Doubt about the equivalence of a particular statement Regarding the second line of the first proof by @JDL in Alternate definitions of almost sure convergence
Are the following two statements equivalent?

$P(\omega:\exists n\in\mathbf{N}, \forall m\in\mathbf{N}, \exists i>m \,\,\,  \text{s.t.} \,\, |X_i(\omega) - X(\omega)| < 1/n)$

AND 

$P(\omega:\forall n\in\mathbf{N}, \exists m\in\mathbf{N},\  \text{s.t.}\  \forall i>m  \,\, |X_i(\omega) - X(\omega)| \ge 1/n)$

 A: The two sets in question don't seem to be equal in general.  Take the counterexample $X(\omega) = 0$ for all $\omega \in \Omega$ and $X_i(\omega) = 1/2$ for all $\omega \in \Omega$ and $i \in \mathbb{N}$.  Then the first set is actually all of $\Omega$ because with $n = 1$ every $m$ and $i > m$ satisfy the inequality.  But for the second set if we again look at $n = 1$ there is no $m$ that could make the rest of the statement true, and so this set is in fact empty.
Edit
The poster is trying to prove that $X_n \to X$ a.s. if and only if $\lim_{n \to \infty} P(\sup_{k \geq n} |X_k - X| > \epsilon) = 0$ for $\epsilon > 0$, but there are much simpler proofs.  Consider the following definition of almost sure convergence:
$$
P(\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}) = 0
$$
for each $\epsilon > 0$.  The idea here is that $\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}$ is the set on which $|X_n - X| > \epsilon$ occurs for infinitely-many $n$.  To see this, note that any $\omega$ for which $|X_n(\omega) - X(\omega)| > \epsilon$ occurs infinitely often will be in every union $\cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}$ and therefore will also be in the infinite intersection $\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}$.  If on the other hand $|X_n(\omega) - X(\omega)| > \epsilon$ only happens for finitely-many $n$, then eventually such an $\omega$ will no longer be in $\cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}$ and hence will not be in $\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}$ either.
If $X_n \stackrel{\text{a.s.}}{\to} X$ then these sets will have zero probability for all $\epsilon$, whereas if $X_n$ does not converge to $X$ then for some $\epsilon > 0$ and $\omega \in A_\epsilon$ with $P(A_\epsilon) > 0$ we will have $|X_n(\omega) - X(\omega)| > \epsilon$ infinitely often and hence $P(\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}) > 0$ for such an $\epsilon$.
Now observe that the sets $\cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}$ are monotone decreasing in $n$ and so we can rewrite the intersection as a limit
$$
\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \} = \lim_{n \to \infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \} .
$$
But the set $\cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}$ is equal to $\{ \sup_{k \geq n} |X_k - X| > \epsilon \}$ and since we can take the limit outside the probability (use the dominated convergence theorem if you like) we end up with
$$
P(\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} \{ |X_k - X| > \epsilon \}) = \lim_{n \to \infty} P(\sup_{k \geq n} |X_k - X| > \epsilon)
$$
and we are done.
