Mathematics RU

Practice · Chapter 53

Supremum and infimum

The supremum and infimum of sequences and of sets given by inequalities — and how the least upper bound differs from the largest element.

How to solve it

The least upper bound ($\sup$) is the smallest number not less than every element of a set. It may belong to the set (then it is the largest element) or not: the interval $(0,\ 1)$ has supremum $1$ but no largest element. Every non-empty set of reals bounded above has a supremum — that is the completeness axiom.

Step by step

  1. A sequence: check whether it is monotonic. For a decreasing one the supremum is the first term and the infimum the limit (if it is not attained).
  2. A set from an inequality: solve it and look at the ends of the interval. A strict inequality leaves the endpoint out of the set, but it is still the bound.
  3. Split a complicated sequence into subsequences (say even and odd indices) and find the bounds of each.
  4. To check that $M$ is the supremum: all elements are $\le M$, and for every $\varepsilon > 0$ some element exceeds $M - \varepsilon$.
The supremum of $A$. $M$ is an upper bound. No smaller bound works: the elements come arbitrarily close to $M$. Example: $a_n = \frac{-n + 5}{3n + 4}$ decreases from $a_1 = \frac{4}{7}$ to the limit $-\frac{1}{3}$: $\sup = \frac{4}{7}$ (the largest element), $\inf = -\frac{1}{3}$ (not attained).

Common mistakes

  • Thinking the supremum must belong to the set. $\{x : x^2 < 8\}$ has supremum $\sqrt{8}$, though $\sqrt{8}$ itself is not in it.
  • Confusing “the largest element” with “the supremum”: the first may not exist, the second always does for a bounded set.
  • Taking the limit as a bound without checking monotonicity: an oscillating sequence may have other bounds.
  • Looking for the bound of a set of rationals among the rationals — it can be irrational.

Example