Exercise 1 Regarding the set \(E = \{\frac{1}{n}|n \in \mathbb{N}\}\), does it have a minimum? Does it have an infimum in \(\mathbb{Q}\)? If the infimum does exist then find it.

 

Showing that \(E\) does not have a minimum is showing that for every \(a \in E\) it will always be possible to find \(a' \in E\) such that \(a' < a\). Consider \(a = \frac{1}{n}\). Consider \(a' = \frac{1}{n+1}\). Indeed we have \(a' < a\). Therefore, \(E\) does not have a minimum.

Consider the set \(L = \{x \in \mathbb{Q} | \forall y \in E, x \leq y \}\). We will claim \(L = \{x \in \mathbb{Q} | x \leq 0\}\). To support this claim let us suppose that \(\exists \gamma > 0\) such that \(\gamma \in L\). That is, we should be unable to find \(n^* \in \mathbb{N}\) such that \(\frac{1}{n^*} < \gamma\).

It is very clear that we can offer any \(n^* \in \{n \in \mathbb{N}|n > \frac{1}{\gamma}\}\) as argument against the existence of a rational lower bound \(\gamma > 0\). If we are urged to offer a particular \(n^*\) then we can provide this one: \(n^* = \lceil \frac{1}{\gamma} \rceil + 1\).

Therefore, the set of lower bounds of \(E\) in \(\mathbb{Q}\) indeed is \(L = \{x \in \mathbb{Q} | x \leq 0\}\). Now, \(\text{inf}E = \text{max}L\). That is, \(\text{inf}E = 0\).

 

Theorem 1  

An upper bound of an ordered set is also its supremum and its maximum if and only if that upper bound belongs to the set itself. Similarly, a lower bound of an ordered set is also its infimum and its minimum if and only if that lower bound belongs to the set itself.

 

Proof. Consider the arbitrary non-empty ordered set \(A\). Let us denote by \(\alpha\) the supremum of \(A\). Since it is stated to also be the maximum of \(A\), and since by definition the maximum of a set is an element of the set itself, \(\alpha\) is included in \(A\)

Now let’s go the other way. Let \(\alpha\) be an upper bound of \(A\). By definition, \(\alpha\) belongs to an ordered superset of \(A\), and \(\nexists x \in A, x > \alpha\). But \(\alpha\) is not just any upper bound of \(A\): it is one that happens to be an actual member of \(A\). In summary, \(\alpha\) is an element of \(A\) such that no other element in \(A\) is larger than it. Hence, \(\alpha\) is the maximum of \(A\).

Regarding whether \(\alpha\) is also the supremum of \(A\), consider \(\alpha' < \alpha\). Note that there exists an element in \(A\) greater than \(\alpha'\) —that element is \(\alpha\)—, and thus \(\alpha'\) is not an upper bound of \(A\). Say that again, this time out loud: \(\alpha\) is a lower bound of \(A\) and no value less than it is an upper bound of \(A\). This is just a verbose way of saying that \(\alpha\) is the supremum of \(A\).

The proof concerning the infimum and the minimum of a set is analogous.

 

Definition 1 (Greatest lower bound property) To say that an ordered set \(S\) satisfies the greatest lower bound property means that for any non-empty subset \(B \subseteq S\) bounded below in \(S\) that we can come up with we will find that the set of lower bounds of \(B\) that are in \(S\) does have a maximum.

 

Example 1 Consider the set \(S = (a, b) \cup [c, d)\), with \(a, b, c, d \in \mathbb{R}, c > b\).

Note that all elements of \([c, d)\) are also elements of \(S\). Hence, \(\boldsymbol{[c, d)}\) is a subset of \(\boldsymbol{S}\). It is not empty, either, as it contains all real numbers greater than or equal to \(c\) and strictly lower than \(d\).

The set of lower bounds of \([c, d)\) that also happen to be elements of \(S\) is \((a, b) \cup \{c\} \neq \emptyset\). Hence, \(\boldsymbol{[c, d)}\) is bounded below in \(\boldsymbol{S}\).

In summary, \([c, d)\) is a non-empty subset of \(S\) that is bounded below in \(S\). And do note that the set \(\boldsymbol{(a, b) \cup \{c\}}\) has a maximum: \(\boldsymbol{c}\). Thus, the set \([c, d)\) is not an example that can allow us to dismiss the claim that \(S\) possesses the greatest lower bound property.

Generalizing, let \(B\) be an arbitrary non-empty subset of \(S\) that is bounded below in \(S\). Let \(K\) be the set of all lower bounds of \(B\) —though we haven’t defined yet what \(\mathbb{R}\) is we do have an informal intuition of what it is and we may as well consider \(K\) to be the set of lower bounds of \(B\) in \(\mathbb{R}\)—. The set of lower bounds of \(B\) in \(S\) is \(K \cap S\). We claim \(\boldsymbol{K = (-\infty, a + \epsilon]}\) —with \(\epsilon > 0\)—. Suppose instead that \(K = (-\infty, a + \epsilon)\). This would imply that \(a + \epsilon\) is not a lower bound of \(B\), and therefore \(\exists a' < a + \epsilon \hspace{2mm} \text{such that} \hspace{2mm} a' \in B\). Note that \(a' \in K\) as well. Thus, by Theorem 1 we conclude that \(a'\) is both the minimum and the infimum of \(B\). To claim that \(a'\) is the infimum of \(B\) is the same as claiming \(\text{max}K\) exists and is equal to \(a'\). Assuming \(K = (-\infty, a + \epsilon)\) has led us to deduce \(K = (-\infty, a']\) —a contradiction—, and therefore \(K\) must be an interval closed in its upper limit.

In order to prove that \(S\) satisfies the greatest lower bound property we need to prove that \(K \cap S\) has a maximum. The only possible way that the claim \(\exists \hspace{1mm} \text{max}K \cap S\) could be thwarted is —given the particular set \(S\) that was defined for this Example— by showing that there is a non-empty set \(B \subseteq S\) bounded below in \(S\) such that \(\epsilon = b - a \Longrightarrow K = (-\infty, b] \Longrightarrow K \cap S = (a, b)\).

Let us denote by \(M\) the set of upper bounds of \(K \cap S\) in \(S\). Note that \(M = (S - K) \cup \{\text{max}K \cap S\}\). Since we’re assuming \(K \cap S = (a, b)\) then \(\{\text{max}K \cap S\} = \emptyset\), and therefore \(M = S - K\). That is, \(M = [c, d)\). Note that \(B \subseteq M\).

The following premises are true.

  1. Every lower bound of \(M\) is a lower bound of \(B\).
  2. \(c\) is the minimum of \(M\) and therefore a lower bound of \(M\).

We conclude that \(c\) is a lower bound of \(B\). Also, \(c\) happens to actually be an element of \(S\). That is, \(c \in K \cap S\). However \(c \notin (a, b)\). This implies \(K \cap S \neq (a, b)\). And with that we have proved that \(S\) satisfies the greatest lower bound property.

\[U \cup\]

 

 

Definition 2 (Least upper bound property) This definition is exactly analogous to the previous one. For pedagogical purposes let us state it in terms of the supremum. To say that an ordered set \(S\) satisfies the least upper bound property means that for any non-empty subset \(B \subseteq S\) bounded above in \(S\) that we can come up with we will find that \(\text{sup}B\) exists in \(S\).

 

Theorem 2 (Supremum of upper bounds)  

Let \(S\) be an ordered set with the least upper bound property. Then \(S\) also has the greatest lower bound property.

Perhaps another way of stating this theorem is the following. Let \(S\) be an ordered set with the least upper bound property. Let \(\boldsymbol{L}\) be the set of lower bounds in \(\boldsymbol{S}\) of an arbitrary non-empty subset \(B \subseteq S\) that is bounded below in \(S\). Then \(\text{inf}B = \text{sup}L\). ```4

 

Proof. Let \(S\) be an ordered set that possesses the least upper bound property. Let \(B\) be an arbitrary non-empty subset of \(S\) that is bounded below in \(S\). The set of lower bounds of \(B\) that are in \(S\) is \(L = \{x \in S | \forall y \in B, x \leq y\}\). We need to prove that \(\boldsymbol{L}\) has a maximum.

Since \(B\) is bounded below in \(S\) then it is safe to claim \(L \neq \emptyset\).

Since every element in \(L\) is also an element in \(S\) then \(L \subseteq S\).

Since \(L\) is a set of lower bounds of \(B\) then \(B\) is a set of upper bounds of \(L\). Furthermore, since \(B \subseteq S\) then at least some element of \(S\) is also an element of \(B\). That is, \(L\) is bounded above in \(S\).

In summary, \(S\) has the least upper bound property, and \(L\) is a non-empty subset of \(S\) that is bounded above in \(S\). Therefore the set of upper bounds of \(L\) in \(S\) has a minimum —let’s denote it \(\alpha\)—. Note that either \(L \cap B = \{\alpha\}\) or \(L \cap B = \emptyset\). We need to prove in both cases that \(\alpha\) is the maximum of \(L\).

Case 1. \(\boldsymbol{L \cap B = \{\alpha\}}\)

In this case \(\alpha\) belongs to \(B\) and it is also a lower bound of \(B\) in \(S\). By Theorem 1 \(\alpha\) is both the minimum of \(B\) and the infimum of \(B\) in \(S\). By definition, claiming that \(\alpha\) is the infimum of \(B\) in \(S\) is the same as claiming \(\alpha = \text{max}L\).

Case 2. \(\boldsymbol{L \cap B = \emptyset}\)

From a skeptical standpoint this case implies either that \(L\) does not have a maximum, or \(B\) does not have a minimum, or both. Upon reflecting on this matter, however, the impossibility of \(L\) not having a maximum becomes apparent. Let us denote by \(M\) the set of upper bounds of \(L\) in \(S\). Note that \(B \subseteq M\) and \(\alpha = \text{min}M\). Hence the truth is that no element in \(B\) is smaller than \(\alpha\), and thus \(\alpha\) is a lower bound of \(B\) —that is, \(\alpha \in L\)—. We conclude \(L \cap M = \alpha\) —that is, \(\alpha\) happens to be both an element of \(L\) and an upper bound of \(L\)—. By Theorem 1 \(\alpha\) is both the supremum and the maximum of \(L\).

Additionally, by Theorem 1 we claim \(\text{max}L = \text{sup}L\). On the other hand, —by definition of the infimum— \(\text{max}L = \text{inf}B\). Therefore, \(\text{sup}L = \text{inf}B\).

 

Theorem 3 (Infimum of upper bounds)  

Let \(S\) be an ordered set with the greatest lower bound property. Then \(S\) also has the least upper bound property.

 

Proof. The steps of the proof are exactly analogous to the steps of the previous proof.

 

Exercise 2 Let \(A\) be a non-empty set of real numbers which is bounded below. Consider \(-A = \{-x | x \in A\}\). Prove that \(\text{inf}A = -\text{sup}(-A)\).

 

The set of lower bounds of \(A\) is \(L = \{\}\).