For $\omega\in\Omega$, define $S(\omega):=\(Y_n(\omega),n\in\mathbb N\)$. We have to show that $P(\omega\mid S(\omega)\mbox{ is relatively compact})=1$.

Considering $\varepsilon=1/j$, we can see that there is $\Omega'$ of probability $1$ such that for each $\omega\in \Omega'$,

- $\sup_n\lVert Y_n(\omega)\rVert<\infty$, and
- for each integer $j$, $\limsup_{n\to +\infty}d(Y_n(\omega),F_j)\leqslant \frac 1j$, where $F_j$ corresponds to the subspace for $\varepsilon=\frac 1j$.

We say that a set $S\subset B$ is *flately concentrated* if for each $
\varepsilon>0$, we can find a finite dimensional subspace $F$ of $B$ such that for all $x\in S$, $d(x,F)<\varepsilon$.

It's linked to relative compactness (using completeness) by the following:

A subset $S$ of the Banach space $B$ has a compact closure if and only $S$ is bounded and flately concentrated .

To conclude here, we have to check that $S(\omega)$ is flately concentrated for each $\omega\in\Omega'$. Argue by contradiction (the problem will come from *infinitely many* $n$).