The supremum and infimum generalize the notions of maximum and minimum, and they are among the fundamental tools for studying sets of numbers.
In the exercises that follow, we will learn how to find upper bounds, lower bounds, the supremum, the infimum, the maximum, and the minimum through worked examples, each explained step by step.
Exercise 1 — level ★☆☆☆☆
Find the set of upper bounds, the set of lower bounds, the supremum, and the infimum of:
\[ A=(2,7). \]
Answer
\[ \sup A=7,\qquad \inf A=2. \]
Set of upper bounds:
\[ [7,+\infty). \]
Set of lower bounds:
\[ (-\infty,2]. \]
Solution
The elements of the set \(A=(2,7)\) are precisely those real numbers lying strictly between \(2\) and \(7\).
By definition, an upper bound of \(A\) is a real number greater than or equal to every element of the set.
Since every element of \(A\) is strictly less than \(7\), the number \(7\) is an upper bound.
Every number greater than \(7\) is an upper bound as well. Hence the set of all upper bounds is:
\[ [7,+\infty). \]
The supremum is the smallest of the upper bounds. Since \(7\) is the least element of the set of upper bounds, it follows that:
\[ \sup A=7. \]
We now turn to the lower bounds.
A lower bound is a real number less than or equal to every element of the set.
Since every element of \(A\) is strictly greater than \(2\), the number \(2\) is a lower bound.
Every number less than \(2\) is a lower bound as well. Consequently, the set of all lower bounds is:
\[ (-\infty,2]. \]
The infimum is the greatest of the lower bounds. Therefore:
\[ \inf A=2. \]
Finally, note that \(2\notin A\) and \(7\notin A\). For this reason the set has neither a minimum nor a maximum, even though it does have an infimum and a supremum.
Exercise 2 — level ★☆☆☆☆
Find the set of upper bounds, the set of lower bounds, the supremum, the infimum, the maximum, and the minimum of:
\[ A=[-3,4]. \]
Answer
\[ \sup A=\max A=4. \]
\[ \inf A=\min A=-3. \]
Set of upper bounds:
\[ [4,+\infty). \]
Set of lower bounds:
\[ (-\infty,-3]. \]
Solution
The elements of the set \(A=[-3,4]\) are all the real numbers between \(-3\) and \(4\), endpoints included.
Hence:
\[ -3\leq x\leq 4 \qquad \forall x\in A. \]
Since every element of the set is less than or equal to \(4\), the number \(4\) is an upper bound of \(A\).
All numbers greater than \(4\) are upper bounds as well. It follows that the set of upper bounds is:
\[ [4,+\infty). \]
The smallest upper bound is \(4\). Therefore:
\[ \sup A=4. \]
Similarly, since every element of \(A\) is greater than or equal to \(-3\), the number \(-3\) is a lower bound.
All numbers less than \(-3\) are lower bounds as well. The set of lower bounds is therefore:
\[ (-\infty,-3]. \]
The greatest lower bound is \(-3\), and so:
\[ \inf A=-3. \]
Now observe that both \(4\) and \(-3\) belong to the set.
Consequently:
\[ \max A=4, \qquad \min A=-3. \]
This example shows that when the supremum belongs to the set it coincides with the maximum, and when the infimum belongs to the set it coincides with the minimum.
Exercise 3 — level ★☆☆☆☆
Find the set of upper bounds, the set of lower bounds, the supremum, the infimum, the maximum, and the minimum of:
\[ A=[0,5). \]
Answer
\[ \sup A=5, \qquad \inf A=\min A=0. \]
The maximum does not exist.
Set of upper bounds:
\[ [5,+\infty). \]
Set of lower bounds:
\[ (-\infty,0]. \]
Solution
The elements of \(A=[0,5)\) satisfy:
\[ 0\leq x<5. \]
Consequently, \(5\) is an upper bound of the set.
Moreover, every number greater than \(5\) is also an upper bound. The set of upper bounds is therefore:
\[ [5,+\infty). \]
No number less than \(5\) can be an upper bound, since the set contains elements arbitrarily close to \(5\).
Therefore:
\[ \sup A=5. \]
As for the lower bounds, every element of the set is greater than or equal to \(0\).
Hence:
\[ (-\infty,0] \]
is the set of all lower bounds.
The greatest of them is \(0\), and therefore:
\[ \inf A=0. \]
Since \(0\in A\), it follows immediately that:
\[ \min A=0. \]
On the other hand, \(5\notin A\). For this reason the set has no maximum.
Exercise 4 — level ★★☆☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{\frac1n:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \sup A=\max A=1. \]
\[ \inf A=0. \]
The minimum does not exist.
Solution
The elements of the set are:
\[ 1,\ \frac12,\ \frac13,\ \frac14,\ldots \]
This is a strictly decreasing sequence of positive numbers.
The largest value is the first one:
\[ 1=\frac11. \]
Therefore:
\[ \sup A=\max A=1. \]
Every element of the set is positive, so \(0\) is a lower bound.
We show that it is the greatest of the lower bounds.
Let \(\varepsilon>0\). By the Archimedean property, there exists \(n\in\mathbb N\) such that:
\[ n>\frac1\varepsilon. \]
From this it follows that:
\[ \frac1n<\varepsilon. \]
We have thus found an element of \(A\) smaller than \(0+\varepsilon\).
By the characterization of the infimum:
\[ \inf A=0. \]
However, \(0\notin A\), so the minimum does not exist.
This is one of the classic examples in which the infimum exists but does not belong to the set.
Exercise 5 — level ★★☆☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{\frac{n}{n+1}:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \inf A=\min A=\frac12. \]
\[ \sup A=1. \]
The maximum does not exist.
Solution
First, observe that:
\[ \frac{n}{n+1} = 1-\frac1{n+1}. \]
The elements of the set are therefore:
\[ \frac12,\ \frac23,\ \frac34,\ \frac45,\ldots \]
The sequence is increasing, since the term \(\frac1{n+1}\) decreases as \(n\) grows.
The first element is:
\[ \frac12. \]
Since the sequence is increasing, no element can be smaller than \(\frac12\).
Moreover, \(\frac12\in A\), and so:
\[ \inf A=\min A=\frac12. \]
For every \(n\) we have:
\[ \frac{n}{n+1}<1. \]
Hence \(1\) is an upper bound of the set.
We show that it is the smallest of the upper bounds.
Let \(\varepsilon>0\).
By the Archimedean property, there exists \(n\) such that:
\[ \frac1{n+1}<\varepsilon. \]
Then:
\[ \frac{n}{n+1} = 1-\frac1{n+1} > 1-\varepsilon. \]
By the characterization of the supremum, it follows that:
\[ \sup A=1. \]
However, \(1\notin A\), so the maximum does not exist.
Exercise 6 — level ★★☆☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{(-1)^n+\frac1n:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \sup A=\max A=\frac32. \]
\[ \inf A=-1. \]
The minimum does not exist.
Solution
It is convenient to distinguish the cases in which \(n\) is even from those in which \(n\) is odd.
If \(n\) is even, then:
\[ (-1)^n+\frac1n = 1+\frac1n. \]
We thus obtain the values:
\[ \frac32,\ \frac54,\ \frac76,\ldots \]
These numbers are all greater than \(1\) and decrease towards \(1\).
The largest is obtained for \(n=2\):
\[ 1+\frac12=\frac32. \]
Therefore:
\[ \sup A=\max A=\frac32. \]
If, on the other hand, \(n\) is odd:
\[ (-1)^n+\frac1n = -1+\frac1n. \]
We obtain:
\[ 0,\ -\frac23,\ -\frac45,\ldots \]
The first value is \(0\), obtained for \(n=1\). For the subsequent odd indices we instead obtain negative values that draw progressively closer to \(-1\) without ever reaching it.
The number \(-1\) is therefore a lower bound of the set.
Moreover, for every \(\varepsilon>0\), by choosing an odd \(n\) large enough we obtain:
\[ -1+\frac1n<-1+\varepsilon. \]
By the characterization of the infimum:
\[ \inf A=-1. \]
Since no element of the set equals \(-1\), the minimum does not exist.
Exercise 7 — level ★★☆☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{x\in\mathbb R:x^2<9\right\}. \]
Answer
\[ \sup A=3, \qquad \inf A=-3. \]
The set has neither a maximum nor a minimum.
Solution
The condition:
\[ x^2<9 \]
is equivalent to:
\[ -3<x<3 \]
Hence:
\[ A=(-3,3). \]
All elements of the set are less than \(3\), so \(3\) is an upper bound.
Moreover, the set contains elements arbitrarily close to \(3\), for instance:
\[ 3-\frac1n. \]
Hence no number less than \(3\) can be an upper bound.
Therefore:
\[ \sup A=3. \]
By an entirely analogous argument, we obtain:
\[ \inf A=-3. \]
Since neither \(3\) nor \(-3\) belongs to the set, there is no maximum and no minimum.
Exercise 8 — level ★★☆☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{x\in\mathbb R:x^2\leq9\right\}. \]
Answer
\[ \sup A=\max A=3. \]
\[ \inf A=\min A=-3. \]
Solution
The inequality:
\[ x^2\leq9 \]
is equivalent to:
\[ -3\leq x\leq3. \]
Hence:
\[ A=[-3,3]. \]
The number \(3\) is an upper bound of the set.
Moreover, it belongs to the set itself.
It follows that:
\[ \sup A=\max A=3. \]
Similarly, the number \(-3\) is a lower bound and belongs to the set.
Therefore:
\[ \inf A=\min A=-3. \]
This exercise highlights the difference between open and closed intervals: by adding the endpoints to the set, a maximum and a minimum automatically appear.
Exercise 9 — level ★★☆☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{x\in\mathbb R:1<x\leq 6\right\}. \]
Answer
\[ \sup A=\max A=6. \]
\[ \inf A=1. \]
The minimum does not exist.
Solution
The set consists of all real numbers greater than \(1\) and less than or equal to \(6\).
We may therefore write:
\[ A=(1,6]. \]
Every element of \(A\) is less than or equal to \(6\), so \(6\) is an upper bound of the set.
Since \(6\in A\), the number \(6\) is also the maximum of the set.
Therefore:
\[ \sup A=\max A=6. \]
We now consider the infimum.
Every element of \(A\) is greater than \(1\), so \(1\) is a lower bound.
Moreover, no number greater than \(1\) can be a lower bound, because the elements of the set can be chosen arbitrarily close to \(1\) from the right.
Therefore:
\[ \inf A=1. \]
However, \(1\notin A\), because the inequality is strict.
So the minimum does not exist.
Exercise 10 — level ★★☆☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{x\in\mathbb R:x>-2\right\}. \]
Answer
\[ \inf A=-2. \]
\[ \sup A=+\infty. \]
The set has neither a maximum nor a minimum.
Solution
The set contains all real numbers greater than \(-2\). Hence:
\[ A=(-2,+\infty). \]
The set is not bounded above. Indeed, whatever real number \(M\) we choose, we can take a number \(x\) greater than both \(M\) and \(-2\). In this way \(x\in A\) and \(x>M\).
Hence \(A\) has no real upper bound.
With the usual convention:
\[ \sup A=+\infty. \]
We now study the lower bounds.
Every element of the set is greater than \(-2\), so \(-2\) is a lower bound.
Moreover, no number greater than \(-2\) can be a lower bound, because the elements of the set can come as close as we like to \(-2\) from the right.
Therefore:
\[ \inf A=-2. \]
Since \(-2\notin A\), the minimum does not exist.
Moreover, since the set is unbounded above, there is no maximum either.
Exercise 11 — level ★★★☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{2-\frac1n:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \inf A=\min A=1. \]
\[ \sup A=2. \]
The maximum does not exist.
Solution
Let us write out a few elements of the set:
\[ 1,\ \frac32,\ \frac53,\ \frac74,\ldots \]
Indeed, for \(n=1\) we obtain:
\[ 2-\frac11=1. \]
As \(n\) grows, the term \(\displaystyle\frac1n\) decreases, so \(2-\displaystyle\frac1n\) increases.
The smallest value is therefore the first one, namely \(1\).
Since \(1\in A\), it follows that:
\[ \inf A=\min A=1. \]
Moreover, for every \(n\geq1\), we have:
\[ 2-\frac1n<2. \]
Hence \(2\) is an upper bound of \(A\).
We show that it is the smallest of the upper bounds.
Let \(\varepsilon>0\). By the Archimedean property, there exists \(n\in\mathbb N\) such that:
\[ \frac1n<\varepsilon. \]
Then:
\[ 2-\frac1n>2-\varepsilon. \]
Hence there is an element of \(A\) greater than \(2-\varepsilon\).
By the characterization of the supremum:
\[ \sup A=2. \]
Since \(2\notin A\), the maximum does not exist.
Exercise 12 — level ★★★☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{\frac{2n+1}{n+1}:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \inf A=\min A=\frac32. \]
\[ \sup A=2. \]
The maximum does not exist.
Solution
We rewrite the general term in a more convenient form:
\[ \frac{2n+1}{n+1} = \frac{2n+2-1}{n+1} = 2-\frac1{n+1}. \]
Hence:
\[ A=\left\{2-\frac1{n+1}:n\in\mathbb N,\ n\geq1\right\}. \]
We compute the first element:
\[ 2-\frac12=\frac32. \]
As \(n\) grows, the term \(\displaystyle\frac1{n+1}\) decreases, so \(2-\displaystyle\frac1{n+1}\) increases.
Therefore the smallest value in the set is:
\[ \frac32. \]
Since this value belongs to the set, we have:
\[ \inf A=\min A=\frac32. \]
Moreover, for every \(n\geq1\), we have:
\[ 2-\frac1{n+1}<2. \]
Hence \(2\) is an upper bound.
To prove that \(2\) is the supremum, we use the \(\varepsilon\)-characterization.
Let \(\varepsilon>0\). We choose \(n\) large enough so that:
\[ \frac1{n+1}<\varepsilon. \]
Then:
\[ 2-\frac1{n+1}>2-\varepsilon. \]
Hence there is an element of \(A\) greater than \(2-\varepsilon\).
It follows that:
\[ \sup A=2. \]
Finally, \(2\notin A\), because \(\frac1{n+1}\) is never equal to \(0\). Therefore the maximum does not exist.
Exercise 13 — level ★★★☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{\frac{n}{n+2}:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \inf A=\min A=\frac13. \]
\[ \sup A=1. \]
The maximum does not exist.
Solution
We rewrite the general term in a more readable form:
\[ \frac{n}{n+2} = \frac{n+2-2}{n+2} = 1-\frac{2}{n+2}. \]
The elements of the set are:
\[ \frac13,\ \frac24,\ \frac35,\ \frac46,\ldots \]
As \(n\) grows, the term \(\frac{2}{n+2}\) decreases, so \(1-\frac{2}{n+2}\) increases.
The smallest value is obtained for \(n=1\):
\[ \frac{1}{1+2}=\frac13. \]
Since \(\frac13\in A\), it follows that:
\[ \inf A=\min A=\frac13. \]
Moreover, for every \(n\geq1\), we have:
\[ \frac{n}{n+2}<1. \]
Hence \(1\) is an upper bound of \(A\).
We show that \(1\) is the smallest of the upper bounds.
Let \(\varepsilon>0\). We want to find an element of \(A\) greater than \(1-\varepsilon\).
Since:
\[ \frac{n}{n+2}=1-\frac{2}{n+2}, \]
it suffices to choose \(n\) such that:
\[ \frac{2}{n+2}<\varepsilon. \]
This is possible by the Archimedean property.
With this choice we obtain:
\[ \frac{n}{n+2} = 1-\frac{2}{n+2} > 1-\varepsilon. \]
By the characterization of the supremum:
\[ \sup A=1. \]
Finally, \(1\notin A\), because the equality \(\frac{n}{n+2}=1\) would imply \(n=n+2\), which is impossible. Hence the maximum does not exist.
Exercise 14 — level ★★★☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{3+\frac{1}{n}:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \sup A=\max A=4. \]
\[ \inf A=3. \]
The minimum does not exist.
Solution
The elements of the set are:
\[ 4,\ \frac72,\ \frac{10}{3},\ \frac{13}{4},\ldots \]
Indeed, for \(n=1\) we obtain:
\[ 3+\frac11=4. \]
As \(n\) grows, the term \(\frac1n\) decreases. Hence \(3+\frac1n\) decreases as well.
The first element is therefore the largest in the set.
Therefore:
\[ \sup A=\max A=4. \]
Moreover, for every \(n\geq1\), we have:
\[ 3+\frac1n>3. \]
Hence \(3\) is a lower bound of \(A\).
We show that it is the greatest of the lower bounds.
Let \(\varepsilon>0\). By the Archimedean property, there exists \(n\in\mathbb N\) such that:
\[ \frac1n<\varepsilon. \]
Then:
\[ 3+\frac1n<3+\varepsilon. \]
We have thus found an element of \(A\) smaller than \(3+\varepsilon\).
By the characterization of the infimum:
\[ \inf A=3. \]
Since \(3\notin A\), the minimum does not exist.
Exercise 15 — level ★★★☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{2+\frac{(-1)^n}{n}:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \inf A=\min A=1. \]
\[ \sup A=\max A=\frac52. \]
Solution
We separate the cases in which \(n\) is even from those in which \(n\) is odd.
If \(n\) is even, then \((-1)^n=1\), so the corresponding elements are:
\[ 2+\frac1n. \]
For \(n=2\) we obtain:
\[ 2+\frac12=\frac52. \]
For the other even values of \(n\), the term \(\frac1n\) is smaller. Hence the largest value among the terms with even index is \(\frac52\).
If \(n\) is odd, then \((-1)^n=-1\), so the corresponding elements are:
\[ 2-\frac1n. \]
For \(n=1\) we obtain:
\[ 2-1=1. \]
For the other odd values of \(n\), the term \(\frac1n\) is smaller, and so \(2-\frac1n\) is greater than \(1\).
It follows that the smallest value in the whole set is:
\[ 1. \]
Since \(1\in A\), we have:
\[ \inf A=\min A=1. \]
The largest value in the set, on the other hand, is:
\[ \frac52. \]
This value too belongs to \(A\), since it is obtained for \(n=2\).
Therefore:
\[ \sup A=\max A=\frac52. \]
Exercise 16 — level ★★★☆☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{x\in\mathbb R:0<x<1\right\}\cup\{2\}. \]
Answer
\[ \inf A=0. \]
\[ \sup A=\max A=2. \]
The minimum does not exist.
Solution
The set is made up of two parts:
\[ (0,1) \]
and the single element:
\[ 2. \]
All elements of the interval \((0,1)\) are less than \(1\), while \(2\) belongs to the set.
The largest value in the set is therefore \(2\).
Consequently:
\[ \sup A=\max A=2. \]
We now turn to the lower end of the set.
All elements of the set are positive, so \(0\) is a lower bound.
Moreover, the elements of the interval \((0,1)\) can be chosen arbitrarily close to \(0\) from the right.
Hence no number greater than \(0\) can be a lower bound.
Therefore:
\[ \inf A=0. \]
Since \(0\notin A\), the minimum does not exist.
The isolated element \(2\) changes the supremum, but it does not change the infimum of the set.
Exercise 17 — level ★★★★☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{x\in\mathbb R:x^2>4,\ -5<x<3\right\}. \]
Answer
\[ \inf A=-5, \qquad \sup A=3. \]
The set has neither a maximum nor a minimum.
Solution
We study separately the two conditions that define the set.
The inequality:
\[ x^2>4 \]
is equivalent to:
\[ x<-2 \qquad\text{or}\qquad x>2. \]
Moreover, we must have:
\[ -5<x<3. \]
Intersecting the conditions, we obtain:
\[ A=(-5,-2)\cup(2,3). \]
The set thus consists of two open intervals.
The smallest value the elements can approach is \(-5\), but \(-5\notin A\). Therefore:
\[ \inf A=-5. \]
The largest value the elements can approach is \(3\), but \(3\notin A\). Hence:
\[ \sup A=3. \]
Since the infimum does not belong to the set, the minimum does not exist.
Since the supremum does not belong to the set, the maximum does not exist.
Notice that the "gap" between \(-2\) and \(2\) changes neither the infimum nor the supremum: these depend only on how far down and how far up the set extends.
Exercise 18 — level ★★★★☆
Find the supremum, infimum, maximum, and minimum of the set:
\[ A=\left\{\frac{(-1)^n n}{n+1}:n\in\mathbb N,\ n\geq1\right\}. \]
Answer
\[ \sup A=1. \]
\[ \inf A=-1. \]
The set has neither a maximum nor a minimum.
Solution
To find the supremum and the infimum, it is convenient to distinguish the terms with even index from those with odd index.
If \(n\) is even, then:
\[ (-1)^n=1. \]
The corresponding elements of the set are therefore:
\[ \frac{n}{n+1}. \]
For \(n=2,4,6,\ldots\) we obtain:
\[ \frac23,\ \frac45,\ \frac67,\ \frac89,\ldots \]
We can rewrite these terms in the form:
\[ \frac{n}{n+1} = 1-\frac1{n+1}. \]
Since \(\frac1{n+1}>0\), each term is strictly less than \(1\).
Moreover, as \(n\) grows, the term \(\frac1{n+1}\) becomes ever smaller and tends to \(0\). Consequently, the values
\[ \frac{n}{n+1} \]
approach \(1\) arbitrarily closely without ever reaching it.
The number \(1\) is therefore an upper bound of the set.
Moreover, for every \(\varepsilon>0\), there is an even index large enough that:
\[ \frac{n}{n+1}>1-\varepsilon. \]
By the characterization of the supremum, it follows that:
\[ \sup A=1. \]
Since no element of the set equals \(1\), the maximum does not exist.
We now consider the odd indices.
If \(n\) is odd, then:
\[ (-1)^n=-1. \]
The corresponding elements of the set are:
\[ -\frac{n}{n+1}. \]
For \(n=1,3,5,\ldots\) we obtain:
\[ -\frac12,\ -\frac34,\ -\frac56,\ -\frac78,\ldots \]
We rewrite these terms as:
\[ -\frac{n}{n+1} = -1+\frac1{n+1}. \]
Since \(\frac1{n+1}>0\), all these values are strictly greater than \(-1\).
Moreover, as \(n\) grows, the term \(\frac1{n+1}\) tends to \(0\), and so the values
\[ -\frac{n}{n+1} \]
approach \(-1\) arbitrarily closely without ever reaching it.
The number \(-1\) is therefore a lower bound of the set.
Moreover, for every \(\varepsilon>0\), there is an odd index large enough that:
\[ -\frac{n}{n+1}< -1+\varepsilon. \]
By the characterization of the infimum, it follows that:
\[ \inf A=-1. \]
Since no element of the set equals \(-1\), the minimum does not exist.
We therefore conclude that:
\[ \sup A=1, \qquad \inf A=-1. \]
while the set has neither a maximum nor a minimum.
Exercise 19 — level ★★★★☆
Verify, using the characterization of the supremum, that:
\[ \sup(2,7)=7. \]
Answer
\[ \sup(2,7)=7. \]
Solution
To prove that \(7\) is the supremum of the set \(A=(2,7)\), we must verify two conditions.
The first condition requires that \(7\) be an upper bound of \(A\).
Indeed, if \(x\in(2,7)\), then:
\[ x<7. \]
A fortiori:
\[ x\leq 7. \]
Hence \(7\) is an upper bound.
The second condition requires that, for every \(\varepsilon>0\), there exist an element \(x\in A\) such that:
\[ 7-\varepsilon<x. \]
So let \(\varepsilon>0\).
If \(0<\varepsilon<10\), consider:
\[ x=7-\frac{\varepsilon}{2}. \]
Then:
\[ 2<x<7 \]
so that \(x\in(2,7)\).
Moreover:
\[ x=7-\frac{\varepsilon}{2}>7-\varepsilon. \]
We have thus found an element of the set greater than \(7-\varepsilon\).
If, on the other hand, \(\varepsilon\geq10\), it suffices to choose \(x=3\).
Indeed:
\[ 3\in(2,7) \]
and
\[ 3>7-\varepsilon. \]
In either case, for every \(\varepsilon>0\), there exists \(x\in A\) such that:
\[ 7-\varepsilon<x. \]
By the characterization of the supremum, we conclude that:
\[ \sup(2,7)=7. \]
Exercise 20 — level ★★★★★
Verify, using the characterization of the infimum, that:
\[ \inf\left\{\frac1n:n\in\mathbb N,\ n\geq1\right\}=0. \]
Answer
\[ \inf\left\{\frac1n:n\in\mathbb N,\ n\geq1\right\}=0. \]
Solution
Set:
\[ A=\left\{\frac1n:n\in\mathbb N,\ n\geq1\right\}. \]
To prove that \(0\) is the infimum of \(A\), we must verify two conditions.
The first condition requires that \(0\) be a lower bound of \(A\).
Indeed, for every \(n\geq1\), we have:
\[ \frac1n>0. \]
Hence:
\[ 0\leq \frac1n \qquad \forall n\geq1. \]
So \(0\) is a lower bound of the set.
The second condition requires that, for every \(\varepsilon>0\), there exist an element of \(A\) smaller than:
\[ 0+\varepsilon=\varepsilon. \]
So let \(\varepsilon>0\).
By the Archimedean property, there exists \(n\in\mathbb N\) such that:
\[ n>\frac1\varepsilon. \]
From this inequality it follows that:
\[ \frac1n<\varepsilon. \]
But \(\frac1n\in A\). Hence, for every \(\varepsilon>0\), we have found an element \(x\in A\) such that:
\[ x<0+\varepsilon. \]
By the characterization of the infimum, it follows that:
\[ \inf A=0. \]
Finally, observe that \(0\notin A\), so \(A\) has no minimum.