A poset $(X,\le) $ is atomic if it has both a smallest and largest element, it is graded ,and every element $x$ of $X$ is the join $x_1\vee \dots\vee x_n$ of some elements of $X$ (also written as $x=\sup\lbrace x_1 ,\ldots x_n\rbrace$). In a product $P\times Q$ of atomic posets $(P,\le)(Q\le)$ is $(p_1\vee\dots\vee p_n,q_1\vee\dots \vee q_m)=(p_1, q_1\vee\dots \vee q_m)\vee\dots \vee(p_n,q_1\vee\dots \vee q_m)=\bigvee _{1\le i\le n;1\le j\le m}(p_i,q_j)?$
I am trying to prove that $P\times Q$ is atomic if $P$ and $Q$ are atomic and the question of equality above is where I get stuck. If this is true can I also do a similar calculation when replacing $\vee$ with $\wedge$?
Let $[n]=\{1,\dots,n\}$, $[m]=\{1,\dots,m\}$, $p=\bigvee_{i=1}^np_i$ and $q=\bigvee_{i=1}^mq_i$. Clearly $\langle p_i,q_j\rangle\le\langle p,q\rangle$ for each $\langle i,j\rangle\in[n]\times[m]$. Now suppose that $\langle p',q'\rangle\in P\times Q$ and $\langle p_i,q_j\rangle\le\langle p',q'\rangle$ for each $\langle i,j\rangle\in[n]\times[m]$. Then $p_i\le p'$ for each $i\in[n]$, and $q_j\le q'$ for each $j\in[m]$, so $p\le p'$, $q\le q'$, and $\langle p,q\rangle\le\langle p',q'\rangle$. It follows that $$\langle p,q\rangle=\bigvee_{\langle i,j\rangle\in[n]\times[m]}\langle p_i,q_j\rangle\;.$$
The argument for infima is similar.