I am reading a proof of the following result : $$X_n \, \xrightarrow{\mathrm{a.s.}} \, X. \iff \forall \varepsilon > 0, \lim_{n \rightarrow + \infty } \mathbb{P} ( \sup_{k \geq n} |X_k - X | \geq \varepsilon ) = 0 $$
to prove $\implies$, we fix a $\varepsilon$ and we define $A_n := \{ \sup_{k \geq n} |X_k - X | \geq \varepsilon \} $ , $C := \{ \omega, \lim X_n (\omega ) = X( \omega) \} $ and $B_n := C \cap A_n$ The proof states that $B_{n+1} \subset B_n$ and this is true because of the construction of $A_n$ and $\mathbb{P}(C) = 1$ which is due to the almost convergence.
Afterwards, the proof stats that $\cap_{n} B_n = \varnothing $ and I don't understand why. Can someone help me understand this point ?
Thank you for your help
If $X_n(\omega) \to X(\omega) $ then $\sup_{k \geq n} |x_n(\omega)-X(\omega)| <\epsilon$ for some $n$ . For that $n$, $\omega \notin A_n$ hence not in $\cap_k A_k$.