbb R$ is a closed set. Suppose $A\subset \mathbb R$ has the property that for every $a\in A$ there is a unique nearest point $f(a)$ of $S$ to $a$. Show that $a\to f(a)$ from $A$ to $S$ is continuous.

Please give some hints on how to proceed. Please dont give complete solution

I don't think that's true, you can definitely always find a 1-1 function from a closed set to another set, that's not continuous. — Gregory Grant, May 12 '15 at 10:28
Is there any relation between $a$ and $f(a)$? How is $f(a)$ defined? — Crostul, May 12 '15 at 10:35
Is this a problem from your book? Which book is it? This statement, exactly as you stated it, is false. — Pedro M., May 12 '15 at 10:35
To make what I said more precise, you can always find a function that's not continuous as long as the domain $A$ does not inherit the discrete topology. — Gregory Grant, May 12 '15 at 10:44

score -1 · Answer 1 · answered May 12 '15 at 23:28

-1

Hint: If $f$ were not continuous, then there would exist a sequence $a_n \to a$ such that $|f(a_n) - f(a)| \geq \epsilon$ for some $\epsilon > 0$. However, at the same time we have $$|f(a) - a| \leq |f(a_n) - a| \leq |f(a) - a_n| \leq |f(a) - a| + |a - a_n|,$$ so that $|f(a_n) - a| \to |f(a) - a|.$

To conclude the argument, take a convergent subsequence of $f(a_n)$ (Question: why does a convergent subsequence exist?).

Can you finish it from here?

answered May 12 '15 at 23:28

Pedro M.

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I will try to finish that and let you know – Learnmore May 13 '15 at 01:37
I have doubts about this (partial) answer. Why "$ |f(a_n) - a| \leq |f(a) - a_n|$"? – Anne Bauval Dec 24 '22 at 09:48

Show that $a\to f(a)$ from $A$ to $S$ is continuous.

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