# Can I prove that Cardinality of the set of all maximum point (for any function $\mathfrak{f}$) is countable or finite by Reductio ad absurdum

I will write my proof in short words: If I write that ths set of maximum points is uncountable so if I have a Polynomial function of n degree so the Polynomial function derivative could have a root with a transcendental number and that is an absurdity. So this is way the set of all maximum point is countable or finite. It is a correct proof ? Thanks

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With maximumum points, you mean points where the local maximum is attained? W.r.t. correctness of the proof: why do you consider a polynomial of degree $n$ only if you are interested in the general case? – Ilya Aug 29 '12 at 8:01
What kind of functions are you talking about? Where do polynomials come in? If you mean by maximum point the highest value archived by a function, there can be at most one maximum point. If you mean the set of points where a maximum is archived, a constant function on the reals has an uncountable number of such points. – Michael Greinecker Aug 29 '12 at 8:01
The set of all maximum points(for any function $\mathfrak{f}$) is either finite or countable. This is what I want to prove. – Hernan Aug 29 '12 at 8:16
Ilya: I limited to myself to a Polynomial – Hernan Aug 29 '12 at 8:18
Transcendency is not relevant. Anyway, $x^2-\pi^2$ has transcendental roots. But it is true that if you have a polynomial $P(x)$ of degree $\ge 1$, then $P'(x)$ will only have finitely many roots. So a polynomial of degree $\ge 1$ has only finitely many local maxima. I don't really hink the problem is about polynomials. – André Nicolas Aug 29 '12 at 8:20

## 1 Answer

Here is an example of a nonconstant continuous function with uncountably many local minima. It's construction follows the construction of the Cantor set.

Begin with $f_0(x):=0$ $\,(0\leq x\leq 1)$. Then replace $f$ in the middle third of $[0,1]$ by a parabolic arc: $$f_1(x):=\cases{\Bigl({2\over3}-x\Bigr)\Bigl(x-{1\over3}\Bigr) &\quad\Bigl({1\over3}<x<{2\over3}\Bigr) \cr f_0(x)&\quad (else)\cr}\qquad,$$ and repeat this middle third replacement ad infinitum.

The resulting function $f$ has value $0$ at all points of the Cantor set (which is uncountable) and is positive otherwise. In any neighborhood of a point $x_0$ where $f$ takes a local minimum there are points where the function is actually greater than at $x_0$.

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On the other hand, it seems that none of these points are strict local minima. – Henning Makholm Aug 29 '12 at 12:58