Worked examples · Sheet 3

Verifying sup and inf with the epsilon characterization

Theory used: Supremum and infimum Epsilon characterization of sup and inf Maximum and minimum

Exercise 1

Prove with the characterization that

sup {1 - \frac{1}{n} : n \in N, n \geq 1} = 1

and determine whether $1$ is the maximum of the set.

1 - \frac{1}{n} < 1 for every n \geq 1

1 - \frac{1}{n} > 1 - ε ⟺ \frac{1}{n} < ε ⟺ n > \frac{1}{ε}

sup A = 1, 1 \in / A : it is not the maximum

The characterization requires two conditions. The first is that $1$ is an upper bound: since $\frac{1}{n} > 0$ for every $n \geq 1$ , every element $1 - \frac{1}{n}$ is less than $1$ , so the condition is satisfied.

The second condition is that, for every $ε > 0$ , the set contains an element greater than $1 - ε$ . The inequality $1 - \frac{1}{n} > 1 - ε$ is equivalent, after cancelling $1$ and changing sign, to $\frac{1}{n} < ε$ , that is to $n > \frac{1}{ε}$ . A natural number greater than $\frac{1}{ε}$ exists by the Archimedean property, so the second condition is satisfied and $sup A = 1$ .

The maximum does not exist: indeed the equality $1 - \frac{1}{n} = 1$ would require $\frac{1}{n} = 0$ , which is impossible, so $1 \in / A$ , and the supremum is a maximum only when it belongs to the set.

Exercise 2

Prove with the characterization that

in f {\frac{3 n + 2}{n} : n \in N, n \geq 1} = 3

and determine whether $3$ is the minimum of the set.

\frac{3 n + 2}{n} = 3 + \frac{2}{n} > 3 for every n \geq 1

3 + \frac{2}{n} < 3 + ε ⟺ \frac{2}{n} < ε ⟺ n > \frac{2}{ε}

in f A = 3, 3 \in / A : it is not the minimum

Term-by-term division rewrites the elements as $3 + \frac{2}{n}$ , and in this form the two conditions of the characterization of the infimum can be checked directly. First condition: since $\frac{2}{n} > 0$ , every element exceeds $3$ , so $3$ is a lower bound.

Second condition: for every $ε > 0$ an element less than $3 + ε$ is needed. The inequality $3 + \frac{2}{n} < 3 + ε$ is equivalent to $\frac{2}{n} < ε$ , that is to $n > \frac{2}{ε}$ , and such a large natural number exists by the Archimedean property. Both conditions are satisfied, so $in f A = 3$ .

The minimum does not exist, because $3$ does not belong to the set; indeed the equality $3 + \frac{2}{n} = 3$ would require $\frac{2}{n} = 0$ , which no natural number achieves.

Exercise 3

Determine the supremum and infimum of

A = {\frac{2 n - 1}{n + 1} : n \in N, n \geq 1},

proving the value of the supremum with the characterization.

\frac{2 n - 1}{n + 1} = \frac{2 ( n + 1 ) - 3}{n + 1} = 2 - \frac{3}{n + 1}

2 - \frac{3}{n + 1} < 2 for every n \geq 1

2 - \frac{3}{n + 1} > 2 - ε ⟺ \frac{3}{n + 1} < ε ⟺ n > \frac{3}{ε} - 1

sup A = 2, in f A = min A = \frac{1}{2}

The numerator can be rewritten as $2 (n + 1) - 3$ , so each element equals $2 - \frac{3}{n + 1}$ . In this form the candidate supremum is $2$ , because the subtracted quantity $\frac{3}{n + 1}$ becomes small as $n$ increases.

Verification with the characterization. First condition: $\frac{3}{n + 1} > 0$ , so every element is less than $2$ and $2$ is an upper bound. Second condition: given $ε > 0$ , the inequality $2 - \frac{3}{n + 1} > 2 - ε$ is equivalent to $\frac{3}{n + 1} < ε$ , that is to $n + 1 > \frac{3}{ε}$ , and a natural number $n > \frac{3}{ε} - 1$ exists by the Archimedean property. Hence $sup A = 2$ ; the maximum does not exist, because no element reaches $2$ .

For the infimum, observe that as $n$ increases the term $\frac{3}{n + 1}$ decreases, so the elements increase: the smallest is obtained with $n = 1$ and equals $2 - \frac{3}{2} = \frac{1}{2}$ . Since $\frac{1}{2}$ is a lower bound and belongs to the set, it is the minimum, and it coincides with the infimum.