Definition 1.2.1.
We say that the number \(B\) is an upper bound for the set \(A\) if \(\forall x\in A, x\leq B\text{.}\)
We say that the number \(b\) is a lower bound for the set \(A\) if \(\forall x\in A, b\leq x\text{.}\)
Section 1.2 Completeness
Proposition 1.1.28 is sometimes described as telling us about a "hole" in \(\mathbb{Q}\text{.}\) We’re going to fill in that hole. The most obvious way to do this might seem to be, as when we constructed \(\mathbb{Q}\) from \(\mathbb{Z}\text{,}\) simply to add some symbols. And indeed, this is a fine thing to do. If you take an abstract algebra course or two, you encounter fields like \(\mathbb{Q}\left(\sqrt{2}\right)\text{,}\) which plug holes corresponding to unsolvable equations by adding a symbol which we declare to be a solution. Galois theory is, essentially, the study of which such symbols to add in order to be able to solve various classes of equations.
But this is an analysis class, so we’re going to do something else entirely.
If we take a look back at Figure 1.1.30, we can frame the problem like this:
Starting at one corner of the unit square and proceding directly toward the opposite corner, how far can we go before leaving the square?
If we, like Pythagoras, take "length" to mean a rational number, then the answer is: any rational number \(r\) with \(r^2\lt 2\text{.}\) On the other hand, if we move a distance \(r\) with \(r^2\gt 2\text{,}\) we’ve definitely gone too far. The sweet spot would be \(r^2=2\text{,}\) but as we saw in Proposition 1.1.28, that can’t happen. Thus we’re led to consider sets like
\begin{equation*}
\left\{r\in\mathbb{Q}\middle\vert r^2\lt 2\right\}
\end{equation*}
or
\begin{equation*}
\left\{r\in\mathbb{Q}\middle\vert r^2\gt 2\right\}
\end{equation*}
and ask: what’s the largest or smallest element of such a set.
Subsection 1.2.1 Maxima, minima, suprema, infima
Let’s start thinking about subsets of an ordered field and elements that relate to those subsets. Let’s fix an ordered field \((S,+,\times,\lt)\) to work with. Throughout, when I say "set", I’ll mean "subset of \(S\)". When I say "number", I’ll mean "element of \(S\)".
Checkpoint 1.2.2.
Show that the set \(A\) is bounded if and only if there is \(M\) with \(\forall x\in A, \lvert x\rvert\leq M\text{.}\)
Example 1.2.3.
The set \(A=\left\{r\in\mathbb{Q}\middle\vert r^2\lt 2\right\}\) has \(B=10\) as an upper bound and \(b=-2\) as a lower bound.
Definition 1.2.4.
We say that \(M\) is a maximum element of the set \(A\) if \(M\in A\) and \(M\) is an upper bound for \(A\text{.}\)
We say that \(m\) is a minimum element of the set \(A\) if \(m\in A\) and \(m\) is a lower bound for \(A\text{.}\)
Example 1.2.5.
It’s important to note that some sets don’t have a maximum.
Checkpoint 1.2.6.
These sets do not have maxima. Explain why in each case.
-
\(\displaystyle \left\{r\in\mathbb{Q}\middle\vert r^2\gt 0\right\}\)
-
\(\displaystyle (-\infty,0)\)
-
\(\displaystyle \left\{r\in\mathbb{Q}\middle\vert r^2\lt 2\right\}\)
-
\(\displaystyle \left\{r\in\mathbb{Q}\middle\vert r^2\leq 2\right\}\)
Checkpoint 1.2.8.
Show that minima and maxima are unique. That is, if a set \(A\) has a minimum, there is only one; and if \(A\) has a minimum, there is only one.
Hint.
As with any uniqueness proof, try to start by supposing you’ve got a set \(A\) with a maximum \(M_1\) and a maximum \(M_2\text{.}\) Then use Definition 1.2.4 to get to \(M_1=M_2\text{.}\)
Remark 1.2.9.
Checkpoint 1.2.8 justifies us using the definite article the to refer to maxima and minima.
Definition 1.2.4 is a bit like the phrase "punk unicorn". We understand what it means, but that’s not reason to imagine one actually exists. As Checkpoint 1.2.6 shows, some sets do not have maxima. (It’s not hard to cook up sets without minima, either.) But bounded sets like \((-2,8)\) do seem to have an edge -- namely, their endpoints. 8 is sort of like a maximum for \((-2,8)\text{;}\) the only problem is, \(8\notin(-2,8)\text{.}\)
Checkpoint 1.2.10.
The set \((-2,8)\) is bounded because it has an upper and a lower bound. Let’s consider two different upper bounds \(B_1\) and \(B_2\text{.}\) If I tell you that \(B_2\) is a "better" upper bound, what do I mean?
Here’s a definition which formalizes this relationship:
Definition 1.2.11.
We call the number \(s\) a least upper bound or supremum for the set \(A\) if
-
\(s\) is an upper bound for \(A\text{,}\) and
-
for any upper bound \(B\) for \(A\text{,}\) \(s\leq B\text{.}\)
We call the number \(i\) a greatest lower bound or infimum for the set \(A\) if
-
\(i\) is a lower bound for \(A\text{,}\) and
-
for any lower bound \(b\) for \(A\text{,}\) \(i\geq b\text{.}\)
Checkpoint 1.2.12.
Show that if \(A\) has a least upper bound, then this least upper bound is unique.
Explain why this result licenses us to refer to the least upper bound for \(A\text{.}\)
Convention 1.2.13.
If \(s\) is a (the!) least upper bound for \(A\text{,}\) we write
\begin{equation*}
s=\sup A\ \ .
\end{equation*}
This is pronounced "soup A".
If \(i\) is a (the!) greatest lower bound for \(A\text{,}\) we write
\begin{equation*}
i=\inf A\ \ .
\end{equation*}
Example 1.2.14.
\begin{gather*}
\inf (-2,8)=-2\\
\sup (-2,8)=8
\end{gather*}
Proof.
\(\inf (-2,8)=-2\).
We have two things to do: we must prove that \(-2\) is a lower bound, and then we must prove it is greatest among such.
Recall the definition of \((-2,8)\text{:}\)
\begin{equation*}
(-2,8)=\left\{x\in\mathbb{Q}\middle\vert -2\lt x\lt 8\right\}
\end{equation*}
So any \(x\in (-2,8)\) has \(-2\lt x\text{.}\) But this immediately implies that \(-2\leq x\text{.}\) So \(-2\) is a lower bound.
Now to show that \(-2\) is greatest amoung the lower bounds for \((-2,8)\text{.}\) Let \(b\) be some lower bound for \((-2,8)\text{.}\) Suppose, by way of contradiction, that \(b\gt -2\text{.}\) Consider the number
\begin{equation*}
b*=\frac{1}{2}\left(-2+b\right)\ \ .
\end{equation*}
\(b^*\lt b\) and \(b^*\gt -2\text{.}\) Therefore \(b^*\in(-2,8)\text{;}\) but that means \(b\) couldn’t have been a lower bound for \((-2,8)\) after all.
Having obtained a contradiction, we see that \(-2\) is greatest.
Checkpoint 1.2.15.
Find the infimum and supremum of the set
\begin{equation*}
\left\{r\in\mathbb{Q}\middle\vert r^2\lt 4\right\}\ \ .
\end{equation*}
Prove that your answer is correct.
Convention 1.2.16.
If a set \(A\) is not bounded above, we write
\begin{equation*}
\sup A=\infty\ \ .
\end{equation*}
If a set \(A\) is not bounded below, we write
\begin{equation*}
\inf A=-\infty\ \ .
\end{equation*}
Subsection 1.2.2 When suprema exist
Example 1.2.17.
The set
\begin{equation*}
A=\left\{r\in \mathbb{Q}\middle\vert r^2\lt 2\right\}
\end{equation*}
has an upper bound, but has no least upper bound.
Proof.
\(A\) has an upper bound.
Let \(r\in A\text{.}\) Since \(1\in A\text{,}\) we may as well take \(r\geq 1\text{,}\) and in particular, \(r\gt 0\text{.}\)
Now suppose \(r\gt 2\text{.}\) Using the ordered field axioms, this would mean
\begin{equation*}
r^2\gt 2r\gt 2\cdot 2=4
\end{equation*}
But by assumption \(r^2\lt 2\text{.}\) So it could not have been that \(r\gt 2\text{,}\) hence 2 is an upper bound.
Observe that this argument applies to any \(B\) with \(B^2\gt 2\text{;}\) we just used \(B=2\) for convenience.
\(A\) has no least upper bound.
Consider a number \(a\text{,}\) which we’ll be more careful about later. We’ll assume that \(a\gt 1\text{.}\)
Consider the number
\begin{equation*}
r=a+\frac{2-a^2}{8a}\ \ .
\end{equation*}
We’ll call \(\frac{2-a^2}{8a}=h\text{.}\)
If \(a\in A\text{,}\) then \(h\gt 0\text{.}\) Further, if \(a\gt 1\text{,}\) then \(2-a^2\lt 1\text{.}\)
Compute \(r^2\text{:}\)
\begin{align*}
r^2&=a^2+2ah+h^2\\
&=a^2+\frac{2-a^2}{4}+\frac{(2-a^2)^2}{64a^2}
\end{align*}
Since \(2-a^2\lt 1\text{,}\) we have \((2-a^2)^2\lt 2-a^2\text{.}\) So we get the inequality:
\begin{align*}
r^2&=a^2+\frac{2-a^2}{4}+\frac{(2-a^2)^2}{64a^2}\\
&\lt a^2+\frac{2-a^2}{4}+\frac{2-a^2}{64}\\
&=a^2+\frac{17}{64}\left(2-a^2\right)
\end{align*}
which shows that \(r^2\) is less than halfway between \(a^2\) and \(2\text{.}\) Hence, \(r\in A\text{.}\) And because \(r=a+h\text{,}\) \(r>a\text{.}\)
This machine we’ve just built takes in \(a\in A\) and produces \(r\in A\) with \(r\gt a\text{.}\) Therefore, no element of \(A\) is an upper bound for \(A\).
Now consider any old upper bound \(B\text{.}\) By the comment above, we need \(B^2\gt 2\text{.}\) Now consider
\begin{equation*}
r=B+\frac{2-B^2}{8B}
\end{equation*}
and, as above, call \(h=\frac{2-B^2}{8B}\text{.}\) Notice that now \(h\lt 0\) and \(r\lt B\text{.}\)
Compute \(r^2\text{:}\)
\begin{align*}
r^2&=B^2+\frac{2-B^2}{4}+h^2\\
&\gt B^2+\frac{2-B^2}{4}
\end{align*}
which means that \(r^2\) is less than \(B^2\text{,}\) but not by more than one-quarter of the distance between \(2\) and \(B^2\text{.}\) So, in particular, \(r^2\gt 2\text{,}\) which means that \(r\) is an upper bound for \(A\text{.}\)
So we’ve build a machine which, given any upper bound \(B\) for \(A\) with \(B\gt 2\text{,}\) finds an upper bound which is less than \(B\text{.}\) Therefore, in order for an upper bound to be least, we need \(B^2\leq 2\text{.}\)
Taking these two into account, we see that in order to be least, a number must have \(B^2\leq 2\text{;}\) while in order to be an upper bound, a number must have \(B^2\geq 2\text{.}\) The only possibility for a least upper bound \(B\text{,}\) therefore, is one with \(B^2=2\text{.}\) But by Proposition 1.1.28, this can’t happen.
Remark 1.2.18.
Compare Checkpoint 1.2.15 and Example 1.2.17. Notice how subtle the difference is between the two sets!
Example 1.2.17 shows that the problem we have with \(\mathbb{Q}\) can be phrased in terms of least upper bounds. That motivates the following definition:
Definition 1.2.19.
We say that an ordered set \(S\) is complete or has the least upper bound property if every subset of \(S\) which
-
is nonempty and
-
has an upper bound
has a least upper bound.
Checkpoint 1.2.20.
We say that an ordered set \(S\) has the greatest lower bound property if every nonempty subset of \(S\) which is bounded below has a greatest lower bound.
Show that an ordered set has the least upper bound property if and only if it has the greatest lower bound property.
Fact 1.2.21.
There is a unique-up-to-isomorphism complete ordered field.
Definition 1.2.22.
We write \(\mathbb{R}\) for the unique complete ordered field.
What good is the completeness property? Let’s first establish that it fills in the gaps we identified in \(\mathbb{Q}\text{.}\)
Proposition 1.2.23.
For any nonnegative \(t\in\mathbb{R}\text{,}\) there is a solution in \(\mathbb{R}\) to the equation
\begin{equation*}
x^2=t\ \ .
\end{equation*}
Proof.
For \(t\gt 0\text{,}\) we consider the set \(A_t=\left\{x\in\mathbb{R}\middle\vert x^2\lt t\right\}\text{.}\) Because \(0\in A_t\text{,}\) \(A_t\) is nonempty. \(B=t+1\) is an upper bound for \(A_t\text{.}\) Therefore, because \(\mathbb{R}\) is complete, there is a least upper bound \(s=\sup A_t\text{.}\)
Because \(s\) is an upper bound for \(A_t\text{,}\) we have \(s^2\geq t\text{.}\) On the other hand, because \(s\) is least, \(s^2\leq t\text{.}\)
Therefore \(s^2=t\text{.}\)
