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Let $X$ be a compact metric space. Does there always exist a continuous $f : X \to \mathbb{R}$ such that $\{ x \in X : f(x) = \| f \|_{\infty} \}$ is at most countable?

Certainly this is true for $[0,1]^n$, take $f(x_1,\ldots,x_n) = x_1 + \cdots + x_n$. It seems it should hold in general but I'm not sure how to attempt to prove that.

someone
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    Since $X$ is compact $m:=\max_{x,y \in X} d(x,y)$ exists. Fix any $x_0 \in X$ and set $f(x)=m-d(x,x_0)$. Then $0 \le f \le m$ on $X$ and $f(x)=m$ iff $x=x_0$. Hence ${x \in X: f(x)=|f|_\infty}={x_0}$. – Gerd Oct 23 '23 at 20:27
  • Actually the difficulty is in the "metric space" condition, i.e. defining a distance: that is probably not possible to get separation on sets $X$ that have a greater cardinal than $\mathbb R$. – Jean-Armand Moroni Oct 23 '23 at 21:10
  • It is always possible in a perfectly normal space, regardless of its cardinality. – freakish Oct 23 '23 at 21:38
  • @freakish You are right, I was dumb. There is the trivial distance $d(x,y)=1$ if $x \ne y$, $d(x,y)=0$ if $x = y$. – Jean-Armand Moroni Oct 24 '23 at 09:20

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Take any point $x$ in $X$ and let $f(y) = \frac 1 {1+d(x,y)}$, where $d$ is the distance of metric space $X$.

Then $f$ is continuous, because $d$ is continuous in $y$ (a distance is always continuous).

$f$ reaches its maximum $1$ only on $x$, because of the separation property of a distance.

The compactness condition is not needed.

Jean-Armand Moroni
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