Perhaps this is a very silly question, and I'm probably just confusing some really obvious property, but suppose that I have a sequence of independent Bernoulli random variables with probability $1/n$. Now, by Borel-Cantelli, I have that $P(\lim\sup X_n =1) = 1$, since $$\sum^\infty_{i=1}P(X_i =1) = \sum^\infty_{i=1}\frac{1}{n} = \infty$$
But, by the same logic, I would have $P(\lim\sup X_n = 0) = 1$. What am I confusing here?
As commented, the problem is that $P(\limsup\{X_n=1\})$ and $P((\limsup X_n)=1)$ are totally different things.
For the example, choose random variables $A_n$ with $$P(\limsup\{A_n=1\})=1$$ and $B_n$ with $$P(\limsup\{B_n=0\})=1,$$and let $$X_n=\begin{cases}A_k,&(n=2k),\\ B_k,&(n=2k+1).\end{cases}$$
Note that $\limsup\{A_n=1\}\subset\limsup\{X_n=1\}$ and $\limsup\{B_n=0\}\subset\limsup\{X_n=0\}$.