Uncover the Hidden Pattern: The Surprising Truth About Nested Intervals

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**Theorem 7** (Nested Intervals Property).: _Let \(I_{1},I_{2},\dots\) be a sequence of nonempty closed and bounded intervals which are nested in the sense that each \(I_{n}\) contains \(I_{n+1}\) for every \(n\in\mathbb{N}\). Then the intersection \(\bigcap_{n=1}^{\infty}I_{n}\neq\emptyset\)._

Proof

: Let us denote the closed and bounded intervals by \(I_{n}=[a_{n},b_{n}]\) for all positive integers \(n\). Then the nesting condition means

\[a_{1}\leq a_{2}\leq a_{3}\leq\dots\leq b_{3}\leq b_{2}\leq b_{1}.\]

Now consider the “left hand point set” \(A=\{a_{1},a_{2},a_{3},\dots\}\). Because the intervals are nested, notice that every \(b_{n}\) is an upper bound of the set \(A\), and thus we can set \(x=\sup A\). Now, consider a particular \(I_{n}=[a_{n},b_{n}]\). Since \(x\) is an upper bound for \(A\), \(a_{n}\leq x\). The fact that each \(b_{n}\) is an upper bound and \(x\) is the least upper bound for \(A\) implies \(x\leq b_{n}\). Thus we have

\[a_{n}\leq x\leq b_{n}\quad\text{for every choice}\quad n\in\mathbb{N},\]

which means \(x\in I_{n}\) for each \(n\), therefore \(x\in\bigcap_{n=1}^{\infty}I_{n}\), and the intersection is not empty.

Note that if the interval’s length shrinks to zero, then the intersection is a single point. In particular, if \(A=\{a_{1},a_{2},a_{3},\dots\}\) and \(B=\{b_{1},b_{2},b_{3},\dots\}\), then by the axiom of completeness \(a=\sup A\) and \(b=\inf B\) lie in all the \(I_{n}\). If the length of \(I_{n}\) shrinks to zero, then \(a=b\).

### The Density of \(\mathbb{Q}\) in \(\mathbb{R}\)

The set of rational numbers \(\mathbb{Q}\) contains the set of natural numbers \(\mathbb{N}\) and both are contained in \(\mathbb{R}\). Now we ask “how do \(\mathbb{Q}\) and \(\mathbb{N}\) fit inside \(\mathbb{R}\)?” This will lead us to the density of rationals in the set of real numbers. We say \(\mathbb{Q}\) is dense in \(\mathbb{R}\) when for every two real numbers \(a\) and \(b\) with with \(a

**Theorem 8**.:
* _Given any real number_ \(M\)_, there exists a positive integer_ \(n\) _such that_ \(n>M\)__(_\(\mathbb{N}\) _is unbounded)._
* _For every positive_ \(\epsilon\)_, no matter how small, there exists a positive integer_ \(n\) _with_ \(0<\dfrac{1}{n}<\epsilon\) _(squeeze in)._ * _Given any real numbers_ \(a\) _and_ \(b\) _with_ \(0b\) _(Archimedean property)._

Proof

: To show \(\mathbb{N}\) is unbounded, we argue for a contradiction. Assume that there is no integer \(n\) with \(n>M\). Then \(n\leq M\) for all \(n\), which means \(M\) is an upper bound for \(\mathbb{N}\). Thus by the completeness principle \(\mathbb{N}\) has a supremum, say \(\lambda=\sup\mathbb{N}\). Since \(\lambda\) is the _least_ upper bound, \(\lambda-1\) is not an upper bound, and so there is some positive integer \(n_{1}\) for which \(n_{1}>\lambda-1\) or equivalently \(n_{1}+1>\lambda\). This is a contradiction to the assumption that no integer exceeds \(\lambda\). The second property (squeeze in) follows from the first, and for the Archimedean property, observe that for positive \(a\) and \(b\)

\[na>b\quad\text{if and only if}\quad n>\dfrac{a}{b},\]

and since \(\mathbb{N}\) is unbounded, \(n>\dfrac{a}{b}\) must hold for sufficiently large \(n\).

The Archimedean property asserts that given a positive number \(a\) as small as one wants, if one takes steps of size \(a\), one will cover and exceed a trip of distance \(b\). In other words, given a positive number \(a\), even if it is very small, its successive multiples \(a,2a,3a,\dots,na\) will eventually exceed any proposed bound \(b\). The next theorem shows that both rational and irrational numbers sit “tightly” inside \(\mathbb{R}\).

**Theorem 9**.: \(\mathbb{Q}\) _is dense in \(\mathbb{R}\). That is, for every two real numbers \(a\) and \(b\) with \(a

Proof

: Assume \(0\leq a

Figure 1.14: Density of \(\mathbb{Q}\) in \(\mathbb{R}\)can pick \(q\in\mathbb{N}\) so that \(\dfrac{1}{q}

\[qa

We have already chosen \(q\), and now we choose \(p\) to be the smallest natural number greater than \(qa\). Equivalently, choose \(p\in\mathbb{N}\) so that

\[p-1\leq qa

Now from \(qa

\[p \leq qa+1,\] \[< q(b-1/q)+1\] \[= qb.\]

Now the inequality \(p

**Corollary 2**.: _The set of irrational numbers \(\mathbb{I}:=\mathbb{R}\setminus\mathbb{Q}\) is also dense in \(\mathbb{R}\). That is, given any two real numbers \(a

Now we introduce a special function, called _the absolute value function_, that is so important that it merits a special notation \(|x|\) in place of the usual notation \(f(x)\).

**Definition 9**.: The absolute value or the magnitude of \(x\in\mathbb{R}\) is a function defined as

\[|x|=\left\{\begin{array}{ll}x&\quad\text{if }x\geq 0\\ -x&\quad\text{if }x<0.\end{array}\right.\]

The distance between two elements \(x\) and \(y\) is then \(|x-y|\).

Certain properties of the absolute value function are worth singling out both for their own value and because of generalizations to abstract spaces later.

**Lemma 2**.: _The absolute value function satisfies the following properties:_

* \(|x|\geq 0\) _for every_ \(x\)_._
* \(|x|=0\) _if and only if_ \(x=0\)_._
* \(|xy|=|x||y|\)_._
* \(|x+y|\leq|x|+|y|\) _(triangle inequality)._

Proof

: The proofs of a), b), and c) are straightforward. For the proof of triangle inequality, consider the different cases that arise when \(x,y,\) and \(x+y\) are positive and negative. For example, if \(x\geq 0,\ y\leq 0,\) and \(x+y\geq 0,\) then \(|x+y|=x+y\leq x+y-y=x=|x|\leq|x|+|y|\). For the proof of reverse triangle inequality, we employ the triangle inequality:

\[|x|=|x+y-y|\leq|x-y|+|y|\]

and thus

\[|x|-|y|\leq|x-y|.\]

By changing the role of \(x\) by \(y\) in the above inequality, we obtain the other half of the inequality necessary in order to claim the reverse triangle inequality.

**Remark 7**.: It turns out that the simple-looking _triangle inequality_ is extremely important and will be frequently used in the following way. Given real numbers \(x,y,\) and \(z,\) we can certainly write

\[|x-y|=|x-z+z-y|.\]

However, if we apply the triangle inequality,

\[|x-z+z-y|\leq|x-z|+|z-y|,\]

thus we obtain

\[|x-y|\leq|x-z|+|z-y|.\]

Now the expression \(|x-z|\) which is equal to \(|z-x|\) is best understood as the distance between \(x\) and \(z\). Similarly \(|z-y|\) is the distance between \(z\) and \(y\), and \(|x-y|\) is the distance between \(x\) and \(y\). When we think of \(x\), \(y\), and \(z\) as points in the plane as opposed to on the real line, we see why it is referred to as the triangle inequality. In this case, the triangle inequality expresses the familiar fact that the sum of lengths of two sides of a triangle is always at least the length of the third side.

**Remark 8**.: Now that we understand the fact that \(|x-r|\) is the distance between \(x\) and \(r\), one can also state the above Theorem 9 as follows:

\[\text{If }x\in\mathbb{R}\text{ and }\epsilon>0,\text{ there is an }r\in\mathbb{Q}\text{ such that }|x-r|<\epsilon.\]

We leave it as an exercise to the reader to check that \(\mathbb{Q}\) is dense in \(\mathbb{R}\) implies that every real number is a limit of rational numbers. However, to prove this, one needs the concept of a limit of a sequence which will be covered in the next section.

### Exercises

1. Compute, without proofs, the suprema and infima of the following sets: 1. \(\{n\in\mathbb{N}:\ n^{2}<20\}\). 2. \(\left\{\frac{n}{n+m}:\ m,n\in\mathbb{N}\right\}\). 3. \(\{r\in\mathbb{Q}:\ r<5\}\). 4. \(\{r\in\mathbb{Q}:\ r^{2}<5\}\). 5. \(\bigcup_{n=1}^{\infty}\left[\frac{1}{n},2-\frac{1}{n}\right]\). 2. Assume that \(A\) and \(B\) are nonempty, bounded above, and satisfy \(B\subseteq A\). Show that \(\sup B\leq\sup A\). What is the relationship between \(\inf A\) and \(\inf B\)? 3. Let \(A\) be a nonempty subset of \(\mathbb{R}\) bounded above. Set \(B=\{-a:\ a\in A\}\). Show that \(B\) is bounded below and \(\inf B=-\sup A\). 4. Let \(x\) be a nonzero rational number and \(y\) be irrational. Prove that \(xy\) is irrational. 5. Show that the product of two irrational numbers may be rational or irrational. 6. Prove that for each \(x\in\mathbb{R}\) and each \(n\in\mathbb{N}\) there exists a rational \(r_{n}\) such that \(|x-r_{n}|<\frac{1}{n}\). 7. Prove that the set of irrational numbers is dense in \(\mathbb{R}\). 8. Show that \(\bigcap_{n=1}^{\infty}\left(0,\frac{1}{n}\right)=\emptyset\) (this shows that intervals in the Nested Intervals Property must be _closed_ for the conclusion of the theorem to hold). 9. Suppose \(A\) and \(B\) are a nonempty subset of \(\mathbb{R}\). Let \[C=\{x+y:\ x\in A\ \text{and}\ y\in B\}.\] If \(A\) and \(B\) have suprema, then show that \(C\) has a supremum and \(\sup C=\sup A+\sup B\). What can you say about \(\inf C\)? 10. Suppose \(A\) and \(B\) are a nonempty subset of \(\mathbb{R}\). Let \[C=A\cdot B=\{ab:\ a\in A,\text{and}\ b\in B\}.\]If \(A\) and \(B\) have suprema, then show that \(C\) has a supremum and \(\sup C=\sup A\sup B\). What can you say about \(\inf C\)?

11. Prove that if \(a\) is a real number, then there exists an integer \(N\) such that \(N-1\leq a

Hint: Use the Dirichlet theorem on rational approximation of any real number.