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Similarity and difference between $\color{red}{\text{least upper bound/greatest lower bound}}$, $\color{blue}{\text{maximal/minimal}}$, $\color{green}{\text{maximum/minimum}}$.

Given a partially ordered set $\left(P,\le\right)$ and a subset $S$ of $P$:

An element $ g \in P$ is the $\color{red}{\text{least upper bound/greatest lower bound}}$ of $S$ if :$$\forall s \in S : s\le g \;\;\text{and for every upper bound u of S} : \;\;\;g\le u \;\;\;\ $$

$$(\forall s \in S : g\le s \;\;\text{and for every lower bound u of S}: \;\;\;u\le g )$$

An element $ g \in S$ is the $\color{blue}{\text{maximal/minimal}}$ of $S$ if :$$\forall s \in S : s\le g \;\;\;\ (s\ge g)$$

Equivalently if there does not exist any $s\in S$ such that:

$$g\le s\ \text{and}\ g\ne s \;\;\;\ (g\ge s\ \text{and}\ g\ne s)$$

An element $ g \in S$ is the $\color{green}{\text{maximum/minimum}}$ of $S$ if :$$\forall s \in S : s\le g \;\;\;\ (s\ge g)$$

The similarity between all of them is that it may happen that neither least upper bound/greatest lower bound nor maximal/minimal nor maximum/minimum exist.

The similarity between maximum/minimum and maximal/minimal is that if they exist then they both belong to the set $S$, although a least upper bound/greatest lower bound may not belong to the set $S$ and their difference is that it may happen to have more than one maximal/minimal but maximum/minimum or least upper bound/greatest lower bound is always unique.

If a set has a $\color{green}{\text{maximum/minimum}}$ then its $\color{blue}{\text{maximal/minimal}}$ are unique and every maximum/minimum is a maximal/minimal but not vice versa, also every maximum/minimum is a least upper bound/greatest lower bound but not vice versa. it can be concluded that a maximum/minimum are a stronger form of maximal/minimal and the condition incomparability never can happen.

are my conclusions right?

1 Answers1

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First thing, some of your definitions seem off to me, for instance I do not see any difference between the definitions you gave of "minimal/maximal" and "minimum/maximum" (and also, the property stated after "equivalently" for minimal/maximal also seems off). I interpret the order relation $\leq$ as meaning "lower or equal". So let me first go through the definitions:

  • $\color{red}{\text{Least upper bound/greatest lower bound}}$ : your definition is perfectly correct. Note that if $g$ is an upper bound/a lower bound of $S$, in particular $g$ is comparable to every element of $S$.
  • $\color{blue}{\text{Maximal/minimal}}$ : an element $g\in S$ is maximal (resp. minimal) if for every $s\in S$ that is comparable to $g$, we have $g\geq s$ (resp. $g\leq s$).
  • $\color{green}{\text{Maximum/minimum}}$ : an element $g\in S$ is the maximum (resp. minimum) if for every $s\in S$, we have $g\geq s$ (resp. $g\leq s$). This implies in particular that $g$is comparable to every element of $S$.

Any of these notions isn't guaranteed to exist. For instance, the natural integers $\mathbb N$ with the natural order, seen as a subset of itself, is a (totally) ordered set without any upper bound (but with a minimum, that is $1$, or $0$ depending on the conventions). The interval $(0,1)$ inside $\mathbb R$ is a subset with both least upper bound, greatest lower bound, but no maximal nor minimal element. The subset $\{2,4,6\}$ of $\mathbb N$ ordered by divisibility (that is, $x\leq y$ if and only if $x$ divides $y$) has to maximal elements $4$ and $6$, but no maximum as $4$ and $6$ are not comparable. It also has a minimum, that is $2$.

As you said, least upper bounds, greatest lower bounds aren't required to be part of the subset $S$, if they exist (but they may be). Maximal/minimal, maximum/minimum elements must be part of the subset $S$. We may have many maximal/minimal elements, but maximum/minimum and least upper bounds, greatest lower bounds are all unique when they exist (that is because they are comparable to every element of $S$, check it !).

If a set has a maximum/minimum, then it is its unique maximal/minimal element. The converse is false : a set can have a unique maximal element that is not the maximum. For instance, consider $\mathbb N$ which has no maximal element, and add an element $e$ that is not comparable to any of the integers. Then $e$ is trivially a maximal (and minimal) element of $\mathbb N \cup \{e\}$, it is the only maximal element, but it is not the maximum (the maximum doesn't exist).

If a subset has a maximum, then this maximum is the least upper bound, and similarly for minimum with respect to greatest lower bound. But again, the converse is easily seen to be false.

In some sort of conclusion, maximum/minimum is the strongest notion to work with. Least upper bound/greatest lower bound and maximal/minimal can be seen, if you want, as two different ways to "generalize" the definition of maximum/minimum. It is weaker, but it is more likely to exist in general.

Suzet
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  • I agree with you about the definition of maximal/minimal –  Feb 15 '20 at 13:23
  • For example if we define $\mathbb S:=\left{\left{1\right},\left{2\right},\left{1,2\right},\left{2,3\right}\right}$, then $\left{1,2\right}$ is a maximal for both $\left{1\right},\left{2\right}$ and there does not exist any other comparable element in $\mathbb S$ such that $\left{1,2\right}$ is not a superset for that element, but clearly it's not maximum, because $ \left{2,3\right}$ is not a subset of $\left{1,2\right}$ (actually $\left{2,3\right}$ is neither maximal nor minimal),and the set has a unique maximal which is not a maximum. –  Feb 15 '20 at 13:23
  • About the other equivalent definition of maximal/minimal you are right, I edited the error, please see it again. For example if we define an interval $I=\left(-2,2\right)$ then $I$ does not have any maximum/minimum but its lub, glb is $2,-2$ respectively, and based on Completeness axiom every nonempty set of real numbers which is bounded above (resp, below) has lub(resp, glb), and any nonempty subset of integers which is bounded above (resp, below) has a lub (resp glb) which is contained in that subset (it means that the set has maximum (resp, minimum)), am I right? –  Feb 15 '20 at 13:23
  • @user715522 I perfectly agree with the corrected version of the equivalent definition for maximal/minimal. I also agree with what you wrote in the third comment. As for the example you gave in the second comment, ${2,3}$ is maximal in $\mathbb S$ actually: the only other element that is comparable to it is ${2}$, which is included in ${2,3}$. So $\mathbb S$ has two maximal elements. It has $2$ minimal elements as well. – Suzet Feb 16 '20 at 00:40
  • An example would be the set of real numbers $\mathbb R \cup \left{i\right}$,where $i=\sqrt{-1}$, then the set of real numbers with this assumption has a unique maximal/minimal but does not have any maximum/minimum. So I guess we can say If maximum (resp. minimum) of a set exists then the maximal (resp. minimal) of that set is unique and the converse is true just when maximum (resp. minimum ) exists. –  Feb 16 '20 at 11:06
  • If I defined a set $\mathbb S:= \left{\left{\right},\left{1\right},\left{2\right},\left{3\right},\left{1,2\right},\left{1,3\right},\left{1,2,3\right}\right}$ then can I say $\left{1,2\right}$ is an upper bound ( maximal, maximum) for the set $\left{\left{1\right},\left{2\right}\right}$, and the only set for which $\left{\right}$ is an upper bound is $\left{\right}$ itself and the only set for which $\left{1,2,3\right}$ is a lower bound is $\left{1,2,3\right}$ itself? –  Feb 16 '20 at 11:06
  • I agree with your first comment ; as for you second one, everything is alright (except ${1,2}$ only is an upper bound for ${{1},{2}}$, it is neither maximal nor maximum as it isn't an element of it). – Suzet Feb 16 '20 at 12:25
  • Thank you for the good post, can you do me a favor and look at this post –  Feb 16 '20 at 13:16