In the last post we wanted to create a system of numbers which have the following property:
If all the numbers be divided into two sets $ L, U$ such that every member of $ L$ is less than any member of $ U$ then there must be a number $ \alpha$ such that all numbers less than $ \alpha$ belong to $ L$ and all the numbers greater than $ \alpha$ belong to $ U$, the number $ \alpha$ may itself lie in one of the sets $ L, U$.
After some considerations of various issues associated with the number $ \sqrt{2}$ we were led to a reasonable conception of a real number system which includes the rationals and the irrationals together. This system has all the properties which the system of rationals possesses. But its now time to think whether we have met our original goal which is stated in bold above. Does the system of real numbers have the above stated property? We had found that the system of rationals didn't have this property and hence we had to develop a new number system. If the newer system also does not have the property desired we might need to generalize further to include new kinds of numbers under consideration.
The fact that the system of real numbers does indeed have this property makes our search complete. The real number system can not be expanded further to include new kinds of numbers using the same process through which rationals were expanded to form the system of real numbers. This is known in technical terms as the completeness of the real number system. It is this property which makes real number system quite a different beast compared to the rational number system. And as we will notice in later posts this is the most important property of the real number system which is specifically never described in the elementary courses on mathematics. We will mention it as a proper theorem in the name of its founder:
Dedekind's Theorem: If the set of real numbers $ \mathbb{R}$ is divided into two non-empty sets $ L, U$ such that
In simpler language a section of the real numbers always leads to a real number which makes this section and thus the system has no gaps whatsoever to fill with numbers of a new kind.
To establish the above result, we make a section of rational numbers which is based on the above section of the real numbers. Let's define $$ L_{1} = L \cap \mathbb{Q},\,\,\, U_{1} = U \cap \mathbb{Q}$$ Thus we have taken the rational members of $ L$ and put them in $ L_{1}$ and similarly put the rationals of $ U$ in $ U_{1}$. It may happen in this case that $ L_{1}$ has a greatest member say $ l$. We will prove that when this is the case $ l$ is also the greatest member of $ L$. For if it were not the case then we would have a number $ \beta \in L$ such that $ \beta > l$. Then there are rational numbers between $ l$ and $ \beta$ which lie in $ L$ because they are less than $ \beta$ and therefore lie in $ L_{1}$. But since they are also greater than the greatest member of $ l$ of $ L_{1}$ they can't lie in $ L_{1}$. This contradiction shows that $ l$ is also the greatest member of $ L$.
In case $ L_{1}$ does not have a greatest member then the sets $ L_{1}, U_{1}$ form a section of the rationals and hence represent a real number $ \alpha$. If $ \alpha \in L$ then we can show that $ \alpha$ is greatest member of $ L$. For if it were not so, then we would have a $ \beta \in L$ such that $ \beta > \alpha$. Now rationals between $ \alpha$ and $ \beta$ would have to belong to $ L$ and therefore to $ L_{1}$ because they are less than $ \beta$ and they would also have to belong to $ U_{1}$ because they are greater than $ \alpha$. It follows from this contradiction that $ \alpha$ is the greatest member of $ L$. If on the other hand $ \alpha \in U$ it can be proved in a similar manner that $ \alpha$ is the least member of $ U$.
What we have achieved here is that whenever we make a section of real numbers, either the lower class has a greatest member or the upper class has a least member. Hence we are able to get the number $ \alpha$ which makes this division and all real number less than $ \alpha$ lie in $ L$ and all real number greater than $ \alpha$ lie in $ U$.
The reasoning we have applied here is truly simple and does not involve any concepts other than those the inequalities and denseness property of real numbers. Readers are requested to go through the proof once again to convince themselves of the fact that it does not require any difficult concepts. However the simplicity of the proof is in quite contrast to the power of this theorem. All the machinery of calculus (and more properly analysis) is based on this single theorem. And it contains the essence of the real number system as is required by the processes of analysis.
In fact many books on analysis take this property as an axiom for the set of real numbers and then develop the whole machinery of calculus. But when we will start to look at some of its applications, the property will cease to look obvious and suitable as an axiom. This completeness property of real numbers is stated in many forms and we will now have a look at them. All of the forms we will discuss are equivalent to each other but not all of them may be suitable to apply in a certain context. Hence it make sense to study all of them. Readers please bear in mind that the following discussion is not difficult but definitely becoming more abstract and I have provided it here for the sake of completeness.
Note that if $ M$ is an upper bound of the set $ S$ then any real number $ M^{\prime} > M$ is also an upper bound of $ S$. Therefore we are normally interested in the least of such upper bounds if any. To be be precise a number $ K$ is said to be the least upper bound or the supremum of a set $ S$ if $ K$ is an upper bound of $ S$, but no real number less than $ K$ is an upper bound of $ S$ and we write $K = \sup S$. To illustrate an example consider the set $ S = \{(n - 1) / n \mid n \in \mathbb{N}\}$. Clearly it is bounded above by say $ 2$. In fact it is also bounded below by $ 0$. It is also bounded above by $ 1$. But if we take any number less than $ 1$ say $ 0.999$ then we can see that $ (n - 1) / n > 0.999$ when $ n > 1000$ so that $ 0.999$ is not an upper bound of $ S$. And this plainly holds for any number less than $ 1$. Therefore $ 1$ is the supremum of $ S$.
The above example was quite simple where we could guess an exact number which is probably going to be the supremum of the set. It might not be so always. For example consider the set $ S = \{ x \mid x \in \mathbb{Q}, x > 0, x^{2} < 2\}$. Clearly all the members of $ S$ are less than $ 2$ so that $ 2$ is an upper bound. But it is not as easy as the previous example to guess the supremum. In fact it is not even clear if there is a supremum. If we are more observant we can see that $ \sqrt{2}$ is the supremum of this set. There could be complicated examples where it might be even difficult to know if a supremum exists or not. The following principle comes to the rescue:
Supremum Principle: If a non-empty set of real numbers is bounded above it has a supremum.
Also by definition of supremum it has to be unique. Now to understand why this principle holds we can assume a non-empty set $ S$ which is bounded above and say $ M$ is an upper bound of $ S$. Now we can divide the real numbers into two sets $ L$ and $ U$ such that $ x \in U$ if $ x$ is an upper bound of $ S$ and $ x \in L$ otherwise. Clearly both the sets are non-empty as $ M \in U$ and any number less than a member of $ S$ is in $ L$. And clearly any member of $ L$ is less than any member of $ U$. By Dedekind's theorem there is a number $ K$ which makes this division between $ L$ and $ U$. This number $ K$ is the supremum of set $ S$.
Clearly if $ K$ lies in $ U$ then it is the least member of $ U$ and hence it is the least upper bound of $ S$. We will now show that $ K$ cannot lie in $ L$. If it were in $ L$ it would be the greatest member of $ L$. And since $ K \in L$ it must be exceeded by some member say $ s \in S$. Now all the numbers between $ K$ and $ s$ have to lie in $ U$ because they are greater than greatest member of $ L$ and they also have to lie in $ L$ because they are less than $ s \in S$. This contradiction shows that $ K$ cannot be in $ L$. And thereby we have established that $ K$ is supremum of $ S$.
In an analogous manner we can define that a real number $ k$ is said to be the greatest lower bound or infimum of a set $ S$ if $ k$ is a lower bound of $ S$, but no real number greater than $ k$ is a lower bound of $ S$ and we write $k = \inf S$. And as before we can prove that if a non-empty set $ S$ is bounded below then it has an infimum. Also the infimum is unique if it exists.
1) Finite intervals:
With this brief introduction to intervals we can mention the nested interval principle as follows:
Nested Interval Principle: If $ \{I_{n}\}$ be a sequence of closed intervals such that $ I_{n + 1} \subseteq I_{n}$ then there is at least one real number $ \alpha$ which lies in all the intervals $ I_{n}$.
The principle holds only for closed intervals and not for open ones. This is very easy to establish. Let us write $ I_{n} = [a_{n}, b_{n}]$ and then it is clear that $ a_{n} \leq a_{n + 1} \leq b_{n + 1} \leq b_{n}$. Hence the set $ A = \{a_{n} \mid n \in \mathbb{N}\}$ is bounded above with supremum say $ a$ and set $ B = \{b_{n} \mid n \in \mathbb{N}\}$ is bounded below with infimum $ b$. We need to understand that $ a \leq b$ because if it were not so, given a real number $ \epsilon > 0$ we could find $ a_{n}, b_{m}$ such that $ a - \epsilon < a_{n}$ and $ b_{m} < b + \epsilon$ so that $ (a - b) - 2\epsilon < a_{n} - b_{n}$. Taking $ 0 < \epsilon < (a - b) / 2$ we would have $ a_{n} - b_{m} > 0$ contrary to the fact that $ a_{n} \leq b_{m}$. Hence we have $ a \leq b$. Clearly from the definition now $ a_{n} \leq a \leq b \leq b_{n}$ for all $ n$ and therefore all points of the interval $ [a, b]$ lie in all of the intervals $ I_{n}$. Also it should be clear from the above argument that for any point $c$ not lying in $[a, b]$ (i.e any point $c$ of the form $c < a$ i.e. $c = a - \epsilon$ or $c > b$ i.e. $c = b + \epsilon$ with $\epsilon > 0$) there will be an interval $I_{n} = [a_{n}, b_{n}]$ which does not contain $c$. Thus $[a, b]$ is in fact the intersection of all intervals $I_{n}$.
In the special case when $ a = b$ then there will be a unique real number $ a$ lying in all the intervals $ I_{n}$ and in practice this turns out to be the most important case. Hence it makes sense to analyze this important case in more detail. Note that $a = b$ implies that for any $\epsilon > 0$ we have positive integers $n_{1}, n_{2}$ such that $a_{n_{1}} > a - \epsilon / 2$ and $b_{n_{2}} < a + \epsilon / 2$ so that $b_{n_{2}} - a_{n_{1}} < \epsilon$. If integer $N = \max(n_{1}, n_{2})$ then for any positive integer $n > N$ we have $0 \leq b_{n} - a_{n} < \epsilon$ and we can see that the intervals $[a_{n}, b_{n}]$ all lie inside $(a - \epsilon, a + \epsilon)$. On the other hand if sequences $\{a_{n}\}, \{b_{n}\}$ satisfy the above condition then it is easy to prove that $a = b$.
In practical applications of this principle the sequence of intervals $\{I_{n}\}$ is normally generated bisecting interval $I_{n}$ into two equal parts and then choosing one of the parts as $I_{n + 1}$. By doing do we reduce the length of intervals $I_{n}$ at each step by a factor of $2$ so that $b_{n} - a_{n} = (b_{1} - a_{1})/2^{n - 1} < (b_{1} - a_{1}/n$ and hence it is easily seen that in this case we always have $a = b$ where $a = \sup \{a_{n} \mid n \in \mathbb{N}\}$ and $b = \inf \{n_{n} \mid n \in \mathbb{N}\}$.
Another principle which might seem strange at first but turns out to be very useful in practice is called the Heine Borel Principle which comes next.
If $ C$ is a collection of intervals covering a set $ S$ and it turns out that a sub-collection $ C^{\prime} \subseteq C$ also covers the set $ S$ then $ C^{\prime}$ can be said to be a subcover of $ S$. Now we are ready to state the Heine Borel Principle.
Heine Borel Principle: Any cover of a closed interval $ [a, b]$ contains a finite subcover.
In other words given any cover $ C$ of a closed interval $ [a, b]$ it is possible to choose a finite number of intervals from $ C$ which are able to cover the given interval $ [a, b]$. To begin with since the point $ a$ is an interior point of some interval $ I \in C$, it follows that there is a number $ x > a$ such that the entire interval $ [a, x]$ is contained in interval $ I \in C$. Hence we can divide the real numbers into two sets $ L$ and $ U$ as follows:
Put all numbers less than $ a$ in $ L$ and all numbers greater than $ b$ in $ U$. Also a number $ x \in [a, b]$ lies in $ L$ if the interval $ [a, x]$ can be covered by a finite number of intervals from $ C$ otherwise $ x \in U$. Clearly both $ L$ and $ U$ are non-empty and form a section of the real numbers. Also $ a$ and some number greater than $ a$ lie in $ L$. Now by Dedekind's theorem there is a number $ \alpha$ which makes this section and clearly in this case $ a < \alpha \leq b$. We shall prove that $ \alpha = b$ and that $ \alpha$ lies in $ L$.
First note that if $ \alpha < b$ then $ \alpha$ is an interior point of $ [a, b]$ and an interior point of some interval $ I \in C$. Clearly then we have an interval of the form $ [\alpha - \epsilon, \alpha + \epsilon], \epsilon > 0$ which is wholly contained in both $ I$ as well as $ [a, b]$. Now $ \alpha - \epsilon \in L$ so the interval $ [a, \alpha - \epsilon]$ can be covered by a finite number of intervals from collection $ C$. It now follows that the interval $ [a, \alpha + \epsilon]$ is also covered by a finite number of intervals from collection $ C$. Therefore $ \alpha + \epsilon$ also lies in $ L$ which is plainly a contradiction as $ \alpha + \epsilon > \alpha$. It thus follows that $ \alpha = b$.
It is easy to establish that $ b \in L$. Clearly $ b$ is an interior point of some interval $ I \in C$ and therefore there is some $ \epsilon > 0$ such that $ [b - \epsilon, b]$ is fully contained in $ I$. Since $ b - \epsilon$ is in $ L$ therefore the interval $ [a, b - \epsilon]$ is covered by a finite number of intervals from collection $ C$. It clearly follows now that the interval $ [a, b]$ is covered by a finite number of intervals from $ C$ and hence $ b \in L$.
Another approach to establish the Heine-Borel Principle is through the Nested Interval Principle. We can assume on the contrary that the interval $ [a, b]$ is not covered by a finite number of intervals from cover $ C$. In that case one of the two intervals $ [a, (a + b)/2]$ and $ [(a + b)/2, b]$ is also not covered by a finite number of intervals from $ C$. Calling this interval $ [a_{1}, b_{1}]$ we can now repeat the procedure and form a sequence of nested intervals $ [a_{n}, b_{n}]$ such that none of these intervals are covered by a finite number of intervals from $ C$.
Since $b_{n} - a_{n} = (b - a)/2^{n}$, we have the particularly important case of Nested Interval Principle here so that there is a unique real number $ c$ which lies in all the intervals $ [a_{n}, b_{n}]$. Also there is a an interval $ I \in C$ such that $ c$ is an interior point of $ I$. Thus we have an $\epsilon > 0$ for which $(c - \epsilon, c + \epsilon) \subset I$. Now (as explained at the end of the proof of Nested Interval Principle) there is an interval $ [a_{n}, b_{n}]$ which is wholly contained in $(c - \epsilon, c + \epsilon)$ and hence in $I$ and this clearly contradicts the assumption that none of the intervals $ [a_{n}, b_{n}]$ is covered by a finite number of intervals from $ C$. This contradiction completes the proof of Heine-Borel Principle.
This principle looks quite abstract in nature, but is very useful in practice. Roughly it is used in the following way. Suppose that for any given $ x \in [a, b]$ there is some useful property which holds in a small interval containing $ x$. Then from all these kinds of intervals we can choose a finite number of intervals which cover $ [a, b]$ and then (because of the finite number of intervals now involved) it becomes quite obvious that the property holds good in the entire interval. In other words the theorem is helpful in deducing some global properties (related to an interval) from local properties (related to a point and a small interval containing it). We will have occasion to see many such applications of this principle in later posts.
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If all the numbers be divided into two sets $ L, U$ such that every member of $ L$ is less than any member of $ U$ then there must be a number $ \alpha$ such that all numbers less than $ \alpha$ belong to $ L$ and all the numbers greater than $ \alpha$ belong to $ U$, the number $ \alpha$ may itself lie in one of the sets $ L, U$.
After some considerations of various issues associated with the number $ \sqrt{2}$ we were led to a reasonable conception of a real number system which includes the rationals and the irrationals together. This system has all the properties which the system of rationals possesses. But its now time to think whether we have met our original goal which is stated in bold above. Does the system of real numbers have the above stated property? We had found that the system of rationals didn't have this property and hence we had to develop a new number system. If the newer system also does not have the property desired we might need to generalize further to include new kinds of numbers under consideration.
The fact that the system of real numbers does indeed have this property makes our search complete. The real number system can not be expanded further to include new kinds of numbers using the same process through which rationals were expanded to form the system of real numbers. This is known in technical terms as the completeness of the real number system. It is this property which makes real number system quite a different beast compared to the rational number system. And as we will notice in later posts this is the most important property of the real number system which is specifically never described in the elementary courses on mathematics. We will mention it as a proper theorem in the name of its founder:
Dedekind's Theorem: If the set of real numbers $ \mathbb{R}$ is divided into two non-empty sets $ L, U$ such that
- $ L \cup U = \mathbb{R},\,\, L \cap U = \Phi$
- $ a \in L, b \in U \Rightarrow a < b$
In simpler language a section of the real numbers always leads to a real number which makes this section and thus the system has no gaps whatsoever to fill with numbers of a new kind.
To establish the above result, we make a section of rational numbers which is based on the above section of the real numbers. Let's define $$ L_{1} = L \cap \mathbb{Q},\,\,\, U_{1} = U \cap \mathbb{Q}$$ Thus we have taken the rational members of $ L$ and put them in $ L_{1}$ and similarly put the rationals of $ U$ in $ U_{1}$. It may happen in this case that $ L_{1}$ has a greatest member say $ l$. We will prove that when this is the case $ l$ is also the greatest member of $ L$. For if it were not the case then we would have a number $ \beta \in L$ such that $ \beta > l$. Then there are rational numbers between $ l$ and $ \beta$ which lie in $ L$ because they are less than $ \beta$ and therefore lie in $ L_{1}$. But since they are also greater than the greatest member of $ l$ of $ L_{1}$ they can't lie in $ L_{1}$. This contradiction shows that $ l$ is also the greatest member of $ L$.
In case $ L_{1}$ does not have a greatest member then the sets $ L_{1}, U_{1}$ form a section of the rationals and hence represent a real number $ \alpha$. If $ \alpha \in L$ then we can show that $ \alpha$ is greatest member of $ L$. For if it were not so, then we would have a $ \beta \in L$ such that $ \beta > \alpha$. Now rationals between $ \alpha$ and $ \beta$ would have to belong to $ L$ and therefore to $ L_{1}$ because they are less than $ \beta$ and they would also have to belong to $ U_{1}$ because they are greater than $ \alpha$. It follows from this contradiction that $ \alpha$ is the greatest member of $ L$. If on the other hand $ \alpha \in U$ it can be proved in a similar manner that $ \alpha$ is the least member of $ U$.
What we have achieved here is that whenever we make a section of real numbers, either the lower class has a greatest member or the upper class has a least member. Hence we are able to get the number $ \alpha$ which makes this division and all real number less than $ \alpha$ lie in $ L$ and all real number greater than $ \alpha$ lie in $ U$.
The reasoning we have applied here is truly simple and does not involve any concepts other than those the inequalities and denseness property of real numbers. Readers are requested to go through the proof once again to convince themselves of the fact that it does not require any difficult concepts. However the simplicity of the proof is in quite contrast to the power of this theorem. All the machinery of calculus (and more properly analysis) is based on this single theorem. And it contains the essence of the real number system as is required by the processes of analysis.
In fact many books on analysis take this property as an axiom for the set of real numbers and then develop the whole machinery of calculus. But when we will start to look at some of its applications, the property will cease to look obvious and suitable as an axiom. This completeness property of real numbers is stated in many forms and we will now have a look at them. All of the forms we will discuss are equivalent to each other but not all of them may be suitable to apply in a certain context. Hence it make sense to study all of them. Readers please bear in mind that the following discussion is not difficult but definitely becoming more abstract and I have provided it here for the sake of completeness.
Least Upper Bound / Supremum Principle
A set $ S$ of real numbers is said to be bounded above if there is a real number $ M$ such that $ x \leq M$ for all $ x \in S$. In this case the real number $ M$ is said to be an upper bound of $ S$. Similarly a set $ S$ of real number is said to be bounded below if there is a real number $ m$ such that $ x \geq m$ for all $ x \in S$. Such a real number $ m$ is said to be a lower bound of $ S$.Note that if $ M$ is an upper bound of the set $ S$ then any real number $ M^{\prime} > M$ is also an upper bound of $ S$. Therefore we are normally interested in the least of such upper bounds if any. To be be precise a number $ K$ is said to be the least upper bound or the supremum of a set $ S$ if $ K$ is an upper bound of $ S$, but no real number less than $ K$ is an upper bound of $ S$ and we write $K = \sup S$. To illustrate an example consider the set $ S = \{(n - 1) / n \mid n \in \mathbb{N}\}$. Clearly it is bounded above by say $ 2$. In fact it is also bounded below by $ 0$. It is also bounded above by $ 1$. But if we take any number less than $ 1$ say $ 0.999$ then we can see that $ (n - 1) / n > 0.999$ when $ n > 1000$ so that $ 0.999$ is not an upper bound of $ S$. And this plainly holds for any number less than $ 1$. Therefore $ 1$ is the supremum of $ S$.
The above example was quite simple where we could guess an exact number which is probably going to be the supremum of the set. It might not be so always. For example consider the set $ S = \{ x \mid x \in \mathbb{Q}, x > 0, x^{2} < 2\}$. Clearly all the members of $ S$ are less than $ 2$ so that $ 2$ is an upper bound. But it is not as easy as the previous example to guess the supremum. In fact it is not even clear if there is a supremum. If we are more observant we can see that $ \sqrt{2}$ is the supremum of this set. There could be complicated examples where it might be even difficult to know if a supremum exists or not. The following principle comes to the rescue:
Supremum Principle: If a non-empty set of real numbers is bounded above it has a supremum.
Also by definition of supremum it has to be unique. Now to understand why this principle holds we can assume a non-empty set $ S$ which is bounded above and say $ M$ is an upper bound of $ S$. Now we can divide the real numbers into two sets $ L$ and $ U$ such that $ x \in U$ if $ x$ is an upper bound of $ S$ and $ x \in L$ otherwise. Clearly both the sets are non-empty as $ M \in U$ and any number less than a member of $ S$ is in $ L$. And clearly any member of $ L$ is less than any member of $ U$. By Dedekind's theorem there is a number $ K$ which makes this division between $ L$ and $ U$. This number $ K$ is the supremum of set $ S$.
Clearly if $ K$ lies in $ U$ then it is the least member of $ U$ and hence it is the least upper bound of $ S$. We will now show that $ K$ cannot lie in $ L$. If it were in $ L$ it would be the greatest member of $ L$. And since $ K \in L$ it must be exceeded by some member say $ s \in S$. Now all the numbers between $ K$ and $ s$ have to lie in $ U$ because they are greater than greatest member of $ L$ and they also have to lie in $ L$ because they are less than $ s \in S$. This contradiction shows that $ K$ cannot be in $ L$. And thereby we have established that $ K$ is supremum of $ S$.
In an analogous manner we can define that a real number $ k$ is said to be the greatest lower bound or infimum of a set $ S$ if $ k$ is a lower bound of $ S$, but no real number greater than $ k$ is a lower bound of $ S$ and we write $k = \inf S$. And as before we can prove that if a non-empty set $ S$ is bounded below then it has an infimum. Also the infimum is unique if it exists.
Nested Interval Principle
We need to define certain sets of real numbers which occur quite often in practice. These we call intervals. A set $ I$ of real numbers is said to be an interval if $ x, y \in S \Rightarrow z \in S$ whenever $ x \leq z \leq y$. When $ x \in I$ we say that $ x$ is a point in the interval $ I$. After some observation it becomes clear that although a formal proof requires a simple application of the supremum principle and reader should try to furnish one) the intervals can be of one of the following forms:1) Finite intervals:
- Closed interval: $ [a, b] = \{x \mid x \in \mathbb{R}, a \leq x \leq b\}$. If $ a = b$ then the interval consists of a single number $ a$ and is said to be a degenerate interval.
- Open interval: $ (a, b) = \{ x \mid x \in \mathbb{R}, a < x < b\}$ and $ a < b$.
- Semi-open/Semi-closed intervals: $ [a, b) = \{x \mid x \in \mathbb{R}, a \leq x < b\},\, a < b$
- Semi-open/Semi-closed intervals: $ (a, b] = \{x \mid x \in \mathbb{R}, a < x \leq b\},\, a < b$
- $ (-\infty, \infty) = \mathbb{R}$
- $ (a, \infty) = \{x \mid x \in \mathbb{R}, x > a\}$
- $ [a, \infty) = \{x \mid x \in \mathbb{R}, x \geq a\}$
- $ (-\infty, b) = \{x \mid x \in \mathbb{R}, x < b\}$
- $ (-\infty, b] = \{x \mid x \in \mathbb{R}, x \leq b\}$
With this brief introduction to intervals we can mention the nested interval principle as follows:
Nested Interval Principle: If $ \{I_{n}\}$ be a sequence of closed intervals such that $ I_{n + 1} \subseteq I_{n}$ then there is at least one real number $ \alpha$ which lies in all the intervals $ I_{n}$.
The principle holds only for closed intervals and not for open ones. This is very easy to establish. Let us write $ I_{n} = [a_{n}, b_{n}]$ and then it is clear that $ a_{n} \leq a_{n + 1} \leq b_{n + 1} \leq b_{n}$. Hence the set $ A = \{a_{n} \mid n \in \mathbb{N}\}$ is bounded above with supremum say $ a$ and set $ B = \{b_{n} \mid n \in \mathbb{N}\}$ is bounded below with infimum $ b$. We need to understand that $ a \leq b$ because if it were not so, given a real number $ \epsilon > 0$ we could find $ a_{n}, b_{m}$ such that $ a - \epsilon < a_{n}$ and $ b_{m} < b + \epsilon$ so that $ (a - b) - 2\epsilon < a_{n} - b_{n}$. Taking $ 0 < \epsilon < (a - b) / 2$ we would have $ a_{n} - b_{m} > 0$ contrary to the fact that $ a_{n} \leq b_{m}$. Hence we have $ a \leq b$. Clearly from the definition now $ a_{n} \leq a \leq b \leq b_{n}$ for all $ n$ and therefore all points of the interval $ [a, b]$ lie in all of the intervals $ I_{n}$. Also it should be clear from the above argument that for any point $c$ not lying in $[a, b]$ (i.e any point $c$ of the form $c < a$ i.e. $c = a - \epsilon$ or $c > b$ i.e. $c = b + \epsilon$ with $\epsilon > 0$) there will be an interval $I_{n} = [a_{n}, b_{n}]$ which does not contain $c$. Thus $[a, b]$ is in fact the intersection of all intervals $I_{n}$.
In the special case when $ a = b$ then there will be a unique real number $ a$ lying in all the intervals $ I_{n}$ and in practice this turns out to be the most important case. Hence it makes sense to analyze this important case in more detail. Note that $a = b$ implies that for any $\epsilon > 0$ we have positive integers $n_{1}, n_{2}$ such that $a_{n_{1}} > a - \epsilon / 2$ and $b_{n_{2}} < a + \epsilon / 2$ so that $b_{n_{2}} - a_{n_{1}} < \epsilon$. If integer $N = \max(n_{1}, n_{2})$ then for any positive integer $n > N$ we have $0 \leq b_{n} - a_{n} < \epsilon$ and we can see that the intervals $[a_{n}, b_{n}]$ all lie inside $(a - \epsilon, a + \epsilon)$. On the other hand if sequences $\{a_{n}\}, \{b_{n}\}$ satisfy the above condition then it is easy to prove that $a = b$.
In practical applications of this principle the sequence of intervals $\{I_{n}\}$ is normally generated bisecting interval $I_{n}$ into two equal parts and then choosing one of the parts as $I_{n + 1}$. By doing do we reduce the length of intervals $I_{n}$ at each step by a factor of $2$ so that $b_{n} - a_{n} = (b_{1} - a_{1})/2^{n - 1} < (b_{1} - a_{1}/n$ and hence it is easily seen that in this case we always have $a = b$ where $a = \sup \{a_{n} \mid n \in \mathbb{N}\}$ and $b = \inf \{n_{n} \mid n \in \mathbb{N}\}$.
Another principle which might seem strange at first but turns out to be very useful in practice is called the Heine Borel Principle which comes next.
Heine Borel Principle
To talk about Heine Borel Principle it is necessary to introduce certain terminology. Let $ S$ be any set of real numbers and let $ C$ be a collection of intervals $ I$ such that each point of the set $ S$ is an interior point of some interval $ I \in C$. In this case $ C$ is said to be a cover of $ S$ or that the collection $ C$ covers $ S$. The collection $ C$ may or may not consists of a finite number of intervals.If $ C$ is a collection of intervals covering a set $ S$ and it turns out that a sub-collection $ C^{\prime} \subseteq C$ also covers the set $ S$ then $ C^{\prime}$ can be said to be a subcover of $ S$. Now we are ready to state the Heine Borel Principle.
Heine Borel Principle: Any cover of a closed interval $ [a, b]$ contains a finite subcover.
In other words given any cover $ C$ of a closed interval $ [a, b]$ it is possible to choose a finite number of intervals from $ C$ which are able to cover the given interval $ [a, b]$. To begin with since the point $ a$ is an interior point of some interval $ I \in C$, it follows that there is a number $ x > a$ such that the entire interval $ [a, x]$ is contained in interval $ I \in C$. Hence we can divide the real numbers into two sets $ L$ and $ U$ as follows:
Put all numbers less than $ a$ in $ L$ and all numbers greater than $ b$ in $ U$. Also a number $ x \in [a, b]$ lies in $ L$ if the interval $ [a, x]$ can be covered by a finite number of intervals from $ C$ otherwise $ x \in U$. Clearly both $ L$ and $ U$ are non-empty and form a section of the real numbers. Also $ a$ and some number greater than $ a$ lie in $ L$. Now by Dedekind's theorem there is a number $ \alpha$ which makes this section and clearly in this case $ a < \alpha \leq b$. We shall prove that $ \alpha = b$ and that $ \alpha$ lies in $ L$.
First note that if $ \alpha < b$ then $ \alpha$ is an interior point of $ [a, b]$ and an interior point of some interval $ I \in C$. Clearly then we have an interval of the form $ [\alpha - \epsilon, \alpha + \epsilon], \epsilon > 0$ which is wholly contained in both $ I$ as well as $ [a, b]$. Now $ \alpha - \epsilon \in L$ so the interval $ [a, \alpha - \epsilon]$ can be covered by a finite number of intervals from collection $ C$. It now follows that the interval $ [a, \alpha + \epsilon]$ is also covered by a finite number of intervals from collection $ C$. Therefore $ \alpha + \epsilon$ also lies in $ L$ which is plainly a contradiction as $ \alpha + \epsilon > \alpha$. It thus follows that $ \alpha = b$.
It is easy to establish that $ b \in L$. Clearly $ b$ is an interior point of some interval $ I \in C$ and therefore there is some $ \epsilon > 0$ such that $ [b - \epsilon, b]$ is fully contained in $ I$. Since $ b - \epsilon$ is in $ L$ therefore the interval $ [a, b - \epsilon]$ is covered by a finite number of intervals from collection $ C$. It clearly follows now that the interval $ [a, b]$ is covered by a finite number of intervals from $ C$ and hence $ b \in L$.
Another approach to establish the Heine-Borel Principle is through the Nested Interval Principle. We can assume on the contrary that the interval $ [a, b]$ is not covered by a finite number of intervals from cover $ C$. In that case one of the two intervals $ [a, (a + b)/2]$ and $ [(a + b)/2, b]$ is also not covered by a finite number of intervals from $ C$. Calling this interval $ [a_{1}, b_{1}]$ we can now repeat the procedure and form a sequence of nested intervals $ [a_{n}, b_{n}]$ such that none of these intervals are covered by a finite number of intervals from $ C$.
Since $b_{n} - a_{n} = (b - a)/2^{n}$, we have the particularly important case of Nested Interval Principle here so that there is a unique real number $ c$ which lies in all the intervals $ [a_{n}, b_{n}]$. Also there is a an interval $ I \in C$ such that $ c$ is an interior point of $ I$. Thus we have an $\epsilon > 0$ for which $(c - \epsilon, c + \epsilon) \subset I$. Now (as explained at the end of the proof of Nested Interval Principle) there is an interval $ [a_{n}, b_{n}]$ which is wholly contained in $(c - \epsilon, c + \epsilon)$ and hence in $I$ and this clearly contradicts the assumption that none of the intervals $ [a_{n}, b_{n}]$ is covered by a finite number of intervals from $ C$. This contradiction completes the proof of Heine-Borel Principle.
This principle looks quite abstract in nature, but is very useful in practice. Roughly it is used in the following way. Suppose that for any given $ x \in [a, b]$ there is some useful property which holds in a small interval containing $ x$. Then from all these kinds of intervals we can choose a finite number of intervals which cover $ [a, b]$ and then (because of the finite number of intervals now involved) it becomes quite obvious that the property holds good in the entire interval. In other words the theorem is helpful in deducing some global properties (related to an interval) from local properties (related to a point and a small interval containing it). We will have occasion to see many such applications of this principle in later posts.
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From a geometrical perspective, I think treating real numbers as shorthands of Dedkind cuts of the set of rational numbers cannot convincingly answer the question - how many points are inside the holes revealed by Dedkind cuts not produced by rational numbers like \(\begin{aligned} L & =\{x \mid x \in \mathbb{Q}, x \leq 0\} \cup\left\{x \mid x \in \mathbb{Q}, x>0, x^2<2\right\} \\ U & =\mathbb{Q}-L=\left\{x \mid x \in \mathbb{Q}, x>0, x^2>2\right\}\end{aligned}\)
To say such a cut corresponds with one and only one point on the number axis is hard to accept, however, which is what required by Cantor-Dedekind axiom. Unless a satisfying proof was given, otherwise it is hard to acknowledge it is an obvious fact.
Philip
January 6, 2023 at 1:48 PM