3.2 Bounds for sets
With this notion of absolute value, we can start asking whether a subset of \(\mathbb{Q}\) contains arbitrarily large or small elements.
Let \(A \subseteq \mathbb{Q}\) be non-empty. We that that \(A\) is:
bounded above (in \(\mathbb{Q}\)) by \(\alpha \in \mathbb{Q}\) if for all \(x\in A\), \(x\leq \alpha\);
bounded below (in \(\mathbb{Q}\)) by \(\alpha \in \mathbb{Q}\) if for all \(x\in A\), \(x\geq \alpha\);
bounded if it is bounded above and below;
If \(A\) is bounded above by \(\alpha\) and below by \(\beta\), then by setting \(\gamma = \max\{|\alpha|,|\beta|\}\) we have \(A\) is bounded by \(\gamma\), i.e., for all \(x\in A\), we have \(|x|\leq \gamma\).
Note that \(\alpha\) is far from unique. For example, take the set \(A = \left\{\frac{1}{n}:n\in \mathbb{Z}_+\right\}\). Then we can see that \(A\) is bounded above by \(1\), but it is also bounded above by \(2\) and by \(100\) etc.
Let \(A \subseteq \mathbb{Q}\) be non-empty. The least (or smallest) upper bound of \(A\) (in \(\mathbb{Q}\)) is \(\alpha \in \mathbb{Q}\) such that:
\(\alpha\) is an upper bound, i.e. for all \(x\in A\), \(x\leq \alpha\);
any rational number \(\beta\) less than \(\alpha\) is not an upper bound, i.e. for all \(\beta \in \mathbb{Q}\) with \(\beta < \alpha\), there exists \(x\in A\) with \(\beta<x\).
The greatest (or largest) lower bound of \(A\) (in \(\mathbb{Q}\)) is \(\alpha \in \mathbb{Q}\) such that:
\(\alpha\) is a lower bound, i.e. for all \(x\in A\), \(x\geq \alpha\);
any rational number \(\beta\) greater than \(\alpha\) is not a lower bound, i.e. for all \(\beta \in \mathbb{Q}\) with \(\beta > \alpha\), there exists \(x\in A\) with \(\beta>x\).
We use our work on negating statements to negate the above definition and say
\(\alpha \in \mathbb{Q}\) is not the least upper bound of \(A\) if:
there exists \(x\in A\) such that \(x>\alpha\) (\(\alpha\) is not an upper bound) or;
there exists \(\beta \in \mathbb{Q}\) with \(\beta < \alpha\) and for all \(x\in A\) we have \(x\leq \beta\) (there is an upper bound lower than \(\alpha\)).
Similarly, \(\alpha \in \mathbb{Q}\) is not the greatest lower bound of \(A\) if:
there exists \(x\in A\) such that \(x<\alpha\) (\(\alpha\) is not a lower bound) or;
there exists \(\beta \in \mathbb{Q}\) with \(\beta > \alpha\) and for all \(x\in A\) we have \(x\geq \beta\) (there is an lower bound greater than \(\alpha\)).
Let \(A = \left\{\frac{1}{n}:n\in \mathbb{Z}_+\right\}\). We show that \(1\) is the least upper bound of \(A\). As remarked before, we have \(1\) is an upper bound since if we take \(x \in A\) then \(x = \frac{1}{n}\) with \(n\in \mathbb{Z}_+\). In particular \(n\geq 1\), so \(x=\frac{1}{n}\leq \frac{1}{1}=1\).
We now show that \(1\) is the least upper bound by showing any number less than \(1\) is not an upper bound. Let \(\beta <1\). By taking \(n=1 \in \mathbb{Z}_+\), we see that \(1=\frac{1}{1} \in A\), hence \(\beta<1\) means \(\beta\) is not an upper bound. Hence \(1\) is the least upper bound.
We show that \(0\) is the greatest lower bound. First we show \(0\) is a lower bound. Let \(x \in A\) then \(x = \frac{1}{n}\) with \(n\in \mathbb{Z}_+\). In particular \(n > 0\), so \(x=\frac{1}{n}> 0\).
We now show that \(0\) is the greatest lower bound by showing any number greater than \(0\) is not a lower bound. Let \(\beta = \frac{a}{b} >0\) [so \(a,b \in \mathbb{Z}_+\)]. Set \(n=b+1 \in \mathbb{Z}\), so \(\frac{1}{n}\in A\). Then \[\frac{1}{n}=\frac{1}{b+1}<\frac{1}{b}\leq a\frac{1}{b} = \frac{a}{b} = \beta.\]
So we have found \(x \in A\) such that \(x<\beta\), so \(\beta\) is not a lower bound.
The argument for \(\beta\) not being a lower bound above seems to come from nowhere. Sometimes it is hard to see where to start a proof, so mathematician first do scratch work. This is the rough working we do as we explore different avenues and arguments, but that is not included in the final proof (so to keep the proof clean and easy to understand). The scratch work for the above proof of might have been along the lines:
We want to find \(n \in \mathbb{Z}_+\) such that \(\beta =\frac{a}{b}>\frac{1}{n}\). Rearranging, this gives \(an>b\) (as both \(n\) and \(b\) are positive), so \(n>\frac{b}{a}\). Since \(a\geq 1\), we have \(\frac{b}{a}\leq b\). Picking \(n=b+1\) would satisfy \(n=b+1>b\geq\frac{b}{a}\).□
Note that in the example above the greatest lower bound of \(A\) is not in \(A\) itself [if \(0\in A\), then there exists \(n\in \mathbb{Z}_+\) such that \(0=\frac{1}{n}\), i.e. \(0=1\), which is a contradiction.]. If a set is a bounded subset of \(\mathbb{Z}\), then we get a different story.
Any non-empty subset of \(\mathbb{Z}_{\geq 0}\) contains a minimal element.
Exercise to be done after we have introduced the completeness axiom.
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Let \(A \subseteq \mathbb{Z}\) be non-empty. If \(A\) is bounded below, it contains a minimal element (i.e. its greatest lower bound is in \(A\)). If it is bounded above, it contains a maximal element (i.e., its least upper bound is in \(A\)).
As we saw above with the set \(A = \left\{\frac{1}{n}:n\in \mathbb{Z}_+\right\}\), this is not the case for general subsets of \(\mathbb{Q}\). However, it is even worst, because a general subset of \(\mathbb{Q}\) might be bounded and not have a greatest lower bound or least upper bound in \(\mathbb{Q}\), as we will see in the next section.