Here it is shown that the min of two stopping is times is a stopping time. They do so by saying $\{\min(\sigma,\tau)\leq n\}=\{\sigma\leq n\}\cup \{\tau \leq n\}$ where $\sigma,\tau$ are stopping times. I understand their argument to why $\{\sigma\leq n\}\cup \{\tau \leq n\}$ belongs to the $\sigma$-algebra $\mathcal{F}_n$. But why is it true that $\{\min(\sigma,\tau)\leq n\}=\{\sigma\leq n\}\cup \{\tau \leq n\}$ or for that matter $\{\max(\sigma,\tau)\leq n\}=\{\sigma\leq n\}\cap \{\tau \leq n\}$? I would think the opposite, namely, $\min$ is with $\cap$ and $\max$ is $\cup$.
I guess I don't see the logic or intuition behind that. I'd appreciate any help!
Best Answer
Note that $$\min(x,y) \leq n \iff x \leq n \quad \text{or} \quad y \leq n.\tag{1}$$ Applying this for $x=\sigma(\omega)$ and $y = \tau(\omega)$ proves \begin{align*} \{\min(\sigma,\tau) \leq n\} &= \{\omega \in \Omega; \min(\sigma(\omega),\tau(\omega)) \leq n\} \\ &\stackrel{(1)}{=} \{\omega \in \Omega; \sigma(\omega) \leq n \, \, \text{or} \, \, \tau(\omega) \leq n\} \\ &= \{\sigma \leq n\} \cup \{\tau \leq n\}.\end{align*} The reasoning for $\max$ is similar; use that $$\max(x,y) \leq n \iff x \leq n \quad \text{and} \quad y \leq n.$$