Among all the upper bounds of a set, one may be smaller than every other. That least upper bound is called the supremum, and it is the sharpest ceiling the set admits. It sometimes belongs to the set and sometimes does not, which is exactly why it succeeds where the notion of a largest member fails.

A crowd presses towards a wall without anyone touching it. There is no closest person, since whoever you name, someone stands nearer. The wall itself is still the sharpest line beyond which nobody stands.

Two conditions define it. The supremum must bound the set above, and no smaller number may do so. The second condition is what makes it least, and it is usually the one that has to be checked.

• $\sup S$ is an upper bound: every member lies at or below it.
• No number below $\sup S$ is an upper bound, so anything smaller is beaten by some member.
• For $S = \{x : x < 1\}$ the supremum is , which is not a member. There is no largest member at all.

Of all those bounds one lies furthest left, and that least one is the supremum. Here $S$ stops short of 2 and $\sup S=2$ all the same — the hollow mark is the point of the definition: the sharpest ceiling need not be a member.

Each candidate is an upper bound, and each is least among upper bounds. Playing those two facts against each other traps the pair between two inequalities that leave only equality.

Assume two of them exist, then show they cannot differ.

Being least means being no larger than any other upper bound.

The same argument with the two swapped.

Both inequalities hold at once.

Two numbers each at most the other are the same number.

Every member is below one, nothing smaller bounds the set, and one is not a member.

$\{x : x < 1\}$ has $\sup = 1$ and no maximum.

Name any member and one halfway to one is larger, so no member is largest.

$10$ bounds $\{x : x < 1\}$ above but is not its supremum.

Two also bounds the set, and two is smaller, so ten was not least.

Three sits below every member, nothing larger does, and three is not itself a member.

Leastness is almost always applied in this shape: pick a tolerance, and find a member of the set within it.

• The supremum is the least upper bound, and a set has at most one.
• It need not belong to the set; when it does, it is the maximum.
• Leastness in use: for any $\varepsilon>0$ some element exceeds $\sup S-\varepsilon$.
• Infima mirror suprema exactly, via $\inf S=-\sup(-S)$.

Great work! You now know how to:

• State the two conditions defining a supremum
• Use leastness in its $\varepsilon$ form
• Tell a supremum from a maximum, and give a set with one and not the other
• Mirror every statement about suprema into one about infima