nested intervals

As stated in the introduction, historic users of mathematics discovered the nesting of intervals and closely related algorithms as methods for specific calculations. Some variations and modern interpretations of these ancient techniques will be introduced here:

When trying to find the square root of a number $x>1$, one can be certain that $1\backslash leq\; \backslash sqrt\{x\}\; \backslash leq\; x$, which gives the first interval $I\_1=[1,\; x]$, in which $x$ has to be found. If one knows the next higher perfect square $k^2\; >\; x$, one can get an even better candidate for the first interval: $I\_1=[1,\; k]$.

The other intervals $I\_n=[a\_n,\; b\_n],\; n\backslash in\backslash mathbb\{N\}$ can now be defined recursively by looking at the sequence of midpoints $m\_n=\backslash frac\{a\_n\; +\; b\_n\}\{2\}$. Given the interval $I\_n$ is already known (starting at $I\_1$), one can define

: $I\_\{n+1\}\; :=\; \backslash left\backslash \{\backslash begin\{matrix\}$

\left[m_n, b_n\right] && \text{if}\;\; m_n^2 \leq x \\

\left[a_n, m_n\right] && \text{if}\;\; m_n^2 > x

\end{matrix}\right.

To put this into words, one can compare the midpoint of $I\_\{n\}$ to $\backslash sqrt\{x\}$ in order to determine whether the midpoint is smaller or larger than $\backslash sqrt\{x\}$. If the midpoint is smaller, one can set it as the lower bound of the next interval $I\_\{n+1\}$, and if the midpoint is larger, one can set it as the upper bound of the next interval. This guarantees that $\backslash sqrt\{x\}\backslash in\; I\_\{n+1\}$. With this construction the intervals are nested and their length $|I\_n|$ get halved in every step of the recursion. Therefore, it is possible to get lower and upper bounds for $\backslash sqrt\{x\}$ with arbitrarily good precision (given enough computational time).

One can also compute $\backslash sqrt\{y\}$, when

To demonstrate this algorithm, here is an example of how it can be used to find the value of $\backslash sqrt\{19\}$. Note that since, the first interval for the algorithm can be defined as$I\_1:=[1,5]$, since $\backslash sqrt\{19\}$ must certainly found within this interval. Thus, using this interval, one can continue to the next step of the algorithm by calculating the midpoint of the interval, determining whether the square of the midpoint is greater than or less than 19, and setting the boundaries of the next interval accordingly before repeating the process:

: $\backslash begin\{aligned\}$

m_1&=\dfrac{1+5}{2}=3 &&\Rightarrow\; m_1^2=9 \leq 19 &&\Rightarrow\; I_2=[3, 5]\\

m_2&=\dfrac{3+5}{2}=4 &&\Rightarrow\; m_2^2=16 \leq 19 &&\Rightarrow\; I_3=[4, 5]\\

m_3&=\dfrac{4+5}{2}=4.5 &&\Rightarrow\; m_3^2=20.25 > 19 &&\Rightarrow\; I_4=[4, 4.5]\\

m_4&=\dfrac{4+4.5}{2}=4.25 &&\Rightarrow\; m_4^2=18.0625 \leq 19 &&\Rightarrow\; I_5=[4.25, 4.5]\\

m_5&=\dfrac{4.25+4.5}{2}=4.375 &&\Rightarrow\; m_5^2=19.140625 > 19 &&\Rightarrow\; I_5=[4.25, 4.375]\\

&\vdots & &

\end{aligned}

: Each time a new midpoint is calculated, the range of possible values for $\backslash sqrt\{19\}$ is able to be constricted so that the values that remain within the interval are closer and closer to the actual value of $\backslash sqrt\{19\}=4.35889894\backslash dots$. That is to say, each successive change in the bounds of the interval within which $\backslash sqrt\{19\}$ must lie allows the value of $\backslash sqrt\{19\}$ to be estimated with a greater precision, either by increasing the lower bounds of the interval or decreasing the upper bounds of the interval.

:

: This procedure can be repeated as many times as needed to attain the desired level of precision. Theoretically, by repeating the steps indefinitely, one can arrive at the true value of this square root.

The Babylonian method uses an even more efficient algorithm that yields accurate approximations of $\backslash sqrt\{x\}$ for an $x>0$ even faster. The modern description using nested intervals is similar to the algorithm above, but instead of using a sequence of midpoints, one uses a sequence $(c\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ given by

: $c\_\{n+1\}:=\backslash frac\{1\}\{2\}\backslash cdot\backslash left(c\_n\; +\; \backslash frac\{x\}\{c\_n\}\backslash right)$.

This results in a sequence of intervals given by $I\_\{n+1\}:=\backslash left[\backslash frac\{x\}\{c\_n\},\; c\_n\backslash right]$ and $I\_1=[0,\; k]$, where $k^2>x$, will provide accurate upper and lower bounds for $\backslash sqrt\{x\}$ very fast. In practice, only $c\_n$ has to be considered, which converges to $\backslash sqrt\{x\}$ (as does of course the lower interval bound). This algorithm is a special case of Newton's method.

{{further|Pi#Polygon approximation era}}

File:Archimedes pi.svg

As shown in the image, lower and upper bounds for the circumference of a circle can be obtained with inscribed and circumscribed regular polygons. When examining a circle with diameter $1$, the circumference is (by definition of Pi) the circle number $\backslash pi$.

Around 250 BCE Archimedes of Syracuse started with regular hexagons, whose side lengths (and therefore circumference) can be directly calculated from the circle diameter. Furthermore, a way to compute the side length of a regular $2n$-gon from the previous $n$-gon can be found, starting at the regular hexagon ($6$-gon). By successively doubling the number of edges until reaching 96-sided polygons, Archimedes reached an interval with . The upper bound $22/7\; \backslash approx\; 3.143$ is still often used as a rough, but pragmatic approximation of $\backslash pi$.

Around the year 1600 CE, Archimedes' method was still the gold standard for calculating Pi and was used by Dutch mathematician Ludolph van Ceulen, to compute more than thirty digits of $\backslash pi$, which took him decades. Soon after, more powerful methods for the computation were found.

Early uses of sequences of nested intervals (or can be described as such with modern mathematics), can be found in the predecessors of calculus (differentiation and integration). In computer science, sequences of nested intervals is used in algorithms for numerical computation. E.g. the bisection method can be used for calculating the roots of continuous functions. In contrast to mathematically infinite sequences, an applied computational algorithm terminates at some point, when the desired zero has been found or sufficiently well approximated.

In mathematical analysis, nested intervals provide one method of axiomatically introducing the real numbers as the completion of the rational numbers, being a necessity for discussing the concepts of continuity and differentiability. Historically, Isaac Newton's and Gottfried Wilhelm Leibniz's discovery of differential and integral calculus from the late 1600s has posed a huge challenge for mathematicians trying to prove their methods rigorously; despite their success in physics, engineering and other sciences. The axiomatic description of nested intervals (or an equivalent axiom) has become an important foundation for the modern understanding of calculus.

In the context of this article, $\backslash mathbb\{R\}$ in conjunction with $+$ and $\backslash cdot$ is an Archimedean ordered field, meaning the axioms of order and the Archimedean property hold.

Source:{{cite book |last1=Königsberger |first1=Konrad |title=Analysis 1 |date=2004 |publisher=Springer |isbn=354040371X |page=11}}

Let $(I\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ be a sequence of closed intervals of the type $I\_n=[a\_n,\; b\_n]$, where $|I\_n|:=b\_n\; -\; a\_n$ denotes the length of such an interval. One can call $(I\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ a sequence of nested intervals, if

1. $\backslash quad\; \backslash forall\; n\; \backslash in\; \backslash mathbb\{N\}:\; \backslash ;\backslash ;\; I\_\{n+1\}\; \backslash subseteq\; I\_n$
2. .

Put into words, property 1 means, that the intervals are nested according to their index. The second property formalizes the notion, that interval sizes get arbitrarily small; meaning, that for an arbitrary constant $\backslash varepsilon\; >\; 0$ one can always find an interval (with index $N$) with a length strictly smaller than that number $\backslash varepsilon$. It is also worth noting that property 1 immediately implies that every interval with an index $n\; \backslash geq\; N$ must also have a length .

Note that some authors refer to such interval-sequences, satisfying both properties above, as shrinking nested intervals. In this case a sequence of nested intervals refers to a sequence that only satisfies property 1.

If $(I\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ is a sequence of nested intervals, there always exists a real number, that is contained in every interval $I\_n$. In formal notation this axiom guarantees, that

:$\backslash exists\; x\backslash in\backslash mathbb\{R\}:\; \backslash ;x\backslash in\backslash bigcap\_\{n\backslash in\backslash mathbb\{N\}\}\; I\_n$.

The intersection of each sequence $(I\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ of nested intervals contains exactly one real number $x$.

Proof: This statement can easily be verified by contradiction. Assume that there exist two different numbers $x,y\backslash in\backslash cap\_\{n\backslash in\backslash mathbb\{N\}\}\; I\_n$. From $x\backslash neq\; y$ it follows that they differ by $|x-y|>0.$ Since both numbers have to be contained in every interval, it follows that $|I\_n|\backslash geq\; |x-y|$ for all $n\backslash in\backslash mathbb\{N\}$. This contradicts property 2 from the definition of nested intervals; therefore, the intersection can contain at most one number $x$. The completeness axiom guarantees that such a real number $x$ exists. $\backslash ;\; \backslash square$

• This axiom is fundamental in the sense that a sequence of nested intervals does not necessarily contain a rational number - meaning that $\backslash cap\_\{n\backslash in\backslash mathbb\{N\}\}I\_n$ could yield $\backslash emptyset$, if only considering the rationals.
• The axiom is equivalent to the existence of the infimum and supremum (proof below), the convergence of Cauchy sequences and the Bolzano–Weierstrass theorem. This means that one of the four has to be introduced axiomatically, while the other three can be successively proven.

By generalizing the algorithm shown above for square roots, one can prove that in the real numbers, the equation $x=y^j,\backslash ;\; j\backslash in\backslash mathbb\{N\},\; x>0$ can always be solved for $y=\backslash sqrt[j]\{x\}=x^\{1/j\}$. This means there exists a unique real number $y>0$, such that $x=y^k$. Comparing to the section above, one achieves a sequence of nested intervals for the $k$-th root of $x$, namely $y$, by looking at whether the midpoint $m\_n$ of the $n$-th interval is lower or equal or greater than $m\_n^k$.

If $A\backslash subset\; \backslash mathbb\{R\}$ has an upper bound, i.e. there exists a number $b$, such that $x\backslash leq\; b$ for all $x\backslash in\; A$, one can call the number $s=\backslash sup(A)$ the supremum of $A$, if

1. the number $s$ is an upper bound of $A$, meaning $\backslash forall\; x\; \backslash in\; A:\; \backslash ;\; x\backslash leq\; s$
2. $s$ is the least upper bound of $A$, meaning

Only one such number $s$ can exist. Analogously one can define the infimum ($\backslash inf(B)$) of a set $B\backslash subset\; \backslash mathbb\{R\}$, that is bounded from below, as the greatest lower bound of that set.

Each set $A\backslash subset\; \backslash mathbb\{R\}$ has a supremum (infimum), if it is bounded from above (below).

Proof: Without loss of generality one can look at a set $A\backslash subset\; \backslash mathbb\{R\}$ that has an upper bound. One can now construct a sequence $(I\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ of nested intervals $I\_n=[a\_n,\; b\_n]$, that has the following two properties:

1. $b\_n$ is an upper bound of $A$ for all $n\backslash in\backslash mathbb\{N\}$
2. $a\_n$ is never an upper bound of $A$ for any $n\backslash in\backslash mathbb\{N\}$.

The construction follows a recursion by starting with any number $a\_1$, that is not an upper bound (e.g. $a\_1=c\; -\; 1$, where

$c\backslash in\; A$ and an arbitrary upper bound $b\_1$ of $A$). Given $I\_n=[a\_n,\; b\_n]$ for some $n\backslash in\backslash mathbb\{N\}$ one can compute the midpoint $m\_n:=\; \backslash frac\{a\_n+b\_n\}\{2\}$ and define

:$I\_\{n+1\}\; :=\; \backslash left\backslash \{\backslash begin\{matrix\}$

\left[a_n, m_n\right] && \text{if}\; m_n \;\text{is an upper bound of}\; A \\

\left[m_n, b_n\right] && \text{if}\; m_n \;\text{is not an upper bound}

\end{matrix}\right.

Note that this interval sequence is well defined and obviously a sequence of nested intervals by construction.

Now let $s$ be the number in every interval (whose existence is guaranteed by the axiom). $s$ is an upper bound of $A$, otherwise there exists a number $x\backslash in\; A$, such that $x>s$. Furthermore, this would imply the existence of an interval $I\_m=[a\_m,\; b\_m]$ with , from which follows, due to $s$ also being an element of $I\_m$. But this is a contradiction to property 1 of the supremum (meaning

Assume that there exists a lower upper bound of $A$. Since $(I\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ is a sequence of nested intervals, the interval lengths get arbitrarily small; in particular, there exists an interval with a length smaller than $s-\backslash sigma$. But from $s\backslash in\; I\_n$ one gets

In two steps, it has been shown that $s$ is an upper bound of $A$ and that a lower upper bound cannot exist. Therefore $s$ is the supremum of $A$ by definition.

As was seen, the existence of suprema and infima of bounded sets is a consequence of the completeness of $\backslash mathbb\{R\}$. In effect the two are actually equivalent, meaning that either of the two can be introduced axiomatically.

Proof: Let $(I\_n)\_\{n\backslash in\backslash mathbb\{N\}\}$ with $I\_n=[a\_n,\; b\_n]$ be a sequence of nested intervals. Then the set $A:=\backslash \{a\_1,\; a\_2,\backslash dots\backslash \}$ is bounded from above, where every $b\_n$ is an upper bound. This implies, that the least upper bound $s=\backslash sup(A)$ fulfills $a\_n\backslash leq\; s\backslash leq\; b\_n$ for all $n\backslash in\backslash mathbb\{N\}$. Therefore $s\backslash in\; I\_n$ for all $n\backslash in\backslash mathbb\{N\}$, respectively $s\backslash in\backslash cap\_\{n\backslash in\backslash mathbb\{N\}\}\; I\_n$.

After formally defining the convergence of sequences and accumulation points of sequences, one can also prove the Bolzano–Weierstrass theorem using nested intervals. In a follow-up, the fact, that Cauchy sequences are convergent (and that all convergent sequences are Cauchy sequences) can be proven. This in turn allows for a proof of the completeness property above, showing their equivalence.

Without any specifying what is meant by interval, all that can be said about the intersection $\backslash cap\_\{n\backslash in\backslash mathbb\{N\}\}\; I\_n$ over all the naturals (i.e. the set of all points common to each interval) is that it is either the empty set $\backslash emptyset$, a point on the number line (called a singleton $\backslash \{x\backslash \}$), or some interval.

The possibility of an empty intersection can be illustrated by looking at a sequence of open intervals

In this case, the empty set $\backslash emptyset$ results from the intersection $\backslash cap\_\{n\backslash in\backslash mathbb\{N\}\}\; I\_n$. This result comes from the fact that, for any number $x>0$ there exists some value of $n\backslash in\backslash mathbb\{N\}$ (namely any $n>1/x$), such that

The situation is different for closed intervals. If one changes the situation above by looking at closed intervals of the type $I\_n=\backslash left[0,\; \backslash frac\{1\}\{n\}\backslash right]\; =\; \backslash left\backslash \{x\backslash in\backslash mathbb\{R\}:0\; \backslash leq\; x\; \backslash leq\; \backslash frac\{1\}\{n\}\backslash right\backslash \}$, one can see this very clearly. Now for each $x>0$ one still can always find intervals not containing said $x$, but for $x=0$, the property $0\backslash leq\; x\; \backslash leq\; 1/n$ holds true for any $n\backslash in\backslash mathbb\{N\}$. One can conclude that, in this case, $\backslash cap\_\{n\backslash in\backslash mathbb\{N\}\}\; I\_n\; =\; \backslash \{0\backslash \}$.

One can also consider the complement of each interval, written as $(-\backslash infty,a\_n)\; \backslash cup\; (b\_n,\; \backslash infty)$ - which, in our last example, is $(-\backslash infty,0)\; \backslash cup\; (1/n,\; \backslash infty)$. By De Morgan's laws, the complement of the intersection is a union of two disjoint open sets. By the connectedness of the real line there must be something between them. This shows that the intersection of (even an uncountable number of) nested, closed, and bounded intervals is nonempty.

In two dimensions there is a similar result: nested closed disks in the plane must have a common intersection. This result was shown by Hermann Weyl to classify the singular behaviour of certain differential equations.

• Bisection
• Cantor's intersection theorem

{{Reflist}}

• {{citation|title=Introductory Analysis: The Theory of Calculus|first=J. A.|last=Fridy|publisher=Academic Press|year=2000|isbn=9780122676550|page=29|url=https://books.google.com/books?id=SaZYs-OKqJcC&pg=PA29|contribution=3.3 The Nested Intervals Theorem}}.
• {{citation|title=Elementary Real and Complex Analysis|series=Dover Books on Mathematics|first=Georgi E.|last=Shilov|publisher=Courier Dover Publications|year=2012|isbn=9780486135007|contribution=1.8 The Principle of Nested Intervals|url=https://books.google.com/books?id=GELCAgAAQBAJ&pg=PA21|pages=21–22}}.
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• {{citation|title=Analysis 1, 6. Auflage (6th edition)|first=Konrad|last=Königsberger|series=Springer-Lehrbuch |publisher=Springer|year=2003|isbn=9783642184901|page=10-15|doi=10.1007/978-3-642-18490-1 |contribution=2.3 Die Vollständigkeit von R (the completeness of the real numbers)|url=https://link.springer.com/book/10.1007/978-3-642-18490-1}}

{{DEFAULTSORT:Nested Intervals}}

Category:Sets of real numbers

Category:Theorems in real analysis