Sets of Real Numbers

Abstract

Most of us believe we have a reasonable understanding of what a real number is. However, sets of real numbers have some deep and surprising properties. We will explore some of these properties in this chapter and in Chap. 4 on cardinality. Of notable interest for future applications are the order completeness of $\mathbb {R},$ the characterization of intervals in terms of the intermediate value property, and the nested interval theorem. Among other results we establish the existence of roots of positive real numbers and we introduce the amazing Cantor set as well as related functions.

Appendices

Problems

1.1 Problems for Sect. 3.1

• 1. Any finite set has a maximum.

• 2. If $A$ has a maximum, then $\sup (A)=\max (A).$

• 3. If $r$ is irrational and $A:=]-\infty ,r[,$ then $\sup (A)=r.$ In particular, $\sup (A)$ is not rational.

• 4. Let $A$ be a subset of $\mathbb {R}.$ Let $u$ be a real numbers such that

  Prove $U=\sup (A),$ if

  • a. $A\cap ]U,\infty [=\emptyset$ and

  • b. $A\cap ]V,\infty [\neq \emptyset ,$ for any $V<U.$

• 5. Let $A$ be a set of real numbers and let $m$ be a real number. Suppose $A\cap \left ]M,\infty \right [=\emptyset$ and

  $$ \forall\varepsilon>0,A\cap\left]M-\varepsilon,\infty\right[\neq\emptyset. $$

  Prove $m$ is the least upper bound of $A.$

• 6. Let $A$ be a set and let $g:A\to [0,\infty [$ be some function. Suppose there exists $M\geq 0,$ such that

  $$ \sum_{b\in B}g(b)\leq M $$

  for all finite subsets $B$ of $A.$ Show there is a real number $L,$ such that

  $$ \forall\varepsilon>0,\exists\text{finite }B\subseteq A,\forall\text{finite }C\subseteq A,B\subseteq C\implies0\leq L-\sum_{c\in C}g(c)<\varepsilon. $$

  This is usually abbreviated as $\sum _{a\in A}g(a)=L.$ [Hint: By assumption the set

  $$ S:=\left\{ \sum_{b\in B}g(b)\mid B\text{ a finite subset of }A\right\} $$

  has $m$ as an upper bound. Let $L$ be the supremum of $S.$]

1.2 Problems for Sect. 3.2

• 1. Prove $A:=]0,1[\cap \mathbb {Q}$ does not have the intermediate value property.

• 2. Let $f:\mathbb {R}\to \mathbb {R}$ be increasing, i.e., $x\leq y\implies f(x)\leq f(y).$ Fix a real number $y_{0}.$ Let

  $$ A:=\left\{ x\in\mathbb{R}\mid y_{0}\leq f(x)\right\} . $$

  Suppose $A$ is nonempty. Prove $A$ is an interval. [Hint: $f$ need not be onto.]

1.3 Problems for Sect. 3.3

• 1. Find two different sequences of closed intervals $I_{k}$ as in (3.1) whose intersection equals $\{1.24\overline {0}\}.$

• 2. Use the Nested Interval Theorem to prove: If $0\leq x$ and $x\leq 1/n$ for all $n\in \mathbb {N},$ then $x=0.$

• 3. Consider the open intervals $I_{n}:=\left ]0,\tfrac {1}{n}\right [.$ Since $\tfrac {1}{n+1}<\tfrac {1}{n}$ the intervals are nested: $I_{n+1}\subset I_{n}.$ Find $\bigcap _{n=1}^{\infty }I_{n}.$

• 4. Consider the closed intervals $I_{n}:=\left [n,\infty \right [.$ Since $n<n+1$ the intervals are nested: $I_{n+1}\subset I_{n}.$ Find $\bigcap _{n=1}^{\infty }I_{n}.$

• 5. Consider the open intervals $I_{n}:=\left ]-\tfrac {1}{n},\tfrac {1}{n}\right [.$ Since $\tfrac {1}{n+1}<\tfrac {1}{n}$ the intervals are nested: $I_{n+1}\subset I_{n}.$ Find $\bigcap _{n=1}^{\infty }I_{n}.$

1.4 Problems for Sect. 3.4

• 1. Fill in the details of the induction in the proof of Theorem 3.4.1.

• 2. Give an example of a function $f:[0,1]\to \mathbb {R}$ that $f$ is continuous on $\left [0,\tfrac {1}{2}\right ]$ and discontinuous on $\left ]\tfrac {1}{2},1\right ]$ or prove that such a function does not exist.

• 3. Give an example of a function $f:[0,1]\to \mathbb {R}$ such that $f$ is continuous on $\left [0,\tfrac {1}{2}\right [$ and discontinuous on $\left [\tfrac {1}{2},1\right ]$ or prove that such a function does not exist.

1.5 Problems for Sect. 3.5

The strategy of our proof that $\sqrt {2}$ is irrational is roughly: Let $x$ be a real number and $m$ be an integer such that $m<x<m+1.$ Suppose $x$ is rational, use this and properties of $x,$ to find a real number $A$ such that $x^{k}A\in \mathbb {N}$ for all $k\in \mathbb {N}.$ Then $\left (\left (x-m\right )^{n}A\right )$ is a null sequence in $\mathbb {N},$ contradiction.

• 1. Use the strategy above to show that $\sqrt [3]{2}=2^{1/3}$ is irrational.

• 2. Why can the strategy above not be used to show that $\frac {5}{3}$ is irrational? More precisely, why does there not exist an $A$ such that $\left (\frac {5}{3}\right )^{k}A\in \mathbb {N}$ for all $k\in \mathbb {N}.$

• 3. Let $n\in \mathbb {N}.$ Show that any $x>0$ has at most one $n^{\mathrm {th}}$ root.

1.6 Problems for Sect. 3.6

Another way to construct the sets $C_{k}$ is to use the functions, $f_{0}(x):=\frac {x}{3}$ and $f_{2}(x):=\frac {x}{3}+\frac {2}{3}.$ Clearly, $C_{k+1}=f_{0}\left (C_{k}\right )\cup f_{2}\left (C_{k}\right ).$ In terms of ternary numbers $f_{m}\left (0\underset {3}{.}d_{1}d_{2}\ldots \right )=0\underset {3}{.}md_{1}d_{2}\ldots$. The iterated function system approach to the Cantor set is based on the first problem below.

• 1. $C=f_{0}\left (C\right )\cup f_{2}\left (C\right ).$

• 2. If $x=0\underset {3}{.}d_{1}d_{2}\ldots$ where $d_{j}\in \{0,2\}$ for all $j.$ For each $k,$ the point $x$ is a left-hand endpoint of one of the intervals in $C_{k}$ iff $d_{j}=0$ for all $j>k.$

• 3. If $x=0\underset {3}{.}d_{1}d_{2}\ldots$ where $d_{j}\in \{0,2\}$ for all $j.$ For each $k,$ the point $x$ is a right-hand endpoint of one of the intervals in $C_{k}$ iff $d_{j}=2$ for all $j>k.$

• 4. Prove $1/4\in C.$ [Hint: $1/4=0\underset {3}{.}\overline {02}.$]

• 5. Prove $1/4$ is not an endpoint of an interval in $C_{k}$ for any $k\in \mathbb {N}.$

• 6. Prove $1/5\notin C.$

• 7. The Cantor function is continuous.

• 8. Prove that any point $x_{0}$ in the Cantor set C is an accumulation point of $C.$

Solutions and Hints for the Exercises

Exercise 3.1.3. (i) States that $u$ is an upper bound for $A.$ (ii) States that any upper bound for $A$ is larger than $u.$ Thus $u$ is the smallest upper bound.

Exercise 3.1.6. Inductively, construct $x_{n}=d_{0}.d_{1}d_{2}\cdots d_{n}$ such that $-x_{n}-1/10^{n}$ is not an upper bound for $A$ and $-x_{n}$ is an upper bound for $A.$

Exercise 3.2.2. If $a=-\infty ,$ then $A$ does not have a lower bound. In particular, $t$ is not a lower bound for $A.$ Hence there is $x\in A,$ such that $x<t.$

Exercise 3.3.3. Recall $\bigcap _{n=1}^{\infty }[a_{n},b_{n}]=\{x\},$ where $x=\sup \{a_{n}\mid n\in \mathbb {N}\}.$ Since $x-\delta <x,$ $x-\delta$ is not an upper bound for $\{a_{n}\mid n\in \mathbb {N}\}.$ Hence, for some $n_{0}$ we have $x-\delta <a_{n_{0}}.$ Similarly, there is an $m_{0}$ such that $b_{m}<x+\delta .$ Setting $N:=\max \{n_{0},m_{0}\}$ completes the proof.

Exercise 3.3.5. The intervals $[a_{k},b_{k}]$ are nested and $\left (b_{k}-a_{k}\right )$ is null. Hence there is real number $x,$ such that $\bigcap _{k=1}^{\infty }[a_{k},b_{k}]=\{x\}.$ Similarly, there is a real number $y,$ such that $\bigcap _{k=1}^{\infty }[c_{k},d_{k}]=\{y\}.$

Exercise 3.5.3. Simplify the right-hand side.

Exercise 3.6.1. Any point $y$ in $[0,1]$ has a binary representation

$$ y=0\underset{2}{.}d_{1}d_{2}\ldots. $$

If $e_{k}:=2d_{k},$ then $x:=0\underset {3}{.}e_{1}e_{2}\ldots$ is a point in $C$ and $f(x)=y.$ Thus $f$ is onto.

That $f$ is not one-to-one is a consequence of the nonuniqueness of the binary representation of some numbers in $[0,1].$ For example, $f\left (0\underset {3}{.}0\overline {2}\right )$ and $f\left (0\underset {3}{.}2\overline {0}\right )$ both equal $0\underset {2}{.}1\overline {0}.$

Exercise 3.6.2. Similar to Exercise 3.6.1.