[pdf]

Consequences of Completeness (Part 1)

MATH 4630 · Chapter 1: The Real Numbers

In the previous handout, we established the Axiom of Completeness (Axiom 3). In this section, we will explore four fundamental consequences of this Axiom. This handout focuses on the first two consequences. We will discuss consequences 3 and 4 in the next handout.

Four Fundamental Consequences:

  1. Archimedean Property: Natural numbers grow without bound
  2. Nested Intervals: Shrinking intervals converge to a point
  3. Density of $\mathbb{Q}$: Rationals are everywhere in $\mathbb{R}$
  4. Existence of $\sqrt{2}$: We can finally prove it exists!

Note: All four properties fail in $\mathbb{Q}$! Completeness is what makes them work.

1. The Archimedean Property

Theorem: Archimedean Property

(i) Given any real number $x\in\mathbb{R}$, there exists $n\in\mathbb{N}$ such that

$$n > x.$$

(ii) Given any real number $\varepsilon>0$, there exists $n\in\mathbb{N}$ such that

$$\frac{1}{n} < \varepsilon.$$

Intuition

Part (i): No matter how large $x$ is, we can find a natural number bigger than $x$. The natural numbers are —they have no upper bound.

Part (ii): No matter how small $\varepsilon>0$ is, we can find $\frac{1}{n}$ smaller than $\varepsilon$. We can make $\frac{1}{n}$ .

Proof of Part (i)

Suppose for contradiction that $\mathbb{N}$ is bounded above. Then by Axiom 3, $\mathbb{N}$ has a ; call it $s=\sup\mathbb{N}$.

Since $s$ is the least upper bound, $s-1$ is an upper bound for $\mathbb{N}$.

Therefore, there exists $n\in\mathbb{N}$ such that: $n > s - 1$.

But then $n+1\in\mathbb{N}$ and: $n+1 > s - 1 + 1 = s$.

This contradicts that $s$ is an for $\mathbb{N}$.

Therefore, $\mathbb{N}$ is . Hence for any $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ with $n>x$. $\square$

Exercise

  1. Use part (i) to prove part (ii).

    Hint: Apply part (i) to $x=\frac{1}{\varepsilon}$.

  2. True or False: For any $x>0$, there exists $n\in\mathbb{N}$ such that $\frac{1}{n}
  • Find $n\in\mathbb{N}$ such that $\frac{1}{n}<0.001$.
    • 2. The Nested Interval Property

    Theorem: Nested Interval Property

    Assume $I_n=[a_n,b_n]$ is a sequence of closed, bounded intervals with

    $$I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$$

    Then the intersection $\displaystyle\bigcap_{n=1}^{\infty}I_n$ is .

    Visually

    Imagine a sequence of nested closed intervals, each shrinking inside the previous one. Their intersection is at least one point.

    Proof Sketch

    Define the set

    $$A=\{a_n:n\in\mathbb{N}\}=\text{(all left endpoints)}.$$

    Step 1: Show that $A$ is nonempty and bounded above.

    Step 2: By Axiom 3, let $x=$.

    Step 3: Prove that $x\in I_n$ for all $n \in \mathbb{N}$:

    Altogether, we have $a_n \le x \le b_n$ for all $n$, and therefore $x\in\bigcap_{n=1}^{\infty}I_n$, so the intersection is nonempty. $\square$

    Exercise: Nested Intervals

    Consider the sequence of intervals $I_n=[0,\tfrac{1}{n}]$ for $n=1,2,3,\ldots$

    1. Verify that $I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$
    2. What is $\bigcap_{n=1}^{\infty} I_n$?
    3. Now consider the nested open intervals $J_n=(0,\tfrac{1}{n})$ for $n=1,2,3,\ldots$
      1. Show that $J_1 \supseteq J_2 \supseteq J_3 \supseteq \cdots$.
      2. Prove that $\bigcap_{n=1}^{\infty} J_n = \emptyset$.
    4. Explain why the Nested Interval Property holds for $(I_n)$ but fails for $(J_n)$.
    ← Back to MATH 4630