Uncountably Many Maxima?

• Given $\delta > 0$ , define $S_{\delta}$ to be the points $x$ such that everything in a $\delta$ -neighborhood of $x$ is strictly smaller than $x$ .

$S_{\delta} = \left \{ x \in \mathbb{R} \mid \forall y \in B_{\delta}(x) : f(y) < f(x) \right \}$

• Now, notice that given any $\delta$ , we could divide up the number line into $\delta$ intervals, and there would be at most one such $\delta$ -maxima in each interval. To be precise, for each $k \in \mathbb{Z}$ , there exists at most one point $x \in [k\frac{\delta}{2},(k+1)\frac{\delta}{2}]$ such that $x \in S_\delta$ . We conclude that, for any $\delta > 0$ , $S_{\delta}$ is atmost countable.

• Notice that if $\delta_1 > \delta_2$ then $x \in S_{\delta_1} \implies x \in S_{\delta_2}$ . Hence, one could express $S$ as a countable union of decreasing $\delta$ 's.

$\bigcup_{n=1}^\infty {S_{\frac{1}{n}}}$

• Because a countable union of countable sets is countable, we can say that $S$ is atmost countable.

---

Inspired by Stack Exchange

I don't actually have a complete solution to this yet. My idea is to pick the neighborhoods for each $x$ in such a way so that no two neighborhoods intersect. If we can do this, then we can map each $x \in S$ to a rational number in $N_x$ ; the map is injective, so $|S| \le |\mathbb{Q}| = \aleph_0$ , proving $S$ cannot be uncountable. Of course, this approach might be entirely incorrect.