Given the set $P=\mathbb{Z}\times\mathbb{Z}$ and the partial order < on P defined by $$(a,b)<(c,d)\text{ if and only if }\operatorname{max}(a,b)\text{ is less than }\operatorname{max}(c,d).$$
Let $S=\mathbb{N}\times\mathbb{N}.$ Describe each of the following.
(i) The minimal elements of $S$.
(ii) The maximal elements of $S$.
(iii) The lower bounds of $S$.
(iv) The upper bounds of $S$.
(v) The lub of $S$.
(vi) The glb of $S$.
I think that (i) $(0,0)$ is the minimal element of $S,$ (ii) the maximal element is infinity, (iii) the lower bounds of $S$ would be $(x<0,y<0),$ (iv) the upper bound of $S$ is infinity, and (vi) there is no glb of $S.$ I don't know what the lub of $S$ would be.
But I am told that $(0,0)$ is included in the lower bound? But then $(0,-1)$ would be a contradiction, since $\operatorname{max}(0,-1)$ not less than $\operatorname{max}(0,0)$?
In my opinion, the best way to go would be to develop a visual understanding for how the partial order works.
Let's start with an arbitrary point of $P,$ say $(a,b).$ We can picture this as being somewhere on the $xy$-plane with integer coordinates. However, given the fact that $\max(a,b)$ comes into play in the definition of the partial order, it will probably make a difference whether our $x$-coordinate is equal to, greater than, or less than our $y$-coordinate. This suggests that we should start by addressing these three cases separately.
First, let's consider the case that our $x$- and $y$-coordinates are equal. That is, we're considering a point on the line $y=x$ with integer coordinates, say $(a,a).$ What would it mean to have $(a,a)<(c,d)$? Well, since $\max(a,a)=a,$ then by definition, we would have $a<\max(c,d),$ but again, that will vary, depending on the relationship between $c$ and $d.$ One possibility is that $\max(c,d)=c,$ so by definition, we have $a<c,$ meaning (visually) that $(c,d)$ is to the right of the line $x=a.$ The other possibility is that $\max(c,d)=d,$ so that $a<d,$ meaning (visually) that $(c,d)$ is above the line $y=a.$
Consequently, if we have a point on the line $y=x,$ then the greater points will be above it and/or to the right of it.
Here's a graph, with the point $(a,a)$ in red. The points above the line $y=a$ and the points to the right of the line $x=a$ will be greater than $(a,a).$ Put another way, the points outside of the region containing $(a,a)$ and enclosed by the dashed lines will be greater than $(a,a).$ The points in black on this graph will all be greater than $(a,a).$ No points incomparable to or less than $(a,a)$ are shown on this graph.
Second, let's consider the case that our $x$-coordinate is greater than our $y$-coordinate. That is, we're considering a point below the line $y=x$ with integer coordinates, say $(a,b).$ What would it mean to have $(a,b)<(c,d)$? Well, $\max(a,b)=a,$ so by definition, we would have $a<\max(c,d).$ If $\max(c,d)=c,$ then by definition, we have $a<c,$ meaning (visually) that $(c,d)$ is to the right of the line $x=a.$ The other possibility is that $\max(c,d)=d,$ so that $a<d,$ meaning (visually) that $(c,d)$ is above the line $y=a.$
Consequently, if we have a point $(a,b)$ with $a>b,$ then the greater points will be above and/or to the right of the point $(a,a).$ Put another way, if $(a,b)$ is below the line $y=x$, then we look to the point on $y=x$ that is directly above $(a,b)$--the points greater than $(a,b)$ will be above and/or to the right of that point.
Finally, let's consider the case that our $y$-coordinate is greater than our $x$-coordinate. That is, we're considering a point above the line $y=x$ with integer coordinates, say $(a,b).$ In much the same way as the second case, we find that the points greater than $(a,b)$ are those above the line $y=b$ or to the right of the line $x=b.$
Consequently, if we have a point $(a,b)$ with $a<b,$ then the greater points will be above and/or to the right of the point $(b,b).$ Put another way, if $(a,b)$ is below the line $y=x$, then we look to the point on $y=x$ that is directly to the right of $(a,b)$--the points greater than $(a,b)$ will be above and/or to the right of that point.
Note that in each case, we wound up finding a point on $y=x$ to make the comparison. This lets us split up the points of $P$ into "levels" of a sort.
Here, we see that our points have been divided up into somewhat L-shaped regions, with each region extending infinitely leftward and downward from the line $y=x.$
The points in any given region will be incomparable to each other--for example, none of the following points will be greater than any of the other following points: $$\dots,(-3,-1),(-2,-1),(-1,-1),(-1,-2),(-1,-3),\dots$$ On the other hand, points in more upward/rightward regions will be greater than points in more downward/leftward regions.
Consequently, we immediately see that any point of $P$ that is below the $x$-axis and to the left of the $y$-axis (that is, having both coordinates negative) will be less than every element of $S,$ so will be a lower bound of $S.$ However, since $(0,0)$ is not greater than the points on the negative $y$-axis or negative $x$-axis, then points on one of the two negative axes will not be lower bounds of $S$.
Now let's take a look at just points of $S.$
Immediately, we see that $(0,0)$ is the least element of $S,$ since it is in the most downward/leftward region. Consequently, it is the only minimal element of $S,$ and is the greatest lower bound--that is, $\operatorname{glb}(S)=(0,0).$ However, the regions extend infinitely upward/rightward from there, so $S$ has no maximal elements, no upper bounds, and no least upper bound. "Infinity" is not an ordered pair of integers, so we cannot say that it is a maximal element of $S,$ nor that it is an upper bound of $S$ in $P.$
To sum up the discussion above, we have the following
See if you can prove this Claim. If you have any questions about what I've written, or want to bounce your ideas off of someone, please leave a comment below.