I think I understand the very basic concepts of these terms, but wanted to check my understanding here.
The max is the largest number in the set.
The supremum is the least upper bound number in the set.
The min is the smallest number in the set.
The infimum is the greatest lower bound in the set.
In a way these all seem like they are saying the same thing, which I know they are not. Can you all help me to understand it via the following examples?
$\{2.9, e, \pi, 2\sqrt{3}, 10/3\}$
If I were to guess I'd say the max is $2\sqrt{3}$, min $e$, and that there is no supremum or infimum.
$\{x \in \mathbb R : x \gt -5\}$
For this one I'd say there is no max, supremum or infimum, but the min is -5.
Any assistance in understanding how to find these is greatly appreciated.
Best Answer
First of all, let me note that the terms minimum, maximum, infimum and supremum do not live in a vacuum. They are properties that a certain element can have with respect to a given order. Now forget about this and let us only focus on the order $(\Bbb R; \le)$ that you seem to be interested in. Given any subset of real numbers $S \subseteq \Bbb R$, we may ask whether or not there is some $r \in \Bbb R$ such that $r$ is bigger then all the elements in $S$. If such an $r$ exists, we call it an upper bound for $S$.
For example: Let $S = (0,1)$. Every element $s \in S$ is smaller than $5$ and hence $5$ is an upper bound for $S$. However, there are smaller upper bounds. $4$ is an upper bound for $S$ as well and so is $\pi$ and $\sqrt{2}$. We may hence ask if there is a least upper bound for $S$ and in fact, we have that $1$ is an upper bound for $S$ and no $r < 1$ is an upper bound for $S$. Hence $1$ is the least upper bound for $S$, which we also call the supremum of $S$.
We designed the real numbers in such a way that once a given subset of reals $S \subseteq \Bbb R$ has any upper bound, it will always have a (unique!) least upper bound, i.e. a supremum. (A supremum is, by definition, the same as a least upper bound.) Let me stress that this is by construction of the reals, it is not true for other orders. For example, in $(\Bbb Q; \le)$, the set $\{ q \in \Bbb Q \mid q^2 < 2\}$ clearly has an upper bound, but it doesn't have a supremum.
Let us return to $(\Bbb R; \le)$. Note that $\{ x \in \Bbb R \mid 0 < x \}$ does not have an upper bound and thus does not have a supremum. Recall that any subset $S \subseteq \Bbb R$ has a supremum if and only if it has an upper bound.
So what about maxima? A maximum of a set of reals $S \subseteq \Bbb R$ is a least upper bound of $S$ that is also an element of $S$. Since suprema are unique (provided they exist), we thus have that a maxima of $S$ is also the supremum of $S$ and that $S$ has at most one maximum.
Let's have a second look at $S = (0,1)$. We already know that $S$ has $1$ as it's suprema, but $1 \not \in (0,1)$. This immediately implies that $S$ does not have a maximum. (By the argument above, if $S$ had a maximum, it would be equal to its supremum. But the supremum of $S$ is not an element of $S$ and hence not a maximum.)
I'll leave the case for infima of $S \subseteq \Bbb R$ (which are greatest lower bounds of $S$) and minima of $S$ (which are greatest lower bounds of $S$ that are also elements of $S$) to you.