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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4Looks 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 $M(x) = \inf (X \cap [x,b]) - x$. – Mateusz Kwaśnicki Jan 13 '21 at 13:42
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1e.g. you can easily fill the holes of the Cantor set with small positive functions, to obtain a smooth function. – Pietro Majer Jan 13 '21 at 16:20
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3Historical 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. – bathalf15320 Jan 13 '21 at 16:23
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See also the questions https://mathoverflow.net/questions/196167/a-result-attributed-to-whitney or https://mathoverflow.net/questions/179445/non-zero-smooth-functions-vanishing-on-a-cantor-set – András Bátkai Jan 22 '21 at 07:06
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As Mateusz Kwaśnicki pointed out in the comments:
Looks like $$ can be an arbitrary closed subset of $[a,b]$: just define $f()=(()())^2$, where $()=−\sup(∩[,])$ and $()=\inf(∩[,])−$.
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.)
Todd Trimble
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