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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.

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    $\begingroup$ Intuitively, the difference between $\max$ and $\sup$ is that, if a nonempty set $S$ bounded above has a $\max$, that is always its $\sup$. But it can also be the case that there is no $\max$, considering, for example, any finite open interval. In this case, the $\sup$ is the largest number that there exist elements $x \in S$ arbitrarily close to. In your example, $-5$ cannot be the minimum of the set since $-5$ is not an element of the set. It is its $\inf$ since for all $\epsilon > 0$, $-5 + \epsilon$ is in the set. $\endgroup$ – MathematicsStudent1122 Jun 16 '16 at 1:27
  • $\begingroup$ the max is the largest value in the set. The supremum does not have to be in the set. e.g. S is the open interval (0,1). S has no maximum, but it does have a supremum = 1. $\endgroup$ – Doug M Jun 16 '16 at 1:51
  • $\begingroup$ @DougM so is there always a max or supremum? $\endgroup$ – hax0r_n_code Jun 16 '16 at 1:54
  • $\begingroup$ There is always a supremum and an inf., there is not always a max or a min. $\endgroup$ – Doug M Jun 16 '16 at 2:25
  • $\begingroup$ @DougM what if the interval is $(-\infty, 5)$ what would the inf be? $\endgroup$ – hax0r_n_code Jun 16 '16 at 3:04
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An upper bound for a non-empty set $S\subset R$ is any $x\in R$ such that $\forall s\in S\;(s\leq x).$ A lower bound for a non-empty $S\subset R$ is any $y\in R$ such that $\forall s\in S\;(s\geq y).$

The supremum is the least upper bound FOR the set , not "the least upper bound IN the set". It is the least of all of the upper bounds for the set. It may or may not be a member of the set. The maximum of the set is the largest member OF the set, if there is such a member.

$\sup S$ belongs to $S$ if and only if $\sup S=\max S.$

The infimum is the greatest lower bound FOR the set, not "the greatest lower bound IN the set". It is the greatest of all of the lower bounds for the set. It may or may not be a member of the set. The minimum of the set is the least member OF the set, if there is such a member.

$\inf S$ belongs to $S$ if and only if $\inf S=\min S.$

The set $S=\{x\in R: x>-5\}$ has no least member, that is, no minimum. The set of lower bounds for $S$ is $\{y\in R:y\leq -5\}$, and the greatest of them is $-5$. So $\inf S=-5.$

It is common to call an upper (lower ) bound FOR a set "an upper (lower) bound OF the set", but this does not mean that it is necessarily a member of the set.

Some modern writers use $glb$ for $\inf$, and $lub$ for $\sup.$

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    $\begingroup$ Based on what you've described, would I be correct in saying that for $\{2.9, e, \pi, 2\sqrt{3}, 10/3\}$; inf = $e$, min=$e$, sup=$2\sqrt{3}$, max=$2\sqrt{3}$? $\endgroup$ – hax0r_n_code Jun 16 '16 at 23:06
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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.

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