Why is the set $\{ W_\cdot(\omega)$ is locally $\alpha$-Hölder at some $s\in[0,1]\}$ contained in ths messy intersection/union of measurable sets? Goal
The goal is to prove that for any $\alpha>\frac{1}{2}$, $\mathbb{P}$-almost all Brownian paths are nowhere on $[0,1]$ locally Hölder-continuous of order $\alpha$.
Let $W$ be standard BM, we look at it as a continuous function for a realisation $\omega$: $W_t(\omega)\in C\big([0,1]\big)$.
To do this we let $M\in\mathbb{N}$ such that $M(\alpha-1/2)>1$.
If $W_\cdot(\omega)$ is locally $\alpha$-Hölder at $s\in[0,1]$, then there exists a constant $C$ with $|W_t(\omega)-W_s(\omega)|\leq C|t-s|^\alpha$ for $t$ near $s$.
Then for large enough $n$, $k/n$ near $s$, and $M$ successive indices $k$, it holds $|W_{k/n}(\omega) - W_{(k-1)/n)}(\omega)|\leq C n^{-\alpha}$.
Therefore the set $\{W_\cdot(\omega)$ is locally $\alpha$-Hölder at some $s\in[0,1]\}$ is contained in the following set:
$$
A:=\bigcup_{C\in\mathbb{N}}\bigcup_{m\in\mathbb{N}}\bigcap_{n\geq m}\bigcup_{k=0,\dots,n-M}\bigcap_{j=1}^M \left\{ \big|W_{(k+j)/n}(\omega)-W_{(k+j-1)/n}(\omega)\big|\leq C\frac{1}{n^\alpha}\right\}
$$
We can then go on proving that $\mathbb{P}[A]=0$, hence $\mathbb{P}$-almost all Brownian paths are nowhere locally $\alpha$-Hölder.

Question
How can I convince myself of the inclusion?
The lecturer glanced over it as if it was trivial but I cannot see why.
I can see that the first two unions (over $C$ and $m$) are there to take into account all possible $C$ in the Hölder inequality and all $m$ that are used as lower bound of $n$.
I cannot make sense of the other intersections/unions.
If someone can give me intuition on all the operations I would be incredibly grateful.
 A: Suppose that  $f \in C[0,1]$ is locally Holder continuous of order $\alpha$ at $s \in [0,1/2]$ (the case where $s>1/2$ is similar).
Then there exists $\delta>0$ and $C_1<\infty$ such that $|f(t)-f(s)\le C_1|t-s|^\alpha$ provided that $|t-s|<\delta$. Therefore, for all $n \ge 2M/\delta$, if $k:=\lceil ns \rceil$ and $1 \le j \le M$, then $s \le (k+j)/n < s+\delta$, so
$$|f((k+j)/n)-f(s)| \le C_1(j/n)^\alpha \,.$$
A similar bound holds when we replace $j$ by $j-1$, and the triangle inequality yields
$$|f((k+j)/n)-f((k+j-1)/n)| \le 2C_1 M^\alpha (1/n)^\alpha \,.$$
Thus for $f=W$, we can see that the local Holder property as some $s \in [0,1/2]$ implies that $\omega \in A$ by taking $C=2C_1 M^\alpha $ and $m=2M/\delta$ and $k:=\lceil ns \rceil$.
See the solution to exercise 1.9 (page 363) in [1] for another explanation of the same result.
See the discussion of slow times, theorem 10.30 page 307 in [1], for a considerable refinement due to Davis[2], as well as    Greenwood and Perkins [3].
[1] https://www.yuval-peres-books.com/brownian-motion/ https://people.bath.ac.uk/maspm/book.pdf
[2]   B. DAVIS. On Brownian slow points. Z. Wahrscheinlichkeitstheorie verw. Gebiete 64,
359–367 (1983).
[3]   P. GREENWOOD and E. PERKINS. A conditioned limit theorem for random walk and
Brownian local time on square root boundaries. Ann. Probab. 11, 227–261 (1983).
