Has the Infimum of a subset necessarily not to be an element of it?

Question

As I went to study the structure of a Lattice I noticed that it consisted on a partially ordered set in which every two elements have a unique supremum and a unique infimum.

Everything until now is okay, but in my teacher textbook it defines supremum/minimum as it follows

An element $c \in A$ is a (upper/lower) bound of a subset $B \subset A$ if there isn't any $b \in B$ which is (subsequent/anteceent) to it. The (supremum/infimum) is the (lowest/highest) of its bounds.

Well, the problem is that if we want to compare the highest or lowest element with other of the set with that definition we can't say that there is any supremum or infimum because any element is subsequent or anteceent to itself.

I interpeted that the infimum/supremum had to be one element in the set $(A-B)$ but as I had said it makes no sense because there couldn't be any Lattice.

Am I interpreting it wrong or it is just that this definition is wrong?

Answer 1

An infimum may ether be or not be an element of the set it's an infimum for.

Consider for example the lattice of positive integers ordered by divisibility.

Then the infimum of $\{12,24\}$ is $12$: The lower bounds (that is, the numbers that divide both $12$ and $24$) are $1, 2, 3, 4, 6, 12$, and the highest (by the divisibility relation) of these is $12$.

On the other hand, the infimum of $\{12,18\}$ is $6$: The lower bounds are $1,2,3,6$, and the highest of these is $6$.

In this lattice the infimum is of course just the greatest common divisor.

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In the case of the infimum of a set with more than two elements, we can see an example of the same just by considering the real numbers with their usual ordering. Then the open interval $(1,2)$ and the closed interval $[1,2]$ each have $1$ as their infimum, but $1$ is an element of the closed interval but not of the open one.

Answer 2

Your interpretation is wrong, sorry.

The concept of infimum was born as a generalization for minimum: some sets of numbers don't have a minimum, but there is an element that is quite like a minimum: the greatest lower bound.

It's a true generalization: if a set $S$ of numbers has a minimum $m$, then it is also the greatest lower bound. Indeed, $m$ is a lower bound by definition; if $x$ is a lower bound of $S$, then in particular $x\le m$.

The notion has revealed useful also for partial orders. And the definition is exactly the same.

If $P,\le$ is a partially ordered set and $S\subseteq P$, then a lower bound $x$ for $S$ is $x\in P$ such that $x\le y$, for all $y\in S$.

It can happen that the set of lower bounds of $S$ has a greatest element $m$. If this happens, then $m$ is called the infimum of $S$. The same argument as before shows that if $S$ has a minimum (that is, an element $m\in S$ with $m\le y$, for all $y\in S$), then $m$ is the infimum for $S$.

A necessary condition for a set $S$ to have an infimum is to be lower bounded. However, this condition is not generally sufficient. It is sufficient in the real numbers, which is a very strong property.

A partially ordered set is “lower complete” if every nonempty subset has an infimum. In particular, the set itself has an infimum, which must be the (global) minimum.

A lattice is a partially ordered set in which every two element subset has an infimum and a supremum. If $x\land y$ denotes the infimum of $\{x,y\}$, then $$ x\le y\qquad\text{if and only if}\qquad x\land y=x $$ so you see that the infimum can be in the set we are computing the infimum of.