While reading on almost sure convergence, I came across this equivalence: $$\{\omega \in \Omega: \lim_n X_n(\omega) = X(\omega)\} \equiv \bigcap\limits_{k=1}^{\infty}\bigcup\limits_{N=1}^{\infty} \bigcap\limits_{n\ge N}^{\infty} \{\omega \in\Omega : |X_n(\omega) - X(\omega)| \le \frac{1}{k}\}.$$ The set on the l.h.s is defined point-wise: fix a $\omega \in \Omega$; if for this "$\omega$" it's the case that $X_n(\omega) \rightarrow X(w)$ as $n \rightarrow \infty$, then include that $\omega$ in the l.h.s set.
The set on the r.h.s, on the other hand, is seemingly defined differently: for a fixed $n\ge N$ and a fixed $k$, find all $\omega$'s for which $|X_n(\omega) - X(\omega)| \le \frac{1}{k}$, then apply the countable union and intersection on these sets generated by varying $k, N$, and $n$.
Is it possible to have an intuitive understanding of this equivalence?
Using the $(\epsilon,\delta)$-definition of limits, $\lim_n X_n(\omega) = X(\omega)$ is equivalent to the following statement: For any $k \in N$ there exists a $N \in \mathbb{N}$ such that \begin{align*} \lvert X_n(\omega)-X(\omega) \rvert \leq \frac{1}{k} \end{align*} for all $n \geq N$. The RHS is exactly the subset of $\Omega$ satisfying this requirement. Getting used to reading $\cap$ as 'for any' and $\cup$ as 'there exists' in these kinds of situations can prove useful.