A bounded set of numbers may have no largest element. The set $(0, 1)$ has numbers as close to $1$ as you like, but $1$ itself is not in it. Analysis needs a name for the number such a set crowds up against, and in Discrete Math you met the set language we use to say it precisely. We also lean on distances throughout.
Key idea
An upper bound of $A$ is a number $u$ with $a \le u$ for every $a \in A$. The supremum $\sup A$ is the least upper bound: it is an upper bound, and no smaller number is one. The infimum $\inf A$ is the greatest lower bound. When $\sup A$ belongs to $A$, it is the maximum; otherwise $A$ has no maximum.
Why it works
Think of the upper bounds of $A$ as a ray on the number line, $[s, \infty)$. Its left end $s$ is the supremum. Every element of $A$ sits at or to the left of $s$ (so $s$ is an upper bound), and every number to the left of $s$ is beaten by some element (so nothing smaller is an upper bound). Whether $s$ itself is an element is a separate question, which is why "sup" and "max" are different words.
Worked examples
Worked example: A sequence that climbs
Find $\sup A$ and $\inf A$ for $A = \left\{1 - \frac{1}{n} : n \ge 1\right\}$.
- The elements are $0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots$: increasing, and always less than $1$.
- $\inf A = 0$, and it is an element (at $n = 1$), so it is the minimum.
- Every element is below $1$, so $1$ is an upper bound. A smaller number $1 - \varepsilon$ is beaten as soon as $\frac{1}{n} < \varepsilon$. So $\sup A = 1$, and $A$ has no maximum.
Answer: $\sup A = 1$ (not attained), $\inf A = 0$ (attained).
Worked example: An interval with an extra point
Find $\sup A$ and $\inf A$ for $A = (-2, 1] \cup \{3\}$.
- The largest element is the isolated point $3$, so $\sup A = 3 = \max A$.
- The elements come as close to $-2$ as you like, but $-2 \notin A$. So $\inf A = -2$, and there is no minimum.
Answer: $\sup A = 3$, $\inf A = -2$.
Worked example: Signs that alternate
Find $\sup A$ and $\inf A$ for $A = \left\{\frac{(-1)^{n}}{n} : n \ge 1\right\}$.
- The elements are $-1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, \ldots$
- The positive ones are largest at $n = 2$, and the negative ones are smallest at $n = 1$.
Answer: $\sup A = \frac{1}{2}$ and $\inf A = -1$, both attained.
Common mistake
Sup is not max. Students often write "$\sup (0, 1)$ does not exist" or "$\sup A$ must be in $A$." The supremum exists for every nonempty set that is bounded above (that is the next topic), but it need not be an element. The opposite slip is giving an upper bound that is too big, like "$\sup (0, 1) = 2$." Check both conditions: is it an upper bound, and is every smaller number beaten by some element?