I'd appreciate a second pair of eyes on a proof. I want to prove that a function $f:\mathbb{R}\to\mathbb{R}$ can have at most countably many strict local maxima. The question has been asked elsewhere on Stack Exchange, but my question is about the validity of the following argument, which isn't discussed.
Assume for contradiction that $f$ has uncountably many strict local maxima. For each $n\in\mathbb{N}$, define $$E_n=\Big\{x\in \mathbb{R}: f(x)>f(y) \hspace{2mm}\text{for all $y$ such that $0<|x-y|<\frac{1}{n}$} \Big\}.$$ For example, if $x\in E_3$, then $x$ provides a strict local maximum on at least an open interval of radius $\frac{1}{3}$. If each $E_n$ was at most countable, then $\bigcup^{\infty}_{n=1}E_n$ would be countable as well, contrary to assumption. Hence, some set, say $E_{n_0}$, is uncountable. Being uncountable, the set $E_{n_0}$ has a limit point, $\xi$.
But now this gives a contradiction. Let $\{x_k\}_{k=1}^{\infty}$ be a sequence in $E_{n_0}$ converging to $\xi$. Let $x_i$ and $x_j$ be terms satisfying $|x_i-\xi|<\frac{1}{2n_0}$ and $|x_j-\xi|<\frac{1}{2n_0}$. Then $|x_i-x_j|<\frac{1}{n_0}$ by the triangle inequality. Since $f(x_i)$ is a strict local maximum on an interval about $x_i$ of radius $\frac{1}{n_0}$, we have $f(x_i)>f(x_j)$. But for the same reason we must have $f(x_j)>f(x_i)$, which is a contradiction.
Thanks in advance for your feedback.
Best Answer
The argument is pretty much fine, but what you're really relying on (without mentioning explicitly) is the fact that an uncountable set in $\mathbb{R}$ must contain a convergent sequence. This property is essentially just the fact that $\mathbb{R}$ cannot contain uncountably many disjoint intervals, which gives you a simpler proof. Your set $E_{n_0}$ would be such an uncountable union, giving a contradiction already.