Consequences of completeness

Section 1.3 Consequences of completeness

Besides the existence of square roots, completeness buys us many other very interesting and useful properties. I’ve collected them all here.

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These consequences differ in flavor, but what they have in common is that their proofs involve one of two strategies:

the direct approach 🔗

defining a subset of \(\mathbb{R}\text{.}\)

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showing that that subset has a supremum or infimum

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showing that the existence of that supremum or infimum implies the statement at hand

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the reductio 🔗

suppose what we want to show were false

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consider some set which has a supremum

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show that that supremum is either not an upper bound or not least

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Subsection 1.3.1 Characterizations of suprema

Showing that a particular number is a least upper bound directly, by showing leastness and upper boundness, is sometimes a bit tricky (cf. Example 1.2.17). So we’ll often use one of the following equivalent characterizations of what it means to be a supremum.

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Proposition 1.3.1. \(\epsilon\)-characterization of supremum.

Let \(A\) be a subset of \(\mathbb{R}\text{.}\) Then \(s=\sup A\) if and only if:

\(s\) is an upper bound for \(A\text{,}\) and

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for any \(\epsilon\gt 0\text{,}\) there is some \(a_\epsilon\in A\) so that \(s-a_\epsilon\lt\epsilon\)

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Proof.

\(s=\sup A\) guarantees (i) and (ii).

(i) is trivial.

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To prove (ii), let \(\epsilon>0\text{.}\) If there were no elements \(a\) of \(A\) with \(s-a\lt \epsilon\text{,}\) then every element of \(A\) would have \(s-a\geq \epsilon\text{,}\) in other words, \(s-\epsilon\geq a\text{.}\) So \(s-\epsilon\) would be an upper bound for \(A\text{.}\) But \(s-\epsilon\) and \(s\) was supposed to be least. So this cannot happen.

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(i) and (ii) guarantee \(s=\sup A\).

Since we assume by (i) that \(s\) is an upper bound for \(A\text{,}\) we need only prove that \(s\) is least.

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Let \(B\) be some upper bound for \(A\text{.}\) If \(B=s\text{,}\) then \(s\leq B\text{.}\) So we may as well assume that \(B\neq s\text{.}\) We need to show that \(B\gt s\text{.}\) Suppose, for a contradiction, that \(B\lt s\text{.}\) Choose \(\epsilon=s-B\text{.}\) Then by assumption (ii), there is a \(a_\epsilon\in A\) so that

\begin{equation*} s-a_\epsilon\lt \epsilon=s-B \end{equation*}

and rearranging we obtain

\begin{equation*} B\lt a_\epsilon\ \ , \end{equation*}

which means that \(B\) wasn’t an upper bound after all. This contradiction establishes that \(B\gt s\text{.}\) Since \(B\) was any upper bound, we’ve shown that \(s\) is least.

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Remark 1.3.2.

This is our first encounter with the letter \(\epsilon\text{,}\) but it won’t be the last. Notice that here, we had an inequality \(B\lt s\text{,}\) which motivated our choice \(\epsilon=s-B\text{.}\) That sort of thing is going to keep happening.

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Checkpoint 1.3.3.

Formulate an \(\epsilon\)-characterization of infimum.

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Adapt the proof of Proposition 1.3.1 to prove your characterization.

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Proposition 1.3.4. Two-\(\epsilon\)-characterization of supremum.

Let \(A\) be a subset of \(\mathbb{R}\text{.}\) Then \(s=\sup A\) if and only if:

for any \(\epsilon\gt 0\) there is \(B_\epsilon\) which is an upper bound for \(A\) and \(B_\epsilon-s\lt \epsilon\text{,}\) and

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for any \(\epsilon\gt 0\) there is \(a_\epsilon\in A\) with \(s-a_\epsilon\lt \epsilon\text{.}\)

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Proof.

You do this one.

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Theorem 1.3.5. Nested Intervals Theorem.

Let \(\mathcal{I}=\left\{I_n\right\}_{n\in\mathbb{N}}\) be a collection of closed intervals with the property that

\begin{equation*} I_1\supseteq I_2\supseteq\cdots I_{k}\supseteq I_{k+1}\cdots \end{equation*}

(such a collection is called nested) so that each \(I_n\) is nonempty.

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Then

\begin{equation*} \bigcap_{n\in\mathbb{N}} I_n \end{equation*}

is nonempty.

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Proof.

Label the endpoints of each closed interval so that

\begin{equation*} I_n=\left[\alpha_k,\beta_k\right]\ \ . \end{equation*}

Observe that the nested condition means that

\begin{equation*} \alpha_1\leq \alpha_k\leq\beta_k\leq\beta_1 \end{equation*}

for all \(k\text{.}\) So the set \(A=\left\{\alpha_k\middle\vert k\in\mathbb{N}\right\}\) is bounded above and \(B=\left\{\beta_k\middle\vert k\in\mathbb{N}\right\}\) is bounded below.

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Set \(\alpha=\sup A\) and \(\beta=\inf B\text{.}\) The claim is any \(p\in\left[\alpha,\beta\right]\) is in the intersection \(\bigcap_{n\in\mathbb{N}} I_n\text{.}\)

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To see this, notice that if \(p\in[\alpha,\beta]\text{,}\) then \(p\geq \alpha\text{.}\) Since \(\alpha\) is an upper bound for \(A\text{,}\) this means \(p\geq \alpha_k\) for all \(k\text{.}\) Similarly, \(p\leq\beta\leq\beta_k\text{.}\) So \(\alpha_k\leq p\leq\beta_k\text{,}\) that is, \(p\in I_k\text{.}\)

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We need to show that \([\alpha,\beta]\) is nonempty, which is equivalent to \(\alpha\leq\beta\text{.}\) Take any \(\beta_k\in B\) and any \(\alpha_n\in A\text{.}\) If \(n\geq k\text{,}\) then

\begin{equation*} \alpha_k\leq\alpha_n\leq\beta_n\leq \beta_k\ \ ; \end{equation*}

on the other hand if \(n\leq k\text{,}\) then

\begin{equation*} \alpha_n\leq\alpha_k\leq\beta_k\leq\beta_n\ \ . \end{equation*}

In either case, \(\alpha_n\leq \beta_k\text{.}\) So each element \(\alpha_n\in A\) is a lower bound for \(B\text{,}\) hence is less than or equal to \(\beta=\inf B\) (because \(\inf B\) is the greatest lower bound). Therefore, \(\alpha_n\leq \beta\) for all \(n\text{.}\) But this makes \(\beta\) an upper bound for \(A\text{.}\) Since \(\alpha=\sup A\) is the least upper bound, this means \(\alpha\leq \beta\text{.}\)

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Checkpoint 1.3.6.

We can actually go further in the proof of Theorem 1.3.5. Show that

\begin{equation*} \bigcap_{n\in\mathbb{N}} I_n=[\alpha,\beta] \end{equation*}

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Remark 1.3.7.

The Nested Intervals Theorem is an example of a trapping theorem. It (and its sequels) will allow us to prove that a desired number exists, by constructing a sequence of intervals \(I_n\) which collapse down onto that number.

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Subsection 1.3.2 Archimedean properties

“Give me a lever long enough and a fulcrum on which to place it and I shall move the world.”―Archimedes of Syracuse (fl. 200s BCE), as quoted in Pappus’ Synagoge VIII (ca. 340 CE)
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Consider the problem of measuring with a ruler. Say we want to measure the length of a meter stick, using a 1-foot ruler. We set the ruler along the meter stick at one end, mark the end of the ruler, slide the ruler down, mark its end, and so on. We’ll be able to lay off the ruler three times, and the fourth time, our 1-foot ruler will stick out beyond the end of the meter stick. So we know that a meter is between 3 and 4 feet long.

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What if our ruler were only 8 inches long? We’d get a different answer but the process is just the same -- we keep laying off copies of the ruler until we go a little past the end of the meter stick.

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And the same will work, at least in principle, if we wanted to measure a football field with a ruler whose length was 4.5 inches. No matter what size ruler we use, or how large the object we want to measure, we know in our guts that we’ll be able to do it (even if it takes a long time and great care).

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Numbers, of course, are about measurement. So we’d better be able to show that, in a complete ordered field, there isn’t a ruler too short or a distance too long to be measured.

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Proposition 1.3.8.

The natural numbers, as a subset of \(\mathbb{R}\text{,}\) are not bounded.

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Proof.

Suppose \(\mathbb{N}\) were bounded. Then there would be \(B=\sup \mathbb{N}\text{.}\) Using \(\epsilon=1\) in the \(\epsilon\)-characterization, there is some natural number \(M\) with \(B-1\lt M\lt B\text{.}\)

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If we add 1 across the board, we get

\begin{equation*} B\lt M+1\lt B+1\ \ . \end{equation*}

But \(M+1\) is a natural number, so \(B\) isn’t an upper bound after all.

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This contradiction establishes the claim.

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Remark 1.3.9.

At first glance, Proposition 1.3.8 seems obvious. We know that the natural numbers just keep going, don’t we? Well, yes. But remember that \(\mathbb{R}\) contains many new elements. What if one of them just happened to be all the way to the right of all the natural numbers? So there actually is something to prove here.

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Proposition 1.3.10.

For any \(x,y\in\mathbb{R}\text{,}\) with \(x\gt 0\text{,}\) there is \(n\in\mathbb{N}\) so that \(nx\gt y\text{.}\)

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Proof.

First consider the case that \(y\leq 0\text{.}\) Then \(n=1\) works. So from now on, we can assume \(y\gt 0\text{.}\)

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If the proposition were false, there would be \(x,y\) so that \(nx\leq y\) for all \(n\in\mathbb{N}\text{.}\) In other words, \(A=\left\{nx\middle\vert n\in\mathbb{N}\right\}\) is bounded above. By completeness, this set has a least upper bound, which we’ll call \(B\text{.}\)

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Consider \(B-x\text{.}\) Since \(x\gt 0\text{,}\) \(B-x\lt B\text{.}\) So there is \(N\in\mathbb{N}\) so that

\begin{equation*} B-x\lt Nx\ \ , \end{equation*}

which we can rearrange to

\begin{equation*} B\lt (N+1)x\ \ . \end{equation*}

But \((N+1)x\in A\text{,}\) which means that \(B\) wasn’t an upper bound to begin with.

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This contradiction establishes the proposition.

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Checkpoint 1.3.11.

Proposition 1.3.10 is what powers us to measure things with rules; and what justifies Archimedes’ confidence in his world-moving abilities.

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Among \(x\text{,}\) \(y\text{,}\) and \(n\text{,}\) which one is the length we’re measuring, which is the length of the ruler, and which is the measurement we report?

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Proposition 1.3.12.

Given any \(\epsilon\gt 0\) there is a natural number \(n\) so that \(\frac{1}{n}\lt \epsilon\text{.}\)

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Proof.

Suppose not. That is, there would be some special \(\epsilon\gt 0\) so that \(\frac{1}{n}\geq \epsilon\) for all natural numbers \(n\text{.}\) But then \(\frac{1}{\epsilon}\geq n\) for all \(n\in \mathbb{N}\text{,}\) i.e. \(\frac{1}{\epsilon}\) would be an upper bound for the natural numbers.

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Proposition 1.3.13. \(\frac{1}{n}\)-characterization of supremum.

Let \(A\) be a subset of \(\mathbb{R}\text{.}\) Then \(s=\sup A\) if and only if:

\(s\) is an upper bound for \(A\text{,}\) and

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for any \(n\in\mathbb{N}\text{,}\) there is \(a_n\in A\) with \(s-a_n\lt\frac{1}{n}\)

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Proof.

This will be a class activity.

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Proposition 1.3.14. density of \(\mathbb{Q}\) in \(\mathbb{R}\).

Given any \(x\lt y\in\mathbb{R}\text{,}\) there is \(r\in\mathbb{Q}\) with \(x\lt r\lt y\text{.}\)

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Proof.

If \(x\lt 0\lt y\text{,}\) then we use \(r=0\text{.}\)

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If \(x=0\lt y\text{,}\) then Proposition 1.3.12 gives a rational of the form \(\frac{1}{n}\text{.}\)

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If \(x\lt y=0\text{,}\) then applying Proposition 1.3.12 gives \(\frac{1}{n}\) with \(0\lt \frac{1}{n}\lt -x\text{,}\) so \(x\lt -\frac{1}{n}\lt 0\text{.}\)

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Consider the case \(0\lt x\lt y\text{.}\) Then \(y-x\gt 0\text{,}\) so Proposition 1.3.12 gives \(n\in\mathbb{N}\) with \(\frac{1}{n}\lt y-x\text{,}\) or equivalently,

\begin{equation*} 1\lt ny-nx\ \ , \end{equation*}

which we’ll rearrange as

\begin{equation*} nx+1\lt ny\ \ . \end{equation*}

Now because \(\mathbb{N}\) is not bounded, there is some natural number \(m\) with \(nx\lt m\text{.}\) In fact, let’s choose the smallest such natural number. Because \(nx\) and \(ny\) are at least 1 apart, we must have \(m\lt ny\text{.}\)

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Thus we have

\begin{equation*} nx\lt m\lt ny\ \ , \end{equation*}

and dividing through, we get

\begin{equation*} x\lt \frac{m}{n}\lt y \end{equation*}

as desired.

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The remaining case is what happens with \(x\lt y\lt 0\text{.}\) But then \(0\lt -y\lt -x\text{,}\) so by the case we just handled, there is \(q\in\mathbb{Q}\) with \(-y\lt q\lt -x\text{.}\) Use \(r=-q\text{.}\)

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Remark 1.3.15.

We say that \(\mathbb{Q}\) is dense in \(\mathbb{R}\text{.}\)

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It’s also true that \(\mathbb{Q}\) is dense in itself -- given \(x,y\in\mathbb{Q}\text{,}\) \(\frac{x+y}{2}\in\mathbb{Q}\text{.}\) By contrast, \(\mathbb{N}\) is not dense in itself, or in \(\mathbb{Q}\text{,}\) or in \(\mathbb{R}\text{.}\)

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Checkpoint 1.3.16.

Show that the irrationals are dense in \(\mathbb{R}\text{.}\)

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Hint.

Given \(x\lt y\text{,}\) we know that \(\sqrt{2}x\lt \sqrt{2} y\text{.}\)

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