Preface: I'm having trouble with the correct solution.
The Original Question: Given that $\mathscr{F}_t$ is a filtration that satisfies all the usual conditions, and given ${\{\mathscr{F}_t\}_\mathbb{R}}_+$ stopping times $\sigma$ and $\tau$, show that $\{min(\sigma,\tau) \leq t\} \in \mathscr{F}_t$.
The Correct Solution: $\{min(\sigma,\tau) \leq t\} = \{\sigma \leq t\}\cup\{\tau \leq t\} \in \mathscr{F}_t$ by the fact that the filtration is closed under countable unions.
My Issues: I don't understand why $\{min(\sigma,\tau) \leq t\} = \{\sigma \leq t\}\cup\{\tau \leq t\}$.
Consider two sample spaces $\Omega_1$ and $\Omega_2$ with $\omega_1 \in \Omega_1$ and $\omega_2 \in \Omega_2$. Then:
$\{ \sigma \leq t\} = \{\omega_1 : \sigma(\omega_1) \leq t\}$
$\{ \tau \leq t\} = \{\omega_2 : \tau(\omega_2) \leq t\}$
Therefore, the RHS of the solution is: $\{\sigma \leq t\}\cup\{\tau \leq t\} = \{\omega_1 : \sigma(\omega_1) \leq t\} \cup \{\omega_2 : \tau(\omega_2) \leq t\} := RHS$
Now let's expand the LHS of the solution: $\{min(\sigma,\tau) \leq t\} = \{ \omega_1 : \sigma(\omega_1) \leq \tau(\omega_2) \leq t\}\cup\{ \omega_2 : \tau(\omega_2) \leq \sigma(\omega_1) \leq t\}:=LHS $
By this, it's obvious that $LHS \subseteq RHS$, and is not necessarily equal to what's stated in the solution.
MY Question: Where have I gone wrong? I'm newb! I know that my $LHS$ is wrong but I'm not sure how to improve my confusion.