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
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
0
$\begingroup$
elementary-set-theory
proof-writing
-
0I think that if you take the open environment of any local extremum point you can find an Injective correspondence with any two rational point one from the left and one from the right. I think that is another way to prove but I am not shure. – 2012-08-29
1 Answers
4
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}
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 $x_0$.
-
0On the other hand, it seems that none of these points are _strict_ local minima. – 2012-08-29