Worked examples · Sheet 2

Supremum and infimum of number sets

Theory used: Upper bound, lower bound, bounded set Maximum and minimum Supremum and infimum Intervals and their notation

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Determine the upper bounds, lower bounds, supremum, infimum, maximum and minimum of $A = (- 2, 5]$ .

upper bounds: [5, + \infty), lower bounds: (- \infty, - 2]

sup A = max A = 5, in f A = - 2, min A does not exist

A number $M$ is an upper bound when it is greater than or equal to every element of the set. Every $M \geq 5$ is an upper bound, because the elements of $A$ are at most $5$ ; no $M < 5$ is one, because $5 \in A$ and $5 > M$ . Since $5$ is an upper bound and belongs to the set, $5$ is the maximum, and the maximum coincides with the supremum.

For lower bounds the reasoning is symmetric: every $m \leq - 2$ is a lower bound, because the elements of $A$ exceed $- 2$ ; a number $m > - 2$ is not one, because between $- 2$ and $m$ the interval contains points less than $m$ . The infimum, which is the largest of the lower bounds, therefore equals $- 2$ . The minimum does not exist, because $- 2 \in / A$ and the minimum of a set must belong to it.

Exercise 2

Determine the supremum, infimum, maximum and minimum of

A = {\frac{n}{n + 1} : n \in N, n \geq 1} .

\frac{n}{n + 1} = 1 - \frac{1}{n + 1}

in f A = min A = \frac{1}{2}, sup A = 1, max A does not exist

The rewriting $\frac{n}{n + 1} = 1 - \frac{1}{n + 1}$ makes the behavior of the elements readable: as $n$ increases the term $\frac{1}{n + 1}$ decreases, so the elements increase. The smallest element is therefore obtained with $n = 1$ and equals $\frac{1}{2}$ : it is a lower bound and belongs to the set, so it is the minimum, and it coincides with the infimum.

The number $1$ is an upper bound, because $\frac{1}{n + 1} > 0$ and therefore every element is less than $1$ . No number $L < 1$ is an upper bound: indeed the inequality $1 - \frac{1}{n + 1} > L$ is equivalent to $\frac{1}{n + 1} < 1 - L$ , that is to $n + 1 > \frac{1}{1 - L}$ , and such a large natural number exists because $N$ is not bounded above. Hence $1$ is the smallest of the upper bounds, that is $sup A = 1$ . The maximum does not exist, because $1 \in / A$ ; indeed the equality $\frac{n}{n + 1} = 1$ would require $\frac{1}{n + 1} = 0$ , which no natural number achieves.

Exercise 3

Determine the supremum, infimum, maximum and minimum of

A = {(- 1)^{n} + \frac{1}{n} : n \in N, n \geq 1} .

n even : 1 + \frac{1}{n} n odd : - 1 + \frac{1}{n}

sup A = max A = \frac{3}{2}, in f A = - 1, min A does not exist

The factor $(- 1)^{n}$ distinguishes two families of elements, so it is convenient to examine them separately. For $n$ even the elements are $1 + \frac{1}{n}$ : the largest is obtained with the smallest even $n$ , that is $n = 2$ , and equals $\frac{3}{2}$ , while the others lie between $1$ and $\frac{3}{2}$ . For $n$ odd the elements are $- 1 + \frac{1}{n}$ , all less than or equal to $0$ , because $\frac{1}{n} \leq 1$ . Every element of the set is therefore less than or equal to $\frac{3}{2}$ , and $\frac{3}{2}$ belongs to $A$ : it is the maximum, therefore it is also the supremum.

For the infimum: every element exceeds $- 1$ , because $(- 1)^{n} \geq - 1$ and $\frac{1}{n} > 0$ , so $- 1$ is a lower bound. No $m > - 1$ is a lower bound: indeed for $n$ odd the inequality $- 1 + \frac{1}{n} < m$ is equivalent to $\frac{1}{n} < m + 1$ , that is to $n > \frac{1}{m + 1}$ , and such a large odd natural number exists. Hence $in f A = - 1$ . The minimum does not exist, because $- 1 \in / A$ ; indeed the equality $- 1 + \frac{1}{n} = - 1$ would require $\frac{1}{n} = 0$ .