Textbook
For a nonempty real set $T$ bounded above, the supremum $M=\sup T$ is an upper bound no larger than any other upper bound. Thus $t\le M$ for every $t\in T$, but no smaller number has that property.
Its sharpness can be stated as follows: for every $\varepsilon>0$, some $t\in T$ satisfies $M-\varepsilon<t\le M$. Values approach arbitrarily closely from below; the boundary need not belong to the set. Existence for a nonempty bounded-above set follows from completeness of the reals.
For a function, $\sup_{x\in S}f(x)$ is the supremum of its values. For $f(x)=x$ on $(0,1)$ it equals 1. One is an upper bound, while any smaller number leaves some admissible values above it.
An attained supremum is a maximum. Since 1 is excluded from this open interval, no maximum exists. For a set unbounded above, there is no finite real supremum; writing $+\infty$ uses an extended convention.
A proposed supremum fails if either requirement fails: an element may lie above it, or a smaller upper bound may exist. These are distinct failures. Two genuinely bounds $(0,1)$ but is not sharp; 0.9 is not even an upper bound. A supremum, when it exists, is unique: two distinct candidates cannot both be least upper bounds because the smaller one would disprove the larger one’s minimality.
Definitions and notation: Basic Analysis I.
Related concepts: A function’s maximum: the largest attained value.
In plain language
A supremum resembles the lowest ceiling that fits above every value. A higher ceiling is allowed, but any downward shift leaves some value outside. For $(0,1)$ the boundary is one, although no tallest tenant exists among its numbers. A finite photograph of heights cannot establish a claim about an entire infinite set; the architect needs a general argument.
Example
For the constructed set $(0,1)$, the example tests upper bounds and approaches 1.
Supremum: the least upper bound
A loose ceiling
Two bounds every value, but 1 gives a smaller upper bound.
Step 1 of 4: A loose ceiling
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A loose ceiling
Two bounds every value, but 1 gives a smaller upper bound.
- Candidates
- 21current
Too low
The number 0.9 fails as an upper bound because 0.95 is higher.
- Candidate; counterexample
- 0.9excluded0.95included
Approaching from below
Values 0.9, 0.99, and 0.999 stay admissible while approaching 1.
- Values
- 0.90.99current0.999included
The exact bound
One is the supremum but not an element of the interval.
- Supremum; membership
- 1included1 ∉ (0,1)excluded
The values 0.9, 0.99, and 0.999 illustrate approach but do not by themselves reject every smaller upper bound. Given any candidate $b<1$, a number strictly between $\max(b,0)$ and one, such as their average, supplies a counterexample. It lies in $(0,1)$ and exceeds $b$, defeating that candidate. Meanwhile, every interval element is below one. These arguments establish both requirements without enumerating infinitely many decimals.
A supremum requires an upper bound that cannot be lowered; a maximum additionally requires attainment.