Let $a < b$ two real numbers and let $f \colon [a,b] \to \mathbb{R}$ a $C^1$ function. Moreover, we consider the set $$ X := \{ x \in [a,b]\mid f(x) = 0 \}. $$ Is it the number of connected components of $X$, at maximum, countable?
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4$\begingroup$ Looks like $X$ can be an arbitrary closed subset of $[a,b]$: just define $f(x) = (m(x)M(x))^2$, where $m(x) = x - \sup (X \cap [a, x])$ and $M(x) = \inf (X \cap [x,b]) - x$. $\endgroup$– Mateusz KwaśnickiCommented Jan 13, 2021 at 13:42
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1$\begingroup$ e.g. you can easily fill the holes of the Cantor set with small positive functions, to obtain a smooth function. $\endgroup$– Pietro MajerCommented Jan 13, 2021 at 16:20
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3$\begingroup$ Historical note: it is an old and celebrated result of Whitney that any closed subset of a smooth manifold is the zero set of an infinitely differentiable function. $\endgroup$– bathalf15320Commented Jan 13, 2021 at 16:23
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$\begingroup$ See also the questions mathoverflow.net/questions/196167/… or mathoverflow.net/questions/179445/… $\endgroup$– András BátkaiCommented Jan 22, 2021 at 7:06
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1 Answer
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As Mateusz KwaΕnicki pointed out in the comments:
Looks like $π$ can be an arbitrary closed subset of $[a,b]$: just define $f(π₯)=(π(π₯)π(π₯))^2$, where $π(π₯)=π₯β\sup(πβ©[π,π₯])$ and $π(π₯)=\inf(πβ©[π₯,π])βπ₯$.
This takes the simpler function $f(x) = \inf \{ |x-y| | y \in X\}$ and fixes the points of non-differentiability (at the boundary of $X$ and at the midpoint of each open interval between two points of $X$).
We can even make it $C^{\infty}$, by taking $f(x) = e^{ - \frac{1}{m(x)M(x)}}$ for $x\notin X$.
We can take $X$ to be a Cantor set, which has uncountably many connected components (every point is its own connected component).
(Note that the complement of $X$ is open, and thus has countably many connected components, since each one contains a rational number.)