Let $A$ be an ordered set and $A = \{(x,y) \in\mathbb{R}^2 | 1 \le x \le 2\}$ is bounded from above by $(3,0)$ in $\mathbb{R}^2$ (with lexicographical order) but $A$ has no supremum. Also, $A$ is bounded from below by $(0,0)$ but $A$ has no infimum.

Now, consider the example: $U \supset A$ and $U = \{(-1,0),(0,0),(1,0),(2,0),(3,0),(3,1),(3,2)\}$.

I understand that $(0,0)$ is a lower bound of $A$ and also $(3,0)$ is an upper bound of $A$, but why $(0,0)$ is not the infimum and $(3,0)$ is not the supremum? For the sake of example, my arguments below are in contradiction with $A$ not having infimum nor supremum.

Let $I$ be the set of lower bounds of $A$.

$I = \{(-1,0),(0,0)\}$

By definition the infimum of $A$ is the greatest element in $U$ that is less than or equal to all elements of $A$.

$\inf(A) = (0,0)$ because $(0,0) \ge (-1,0)$ in the lexicographical order.

Similarly with the supremum:

Let $S$ be the set of upper bounds of $A$.

$S = \{(3,0),(3,1),(3,2)\}$

By definition the supremum of $A$ is the least element in $U$ that is greater than or equal to all elements $A$.

$\sup(A) = (3,0)$ because $(3,0) \le (3,1) \le (3,2)$ in the lexicographical order.

  • $\begingroup$ The definition of infimum of $A$ is the greatest lower bound of $A$. Which means that you want to find the set of all lower bounds of $A$ and then take the greatest element. You are choosing $I$, a subset of the set of all lower bounds, but you must take all lower bounds into account when taking the greatest element. $\endgroup$ – The Phenotype Jul 1 '18 at 18:40

Let's look at the upper bounds of $A = \{(x,y) \in\mathbb{R}^2 | 1 \le x \le 2\}$. You already saw that $(3,0)$ is an upper bound, because $3>2$, but so are each $(3,-1)>(3,-2)>(3,-3)>\ldots$, so we need to find smaller upper bounds. The same holds if we replace $3$ above with some $a>2$.

Trying $a=2$ doesn't work, since for any $b\in\mathbb{R}$, we have that $(2,b)<(2,b+1)\in A$, so $(2,b)$ is not a canditate for the supremum. Obviously the there are no upper bounds with $a<2$.

So we are left with the set $\{(a,b)\mid a>2,b\in\mathbb{R}\}$ of upper bounds of $A$. The problem here is that we cannot find a smallest element in this set, as you can always choose $2+\frac{a-2}{2}$, which is smaller than $a$ (even if you could find this $a$, then we could still make $b$ approach $-\infty$, meaning that there is still no smallest element).

Analogously, $A$ doesn't have an infimum.

  • $\begingroup$ So $A$ doesn't have an supremum because we cannnot find the smallest upper bound as we can always find a smaller element than the one we could choose? So that happens with $\mathbb{R}$ but not with $\mathbb{N}$? In $\mathbb{R}$ there are no gaps (completeness) that we could find in $\mathbb{N}$ for example we have the number $1/2$ that we can make it smaller and smaller. $\endgroup$ – adriana634 Jul 1 '18 at 19:08
  • $\begingroup$ @adriana634 Yes; Yes; I don't understand why you're considering $\mathbb{N}$, but if we change $\mathbb{R}^2$ to $\mathbb{N}^2$, then it has supremum $(2,0)$ and infimum $(0,0)$. If we change $\mathbb{R}^2$ to $\mathbb{Z}^2$, then it does not have a supremum and not an infimum as now the argument that I wrote between brackets applies (we can always make $b$ infinitely smaller). $\endgroup$ – The Phenotype Jul 1 '18 at 19:32
  • $\begingroup$ I am considering $\mathbb{N}$ just for the sake of example because for $\mathbb{N}^2$ we have supremum and infimum, for understanding when we have these properties, due to the gaps of $\mathbb{N}$ in the real line. $\endgroup$ – adriana634 Jul 1 '18 at 19:49

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