Let $E$ non empty set. Let $\tau_1$ and $\tau_2$ both are $\sigma$-algebra on E, and $A\in\tau$. If $A\in \tau_1$ and $A\in \tau_2$. Prove that $\tau$ is also $\sigma$-algebra.
I have try to answer it.
$\tau_1$ is $\sigma$-algebra, so $\bigcup\limits_{n=1}^\infty P_n\in \tau_1, \forall P_n\in \tau_1$.
$\tau_2$ is $\sigma$-algebra, so $\bigcup\limits_{n=1}^\infty Q_n\in \tau_2, \forall Q_n\in \tau_2$.
Because of $A\in \tau_1$ and $A\in \tau_2$, $A\in \tau_1\cup\tau_2$. $\tau_1\cup\tau_2$ is also $\sigma$-algebra.
Because $A\in \tau$ and $A\in \tau_1\cup\tau_2$, so $\tau$ is also $\sigma$-algebra.
But I'm not sure with my answer. Please anyone correct my answer if wrong.. Thanks