# Is this poset a lattice?

Given the set $\Bbb Z^+\times\Bbb Z^+$ and the relation

\begin{align*} (x_1,x_2)\,R\,(y_1,y_2)\iff &(x_1+x_2 < y_1 + y_2)\\ &\text{ OR }(x_1 + x_2 = y_1 + y_2\text{ AND }x_1 \le y_1)\;: \end{align*}

I know for a poset to be a lattice, every subset in $\Bbb Z^+\times\Bbb Z^+$ has to have a LUB and GLB, but how do I determine whether this is true for all subsets?

Also, is it correct that this relation is a total order?

Much appreciated

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Small note: lattices only need lubs and glbs for finite sets. Also, if it is a total order then it definitely must be a lattice. – Eric Stucky Oct 28 '13 at 11:22

HINT: You may find a picture helpful. For any real number $\alpha$ let $$L_\alpha=\{\langle x,y\rangle\in\Bbb R^2:x+y=\alpha\}\;;$$ this is just the graph of $x+y=\alpha$ and is a straight line of slope $-1$ through the points $\langle\alpha,0\rangle$ and $\langle 0,\alpha\rangle$. Each $\langle a,b\rangle\in\Bbb R^2$ is in exactly one of these sets $L_\alpha$, namely, $L_{a+b}$. Let $p$ and $q$ be distinct points in $\Bbb R^2$, and suppose that $p\in L_\alpha$ and $q\in L_\beta$. Then $p\,R\,q$ if and only if
• $L_\alpha$ lies below $L_\beta$ (i.e., if $\alpha<\beta$), or
• $L_\alpha=L_\beta$, and $p$ lies to the left of $q$ (i.e., the $x$-coordinate of $p$ is less than that of $q$).
In order for $\langle\Bbb Z^+\times\Bbb Z^+,R\rangle$ to be a lattice, it’s not necessary that every subset have a supremum and infimum: the requirement is that every pair of elements have a supremum and infimum. This is automatic if $\langle\Bbb Z^+\times\Bbb Z^+,R\rangle$ is a linear (or total) order, which in fact it is. This is very straightforward to prove directly: given $p_1=\langle x_1,y_1\rangle$ and $p_2=\langle x_2,y_2\rangle$ in $\Bbb Z^+\times\Bbb Z^+$, show that if it’s not the case that $p_1\,R\,p_2$, then $p_2\,R\,p_1$. You may find the geometric interpretation above helpful in thinking about this.