Let us consider $\mathbb{R}^n, n\geq 2$. For simplicity, let's say $n=2$ (but I'm wondering about when $n>2$ as well). lets consider the set $$ Z=\{ (x,y)\mid x+y\leq 2\} $$
Does this set have a supremum? If so, what is it?
I am confused because there are point that are neither $\leq$ or $\geq$ other points. For example, $(2,0)$ has nothing greater than it. However, $(2,0)$ is neither $\leq$ or $\geq (0,2)$. So would these both be supremums, along with $(1,1), (1.5,.5)$ etc.? I guess the answer may be that $(2,2)$ is the supremum, which is not in this set, but then it seems that in higher dimensions the supremum would often not be in the set...
I guess, upon a little further thinking, my question is are points like $(2,0)$ that are in the boundary, but not greater than or equal to everything (because we are in higher dimensions) supremum or maximum?
Thanks.
First you have do define an order on $\Bbb R^n$. In $\Bbb R^1$ we have the usual order, but in $\Bbb R^n$ there isn't one. You could use lexicographic order, where $(a,b) \gt (c,d)$ if $a \gt c$ or $a=c, b \gt d$. You also seem to be assuming that $x \ge 0$ and $y \ge 0$. In that case your $Z$ is an isosceles right triangle with corners $(0,0), (2,0), (0,2)$ In the lexicographic order, $(2,0)$ is the greatest point in or on the triangle and is the supremum. Alternately you could define a partial order as $(a,b) \gt (c,d)$ if $a \gt c$ and $ b \gt d$ In this case the supremum would be $(2,2)$ as that is the least point greater than all the points of the triangle. If you don't restrict to $x \ge 0, y \ge 0$ your set is the half plane below $x+y=2$ It does not have a supremum in either of these orders. The same thing happens in higher dimensions.