Let $\tau$ and $\nu$ be stopping times. Show that $F_{\tau \land \nu}=F_{\tau} \cap F_{\nu} $ 
Let $\tau $,$ \nu$ be stopping times. Show $F_{\tau \land \nu}=F_{\tau} \cap F_{\nu}$.

Definitons I am using are given by:
$$F_\nu := \{A \in F_{\infty} \mid A \cap \{\nu =n \} \in F_n \forall n\} $$
and
Let $\{F_n\}$ be a filtration on $(\Omega,F)$, then $\tau : \Omega \to \mathbb N \cup \{\infty\}$ is called an $F$ stopping time if for all $n \in \mathbb N_0$ we have $\{\tau=n \} \in F_n.$
My attempt:
\begin{align}
F_{\tau \land \nu}&=\{A \in F_{\infty} \mid A \cap \{\tau \land\nu =n \} \in F_n \forall n\}\\
&=\{A \in F_{\infty} \mid A \cap (\{\tau =n \}\cup\{\nu=n\}) \in F_n \forall n\}\\
&=\{A \in F_{\infty} \mid A \cap \{\tau =n \} \in F_n \forall n\} \cup \{A \in F_{\infty} \mid A \cap \{\nu =n \} \in F_n \forall n\}\\
&= F_{\nu }\cup F_{\tau}.
\end{align}
I know that $ \{\tau \land\nu =n \}= \{\tau=n \}\cup \{\nu=n \}$, but obviously  I am doing something wrong. Any help is much appreciated!
 A: Very often if one should show that two sets are equal it's easier to do it in two steps: 
First show "$\subseteq$" and then "$\supseteq$".
For "$\subseteq$" we take an $A \in F_{\tau \wedge \nu}$ so it's given that $$A \cap \{\tau \wedge \nu = n\} \in F_n$$ for all $n\in \Bbb N$ and we have to show that $A \in F_\tau$ as well as $A \in F_\nu$.
We know that $\tau$ is a stopping time hence $\{\tau = n\},\{\tau \le n\}, \{\tau > n\}$ and $\{\tau \ge n\}$ are in $F_n$
$F_n$ is a $\sigma$-algebra so also the intersection of sets of it are contained in $F_n$, now intersect $\{\tau > n\}$ with $A \cap \{\tau \wedge \nu = n\}$ and simplify (remember: $\tau \wedge \nu$ is the minimum of $\tau$ and $\nu$).
For "$\supseteq$" we assume $$A \cap \{\tau = n\} \in F_n$$
as well as $$A \cap \{\nu = n\} \in F_n$$
Again we know $\{\nu \ge n\}$ and $\{\tau \ge n\}$ are in $F_n$ so intersect the first set with the first expression and the second set with the second expression and unify the resulting sets. 
