[proofplan] We prove the stopping-time condition directly from the definition. For each integer time $n\geq 0$, the event that the minimum has occurred by time $n$ is the union of the two events that $\tau$ or $\sigma$ has occurred by time $n$. Similarly, the event that the maximum has occurred by time $n$ is the intersection of those two events. Since both events are already in $\mathcal F_n$ and $\mathcal F_n$ is a $\sigma$-algebra, closure under finite unions and intersections gives the result. [/proofplan] [step:Express the minimum event as a finite union of stopping-time events] Fix an integer $n\in\mathbb Z_{\geq 0}$. Define the event \begin{align*} A_n:=\{\omega\in\Omega:(\tau\wedge\sigma)(\omega)\leq n\}. \end{align*} By the pointwise definition of $\tau\wedge\sigma$, for every $\omega\in\Omega$, \begin{align*} (\tau\wedge\sigma)(\omega)\leq n \iff \tau(\omega)\leq n \text{ or } \sigma(\omega)\leq n. \end{align*} Therefore \begin{align*} A_n=\{\omega\in\Omega:\tau(\omega)\leq n\}\cup\{\omega\in\Omega:\sigma(\omega)\leq n\}. \end{align*} Since $\tau$ and $\sigma$ are stopping times, both events on the right-hand side belong to $\mathcal F_n$. Because $\mathcal F_n$ is a $\sigma$-algebra, it is closed under finite unions, so $A_n\in\mathcal F_n$. [guided] Fix an integer $n\in\mathbb Z_{\geq 0}$. To prove that $\tau\wedge\sigma$ is a [stopping time](/page/Stopping%20Time), we must verify the defining condition at this time $n$: the event \begin{align*} A_n:=\{\omega\in\Omega:(\tau\wedge\sigma)(\omega)\leq n\} \end{align*} must lie in $\mathcal F_n$. The pointwise minimum is at most $n$ exactly when at least one of the two values is at most $n$. Thus, for each $\omega\in\Omega$, \begin{align*} (\tau\wedge\sigma)(\omega)\leq n \iff \min\{\tau(\omega),\sigma(\omega)\}\leq n. \end{align*} This is equivalent to \begin{align*} \tau(\omega)\leq n \text{ or } \sigma(\omega)\leq n. \end{align*} Therefore the event $A_n$ can be written as \begin{align*} A_n=\{\omega\in\Omega:\tau(\omega)\leq n\}\cup\{\omega\in\Omega:\sigma(\omega)\leq n\}. \end{align*} Now we use the stopping-time hypotheses. Since $\tau$ is a stopping time, the event $\{\omega\in\Omega:\tau(\omega)\leq n\}$ belongs to $\mathcal F_n$. Since $\sigma$ is a stopping time, the event $\{\omega\in\Omega:\sigma(\omega)\leq n\}$ also belongs to $\mathcal F_n$. Finally, $\mathcal F_n$ is a $\sigma$-algebra, hence it is closed under finite unions. Consequently \begin{align*} A_n\in\mathcal F_n. \end{align*} This verifies the stopping-time condition for $\tau\wedge\sigma$ at the arbitrary time $n$. [/guided] [/step] [step:Express the maximum event as a finite intersection of stopping-time events] Fix an integer $n\in\mathbb Z_{\geq 0}$. Define the event \begin{align*} B_n:=\{\omega\in\Omega:(\tau\vee\sigma)(\omega)\leq n\}. \end{align*} By the pointwise definition of $\tau\vee\sigma$, for every $\omega\in\Omega$, \begin{align*} (\tau\vee\sigma)(\omega)\leq n \iff \tau(\omega)\leq n \text{ and } \sigma(\omega)\leq n. \end{align*} Therefore \begin{align*} B_n=\{\omega\in\Omega:\tau(\omega)\leq n\}\cap\{\omega\in\Omega:\sigma(\omega)\leq n\}. \end{align*} Since $\tau$ and $\sigma$ are stopping times, both events on the right-hand side belong to $\mathcal F_n$. Because $\mathcal F_n$ is a $\sigma$-algebra, it is closed under finite intersections, so $B_n\in\mathcal F_n$. [/step] [step:Conclude the stopping-time property for both closure operations] The integer $n\in\mathbb Z_{\geq 0}$ was arbitrary. The first step shows that \begin{align*} \{\omega\in\Omega:(\tau\wedge\sigma)(\omega)\leq n\}\in\mathcal F_n \end{align*} for every $n\in\mathbb Z_{\geq 0}$, so $\tau\wedge\sigma$ is a stopping time. The second step shows that \begin{align*} \{\omega\in\Omega:(\tau\vee\sigma)(\omega)\leq n\}\in\mathcal F_n \end{align*} for every $n\in\mathbb Z_{\geq 0}$, so $\tau\vee\sigma$ is a stopping time. Thus stopping times with respect to the same discrete-time filtration are closed under pointwise minimum and pointwise maximum. [/step]