# Difference between supremum and maximum

Referring to this lecture , I want to know what is the difference between supremum and maximum. It looks same as far as the lecture is concerned when it explains pointwise supremum and pointwise maximum

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Say that, $\text{sup}(A)=\alpha$, if $\alpha\in A$, then $\alpha$ considered to be $\text{max}$ of $A$. –  Salech Alhasov Jun 19 '12 at 19:11
Generally, a maximum is attained, but a supremum might not. For example, the set $[0,1)$ has a supremum ($1$), but no maximum (if $x\in [0,1)$, then there is always a point $y \in [0,1)$ such that $x <y$). The set $[0,1]$ has both a maximum & a supremum ($1$). –  copper.hat Jun 19 '12 at 19:17
@copper.hat: The two sentences seem to contradict each other. Correct me if I'm wrong, but first you say maximum is attained and supremum might not, and then you give an example where supremum is attained but maximum isn't. Maybe I misunderstood your point? –  Thomas E. Jun 19 '12 at 19:22
Both sets have suprema. The first example has no maximum, the second does. –  copper.hat Jun 19 '12 at 19:43
My wording was confusing. By 'generally' I meant that the usual definition is that a maximum is attained. But a supremum need not be attained. –  copper.hat Jun 19 '12 at 19:45

A maximum of a set must be an element of the set. A supremum need not be.

Explicitly, if $X$ is a (partially) ordered set, and $S$ is a subset, then an element $s_0$ is the supremum of $S$ if and only if:

1. $s\leq s_0$ for all $s\in S$; and
2. If $t\in X$ is such that $s\leq t$ for all $s\in S$, then $s_0\leq t$.

By contrast, an element $m$ is the maximum of $S$ if and only if:

1. $s\leq m$ for all $s\in S$; and
2. $m\in S$.

Note that if $S$ has a maximum, then the maximum must be the supremum: indeed, if $t\in X$ is such that $s\leq t$ for all $s\in S$, then in particular $m\in S$, so $m\leq t$, proving that $m$ satisfies the conditions to be the supremum.

But it is possible for a set to have a supremum but not a maximum. For instance, in the real numbers, the set of all negative numbers does not have a maximum: there is no negative number $m$ with the property that $n\leq m$ for all negative numbers $n$. However, the set of all negative numbers does have a supremum: $0$ is the supremum of the set of negative numbers. Indeed, $a\leq 0$ for all negative numbers $a$; and if $a\leq b$ for all negative numbers $a$, then $0\leq b$.

The full relationship between supremum and maximum is:

1. If $S$ has a maximum $m$, then $S$ also has a supremum and in fact $m$ is also a supremum of $S$.
2. Conversely, if $S$ has a supremum $s$, then $S$ has a maximum if and only if $s\in S$, in which case the maximum is also $s$.

In particular, if a set has both a supremum and a maximum, then they are the same element. The set may also have neither a supremum nor a maximum (e.g., the rationals as a subset of the reals). But if it has only one them, then it has a supremum which is not a maximum and is not in the set.

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In terms of sets, the maximum is the largest member of the set, while the supremum is the smallest upper bound of the set.

So, consider $A=\{1,2,3,4\}$. Assuming we're operating with the normal reals, the maximum is 4, as that is the largest element. The supremum is also 4, as four is the smallest upper bound.

However, consider the set $B=\{x | x < 2\}$. Then, the maximum of B is not 2, as 2 is not a member of the set; in fact, the maximum is not well defined. The supremum, though is well defined: 2 is clearly the smallest upper bound for the set.

You will find many points in real analysis (har har har) where it is easier and more profitable to consider the supremum of a given set (or construct the supremum) than it would be to consider or construct the maximum.

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@user31820 $-000\dots 001$ makes no sense. –  Pedro Tamaroff Jun 19 '12 at 22:03