Let $f(n)$ be the number of subsets of $\{1, 2, \ldots n\}$ containing no two consecutive numbers. Then take such a subset of $\{1, 2, \ldots n+2\}$. Either it doesn't contain $n+2$, in which case it is such a subset for $n+1$ and is counted once in $f(n+1)$, or it contains $n+2$, in which case removing $n+2$ leaves such a subset of $n$, and is hence sounted in $f(n)$. So $f(n+2)=f(n+1)+f(n)$. $f(1)=2$, $f(2)=3$ so $f(n)=F_{n+2}$ and $f(11)=F_{13}=233$.