Borel-Cantelli Lemma I have some difficulties understanding the following: 
Let $(X_n)$ be a sequence of independent random variables s.t.
$P[X_n=1]=1−P[X_n=0]=\frac{1}{n}$
After using the Borell Cantelli lemma, I could show that:
$P(\limsup [X_n=0])=1$
$P(\limsup [X_n=1])=1$
Now using the fact that $\operatorname{[\limsup F_n]}^c  = [\liminf \operatorname{(F_n)}^c]$ I concluded that 
$P(\liminf [X_n=1])=0$
$P(\liminf [X_n=0])=0$         as $P(A)+P(A^C)=1$.
This implies that as $X_n$ is either $0$ or $1$, $\liminf X_n$ is never constant with probability $1$, contradicting some of the implications of the Kolmogorov $0$-$1$ law. 
Could someone please clarify this?
Much appreciated. 

Consider the events $A_n=[X_n=0]$. By definition, the event $\limsup A_n$ happens if and only if $A_n$ happens for infinitely many $n$, hence the general formula 
$$
\limsup A_n=\bigcap\limits_n\bigcup\limits_{k\geqslant n}A_k.
$$
In our case, consider $N=\{n\mid X_n=0\}$, then 
$$
\limsup A_n=[X_n=0\ \text{for infinitely many}\ n]=[N\ \text{is infinite}].
$$
Consider $Y=\liminf X_n$. This is a random variable $Y$ characterized by the fact that $Y=1$ if $X_n=1$ for every $n$ large enough, that is,
$$
[Y=1]=[\exists n,\ \forall k\geqslant n,\ X_k=1].
$$
and that $[Y=0]=[N\ \text{is infinite}]$.
Finally, for $\{0,1\}$-valued random variables $(X_n)$,
$$
\limsup[X_n=0]=[\liminf X_n=0].
$$
Likewise, again for $\{0,1\}$-valued random variables $(X_n)$,
$$
\liminf[X_n=0]=[(X_n)\ \text{converges and}\ \lim X_n=0]=[\limsup X_n=0],
$$
$$
\liminf[X_n=1]=[(X_n)\ \text{converges and}\ \lim X_n=1]=[\liminf X_n=1],
$$
and
$$
\limsup[X_n=1]=[X_n=1\ \text{for infinitely many}\ n]=[\limsup X_n=1].
$$
