If an event $\{X_n = 1\}$ happens infinitely often with probability 1, why can't $X_n \overset{a.s.}\to 1$? I saw an example in a book where $X_n \sim^{iid} Bern(\frac{1}{n})$. The book claims that since $\sum_{n=1}^{\infty}P(\{X_n = 1\}) = \infty$, the event $\{X_n = 1\}$ happens infinitely often with probability 1 (by Borel-Cantelli). Why can't $X_n \overset{a.s.}\to 1$? It seems very obvious to me but I cannot get it to work by the book's definition of almost sure convergence, that $X_n \overset{a.s.}\to X$ means:
$$
P(\{\omega \in \Omega: \lim_{n \to \infty}X_n(\omega) = X(\omega)\})=1.
$$
Is there a way to show this using this very definition of almost sure convergence? Thanks!
 A: Assuming the $X_n$ are independent, then it follows from the second Borel-Cantelli lemma that
$$\mathbb P\left(\limsup_{n\to\infty} \{X_n=1\}\right)=1. $$
(See for example here for a proof of the Borel-Canelli lemmas.) However,
$$\mathbb P\left(\liminf_{n\to\infty}\{X_n=1\} \right) = \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{X_k=1\}\right)=0, $$
since for any $n$,
$$\mathbb P\left(\bigcap_{k=n}^\infty \{X_k=1\} \right)=\prod_{k=n}^\infty\mathbb P(X_k=1)=\prod_{k=n}^\infty\frac1k=0 $$
and hence
$$P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{X_k=1\}\right)\leqslant \sum_{n=1}^\infty\mathbb P\left(\bigcap_{k=n}^\infty \{X_k=1\}\right)=0. $$
Since $$\mathbb P\left(\liminf_{n\to\infty} \{X_n=1\}\right) \ne \mathbb P\left(\limsup_{n\to\infty}\{X_n=1\}\right), $$
it is clear that
$$\mathbb P\left(\lim_{n\to\infty}\{X_n=1\}\right) $$
does not exist, and so $X_n$ does not converge almost surely to $1$.
By similar computations we find that
$$\mathbb P\left(\liminf_{n\to\infty}\{X_n=0\}\right)=0<1=P\left(\limsup_{n\to\infty}\{X_n=0\}\right), $$
so $X_n$ does not converge almost surely to $0$.
Footnote: Recall that

$$X_n\stackrel{\mathrm{a.s.}}\longrightarrow X\iff \mathbb P\left(\liminf_{n\to\infty} \{|X_n-X|<\varepsilon\}\right)=1$$
for all $\varepsilon>0$.

Since in this case $\mathbb P(X_n\in\{0,1\})=1$ for all $n$, taking $\varepsilon<\frac12$ justifies the use of e.g. $\limsup_{n\to\infty}\{X_n=1\}$ as opposed to $\{\limsup_{n\to\infty} X_n\}=1$.
A: By the second Borel-Cantelli lemma we have that $X_n=1$ infinitely often with probability 1, as the book states.
The definition given says that convergence occurs a.s. if the probability that the limit equals $0$ is $1$. This means that the for every $\epsilon$, there exists an $N$ such that $n>N$ implies $|X_n(\omega)|<\epsilon$ for a.e. $\omega$.
But we know that $X_n(\omega)=1$ infinitely often for a.e. $\omega$, and hence you cannot find any such $N$. So in fact the relevant probability is $0$ and not $1$, and so the a.s. convergence to $0$ does not occur.
Since $\sum_{n=1}^\infty(1-\frac1n) = \infty$ also, this means that $X_n=0$ infinitely often as well. Then you can repeat the above argument to conclude that $X_n$ does not converge a.s. to $1$.
A: $$\liminf X_n$$
$$ = \sup [\inf_{n \ge 1} X_n, \inf_{n \ge 2} X_n, ... ]$$
$$ = \sup [0, 0, ... ] = 0$$
$$\limsup X_n$$
$$ = \inf [\sup_{n \ge 1} X_n, \sup_{n \ge 2} X_n, ... ]$$
$$ = \inf [1, 1, ... ] = 1$$
Since
$$\limsup X_n \ne \liminf X_n,$$
$\lim X_n$ does not exist.

Note that the reason we say that $\inf_{n \ge m} X_n = 0 \ \forall m \ge 1$ is $X_n = 0 \ \text{i.o.}$ because $X_n = 0 \ \text{i.o.}$ means that $\forall m \ge 1, \exists n \ge m$ s.t. $X_n = 0$.
///ly, the reason we say that $\sup_{n \ge m} X_n = 1 \ \forall m \ge 1$ is $X_n = 1 \ \text{i.o.}$ because $X_n = 1 \ \text{i.o.}$ means that $\forall m \ge 1, \exists n \ge m$ s.t. $X_n = 1$.
