Using these definitions, it is also possible to show that the set of all real numbers is a field. All this means is that the set obeys certain axioms which relate to the two operations defined on the set ($*$ and $+$), although this will be left as an exercise to the reader. Much of this post will be left as an exercise to the reader, as many facts about the real numbers follow from the definitions that will be given in this post. Also, I would like to move on to proving completeness as quickly as possible as it will enable us to start talking about sequences and series of real numbers.
Lets start with defining the 'less than' relation:
$Definition \ 1 - (Less \ Than) - $ For two real numbers $[\alpha], [\beta]$, we say $[\alpha] < [\beta]$ if and only if there exists a positive rational number $\epsilon > 0$ such that for some constructive interval sequence $\{I_n\}=[a_n,b_n]$ in $[\alpha]$ and $\{K_n\}=[c_n,d_n]$ in $[\beta]$, $c_n - a_n > \epsilon$ for all $n>M$ where $M$ is some natural number.
Although Definition 1 does not require that for any two constructive interval sequences $\{I_n\}$ and $\{K_n\}$ in $[\alpha]$ and $[\beta]$ respectively, there must exist a positive rational number $\epsilon > 0 $ such that for all $n>M$ where $M$ is some natural number, $c_n - a_n > \epsilon$, this is implied by Definition 1 as $[\alpha]$ and $[\beta]$ are both defined to be equivalence classes of constructive interval sequences. The actual proof of this will be left as an exercise to the reader, but essentially it is a consequence of the fact that any two sequences in an equivalence class of nested interval sequences can be made 'arbitrarily close together'.
The definition of 'less than or equal to' will literally be two numbers are 'less than or equal to' each other if and only if they are 'less than' each other xor 'equal to' each other.
We will now prove the law of trichotomy of the real numbers:
$Theorem \ 1 \ (Law \ of \ trichotomy) - $ For any two real numbers , $[\alpha]$ and $[\beta]$, one and only one of the following statements is true:
(i) $[\alpha] < [\beta]$
(ii) $[\beta] < [\alpha]$
(iii) $[\beta] = [\alpha]$
$Proof \ (Theorem \ 1)$ -
Let $\{I_n\} = [a_n,b_n]$ and $\{K_n\} = [c_n,d_n]$ be two constructive interval sequences in $[\alpha]$ and $[\beta]$ respectively.
Suppose $[\beta] = [\alpha]$ were false. Then there exists some positive rational number $\epsilon$ such that for all natural numbers $M$, there exists an $n>M$ such that $|c_n - a_n| \geq \epsilon$
Now, as $[a_n,b_n]$ and $[c_n,d_n]$ are two constructive interval sequences, we can make $b_n - a_n < \frac{\epsilon}{2}$ and $d_n - c_n < \frac{\epsilon}{2}$ for all $n>M$.
But we know that for some $n>M$, $|c_n - a_n| \geq \epsilon$.
This implies that either $c_n - a_n \geq \epsilon$ or $a_n - c_n \geq \epsilon$.
Suppose $c_n - a_n \geq \epsilon$. However we know that $b_n - a_n < \frac{\epsilon}{2}$.
Therefore, $c_n - b_n > \frac{\epsilon}{2}$.
Now $c_{n+i} \geq c_n$ for all $i$, and $a_{n+i} < b_n$ for all $i$.
This implies that $c_{n+i} - a_{n+i} > c_n - b_n > \frac{\epsilon}{2}$ for all $i$.
Thus $[\alpha] < [\beta]$.
We can similarly prove that if $a_n - c_n \geq \epsilon$, $[\beta] < [\alpha]$.
$\blacksquare$
In the hopes of defining multiplication successfully, I will now present another definition and a lemma about $0$ (and all rational numbers) in the 'real number' sense.
$Definition \ 2- $ The rational number $a$ is equivalent to the equivalence class the constructive interval sequence $[a,b_n]$ belongs to. We denote $a$ in the real number sense, $[a]$.
$Lemma \ 1 - $ If $[\alpha] < [0]$, there exists a constructive interval sequence $I_n =[a_n,b_n]$ in $[\alpha]$ such that $a_n<0$ and $b_n<0$ for all natural $n$.
$Proof \ (Lemma \ 1) - $ By definition, if $[\alpha] < [0]$, we must be able to find a natural number $M$ such that for all $n>M$ , $0 - a_n > \epsilon$ for some positive rational number $\epsilon$, where $[a_n,b_n]$ is in $[\alpha]$.
Therefore, for all $n>M$, $a_n < -\epsilon < 0$.
As $[a_n,b_n]$ is a constructive interval sequence, it is also true that $lim_{n \to \infty} b_n - a_n = 0$, which implies that for $n$ large enough, $b_n < -\epsilon < 0$.
Thus for $n$ large enough, say for $n$ larger or equal to than $M_1$, we can ensure $a_n < 0 $ and $b_n < 0$.
Take another constructive interval sequence $[a^{*}_n,b^{*}_n]$, where $a^{*}_n = a_{M_1 + n}$ and $b^{*}_n = b_{M_1 + n}$.
This is simply a sub-sequence of the aforementioned sequence, and also must belong to $[\alpha]$, as $|a_{M_1 + n} - a_n| < b_n - a_n$ and $\displaystyle lim_{n \to \infty} b_n - a_n = 0$.
Thus Lemma 1 has been proved.
$\blacksquare$
$Lemma \ 2 - $ If $[0] < [\alpha]$, there exists a constructive interval sequence $I_n =[a_n,b_n]$ in $[\alpha]$ such that $0<a_n$ and $0<b_n$ for all natural $n$.
Lemma 2 can be proved similarly.
$\square$
Now we will define the operations $+$ and $*$ on the real numbers.
$Definition \ 3 \ (Addition) - $ For any two real numbers $[\alpha]$, $[\beta]$, we define $[\alpha] + [\beta]$ as the equivalence class the constructive interval sequence $[a_n + c_n , b_n + d_n]$ belongs to, where $[a_n,b_n]$ is in $[\alpha]$ and $[c_n,d_n]$ is in $[\beta]$.
$Definition \ 4 \ (Multiplication) - $ Let $[\alpha] \geq 0$ and $[\beta] \geq 0$ , then by Lemma 2, there exist constructive interval sequences, $\{I_n\}=[a_n,b_n]$ and $\{K_n\}=[c_n,d_n]$ in $[\alpha]$ and $[\beta]$ respectively such that $a_n, b_n,c_n,d_n \geq 0$ for all natural $n$.
We define $[\alpha]*[\beta]$ as the equivalence class of the constructive interval sequence $[a_nc_n,b_nd_n]$. (It follows from this definition that (at least for $[\alpha] > 0$ and $[\beta] > 0$, at this point) multiplication is commutative.
Now suppose $[\alpha]<0$ and $[\beta]\geq0$, then by Lemma 1 and Lemma 2, we know there must exist constructive interval sequences, $\{I_n\}=[a_n,b_n]$ and $\{K_n\}=[c_n,d_n]$ in $[\alpha]$ and $[\beta]$ respectively such that $a_n, b_n < 0 $ and $c_n, d_n > 0 $ for all natural $n$.
We define $[\alpha]*[\beta]$ in this case as the equivalence class of the constructive interval sequence $[-|a_n|d_n,-|b_n|c_n]$.
For $[\alpha] \leq 0$ and $[\beta] \leq 0$ we know there must exist constructive interval sequences, $\{I_n\}=[a_n,b_n]$ and $\{K_n\}=[c_n,d_n]$ in $[\alpha]$ and $[\beta]$ respectively such that $a_n, b_n ,c_n,d_n \leq 0$for all natural $n$.
We define $[\alpha]*[\beta]$ as the equivalence class of the constructive interval sequence $[|b_n||d_n|,|c_n||a_n|]$.
Using these definitions it is now possible to prove all of the field axioms of the real numbers: http://mathworld.wolfram.com/FieldAxioms.html
$\square$
We will now prove that the real numbers are complete.
$Property \ 1 \ (Completeness) - $ Let $S$ be some non-empty subset of $\mathbb{R}$ that is bounded above. There exists a real number $\xi \in \mathbb{R}$ such that $\xi$ is the smallest upper bound of $S$.
$Proof \ (Property \ 1)$ - $S \subset \mathbb{R}$ and $S$ is bounded above. Therefore there must exist some $[\beta] \in \mathbb{R}$ such that for all $[\alpha] \in S$, $[\alpha] \leq [\beta]$.
Now let $\{I_n\} \in [\alpha] \in S$ and $\{K_n\} \in [\beta]$ where $I_n = [a_n,b_n]$ and $K_n = [c_n,d_n]$.
Suppose, for some $N \in \mathbb{N}$, $a_N > d_N$.
We will show that this is not possible.
If $a_N > d_N$, then $a_{N+i} \geq a_N > d_N \geq d_{N+i}> c_{N+i}$
This implies $a_{N+i} - c_{N+i} > a_N - c_N$ for all natural $i$.
Thus by definition, $[\alpha] > [\beta]$.
This is clearly a contradiction. (By the law of trichotomy)
Therefore, our initial supposition was false; and so for all natural $n$, $a_n <\leq d_n$.
Now this also means that the real number $[d_N]$ for some fixed natural $N$ (where $d_n$ is the same $d_n$ as mentioned above, that is, $d_n$ is in general a right endpoint of the constructive interval sequence $\{K_n\} \in [\beta]$ where $[\beta]$ is some upper bound of $S$) is an upper bound of $S$, as if $[\alpha] > [d_N]$, this implies that for all $n>M$, $a_n - d_N > \epsilon > 0$, which contradicts the fact we proved earlier. Thus the fact that $[\alpha] \leq [d_N]$ follows from the law of trichotomy.
Note that $d_N$ is also a rational number, as it is one of the right endpoints of $\{K_n\} \in [\beta]$
We will now go on to prove a useful lemma:
$Lemma \ 1 - $ For all $n \in \mathbb{N}$, there exists a left endpoint $a_n \in \{I_n\} \in [\alpha] \in S$ such that $a_n + \frac{1}{n}$ (which is a rational number) is an upper bound of $S$.
$Proof \ (Lemma \ 1) - $ Suppose there exists a natural number $m$ where for any left endpoints $a_m \in \{I_n\} \in [\alpha] \in S$ (where $[\alpha]$ is any element in $S$, $\{I_n\}$ is any constructive interval sequence in $[\alpha]$), $a_m + \frac{1}{m}$ is NOT an upper bound of $S$.
This means that there must exist some real number $[\alpha_1] \in S$ such that $[a_m + \frac{1}{m}] < [\alpha_1]$.
This means that for a constructive interval sequence $[y_i,t_i]$ in $[\alpha_1]$,
$y_i - (a_m + \frac{1}{m}) > 0$ for all $i \geq N$ where $N$ is some natural number.
Which means that in particular, $y_N - (a_m + \frac{1}{m}) > 0$.
Now $y_N$ is some left end point of $[y_i,t_i] \in [\alpha_1] \in S$, and therefore using the exact same argument as before, $y_N + \frac{1}{m}$ may not be an upper bound of $S$, which means that $a_m + \frac{2}{m}$ is not an upper bound of $S$ either.
Continuing this process, we realize that we will eventually conclude that $a_m + \frac{k}{m}$ can never be an upper bound of $S$ for any natural $k$, we know this must yield a contradiction, as at some point $a_m + \frac{k}{m} \geq d_N$ for some $k$, due to the fact that the natural numbers are unbounded in $\mathbb{Q}$
(This can be proven formally with induction)
$\blacksquare$
Now with the help of Lemma 1, we will begin our construction of the 'least upper bound' or 'supremum' of $S$.
We will do so by first considering an increasing sequence $\{a_n\}$ where for each natural $n$ (we start indexing at n=1), $a_n + \frac{1}{n}$ is an upper bound of $S$. By Lemma 1, these $a_n$ exist as left endpoints of constructive interval sequences within numbers $[\alpha]$ in $S$.
The only thing we need to show is that we can find an INCREASING sequence of such numbers.
Suppose for the sake of contradiction, $a_{n+1} \leq a_n$. Then $a_{n+1} + \frac{1}{n+1} \leq a_n + \frac{1}{n+1}$. By definition, $a_{n+1} + \frac{1}{n+1}$ is an upper bound of $S$ and therefore $a_n + \frac{1}{n+1}$ is also an upper bound of $S$. Thus in order to fix this issue, we can just swap indexes. That is, rather than call $a_{n+1}$, $a_{n+1}$, we simply call it $a_n$ and call $a_n$, $a_{n+1}$. In this way, we have straightened out the issue, as $a_n < a_{n+1}$.
Thus we can create such a sequence.
We will now consider the opposite side, that is, we want to produce a decreasing sequence of upper bounds, $b_n$, such that $b_n - a_n \leq \frac{1}{n}$ for all $n$.
First we need to show this is indeed possible.
Suppose $a_{n+1} + \frac{1}{n+1} \geq a_n + \frac{1}{n}$.
If this is the case, choose some real number $[\beta_{n+1}] < [a_n + \frac{1}{n}]$ such that $[\beta_{n+1}]$ is an upper bound of $S$. If such a real number does not exist, then this implies that $[a_n + \frac{1}{n}]$ is the least upper bound of $S$ and we would be done.
Before we continue, let us prove one more useful lemma:
$Lemma \ 3 \ - $ $[\alpha] < [\beta]$ if and only if for some $\{I_n\} = [a_n,b_n]$ and $\{K_n\} = [c_n,d_n]$ in $[\alpha]$ and $[\beta]$ respectively, $d_n - b_n > \epsilon$ for all $n>M$ where $M$ is some natural number and $\epsilon>0 \in \mathbb{Q}$
$Proof \ (Lemma \ 3) - $ Let $[\alpha] < [\beta]$. Then for $\{I_n\} = [a_n,b_n]$ and $\{K_n\} = [c_n,d_n]$ there exists a natural number $M$ such that for all $n>M$, $c_n - a_n > \epsilon_1 > 0$ for some $\epsilon_1 \in \mathbb{Q}$.
Now, as $[a_n,b_n]$ and $[c_n,d_n]$ are both constructive interval sequences, there exists a natural number $P$ such that for all $n>P$, $|b_n - a_n| < \frac{\epsilon_2}{2}$ and $|d_n - c_n| < \frac{\epsilon_2}{2}$, where $\epsilon_2 < \epsilon_1$. Therefore, $ -\frac{\epsilon_2}{2} < a_n - b_n < \frac{\epsilon_2}{2}$ and $-\frac{\epsilon_2}{2} < d_n - c_n < \frac{\epsilon_2}{2}$.
Adding together both inequalities, we have $ -\epsilon_2 <d_n - b_n + a_n - c_n < \epsilon_2$, which impliess $ \epsilon_1 - \epsilon_2 < d_n - b_n$ for all $n>P$.
Now the proof of the converse is symmetric.
$\blacksquare$
Continuing on, we will choose some $\{M_i\} \in [\beta_{n+1}]$ where $\{M_i\} = [t_i,b_i]$. Note that the constructive interval sequence $[p_i,a_n + \frac{1}{n}]$ where $\displaystyle \lim_{i \to \infty} (a_n + \frac{1}{n}) - p_i = 0$, certainly belongs to $[a_n + \frac{1}{n}]$. Then by definition, for all $i > N$, $ a_n + \frac{1}{n} - b_i > \epsilon > 0$ for some rational $\epsilon$ (this is a consequence of Lemma 3). In particular, we can find a rational number, which we will denote $b_{n+1}$ such that $a_n + \frac{1}{n} - b_{n+1} > 0$ and $b_{n+1}$ is a rational upper bound of $S$.
The fact that $b_{n+1}$ is an upper bound of $S$ follows from the discussion above pertaining to the fact that for any $[\alpha] \in S$ and any upper bound $[\beta]$, a left endpoint of a constructive interval sequence in $[\alpha]$ may never be greater than a right end-point of a constructive interval sequence of $[\beta]$.
It thus follows that if the event where $a_{n+1} + \frac{1}{n+1} \geq a_n + \frac{1}{n}$ occurs, we are always able to find a rational upper bound $b_{n+1} < a_n + \frac{1}{n}$, this also implies that $b_{n+1} - a_{n+1} < \frac{1}{n+1}$.
We can thus inductively build a constructive interval sequence $[a_n,b_n]$ where $b_n - a_n < \frac{1}{n}$ for all natural $n$ (we start indexing at $n=1$), where $b_n$ is a rational upper bound of $S$, and $a_n$ is some left end point of some constructive interval sequence of some $[\alpha]$ in $S$.
Clearly, this interval sequence fulfills the criterion for a constructive interval sequence, and thus belongs to an equivalence class.
Let us denote the equivalence class this sequence belongs to $[\xi]$.
We will show that $[\xi]$ is the least upper bound of $S$.
Consider some $[\alpha] \in S$. Take $[c_n,d_n] \in [\alpha] \in S$.
Case 1 - $d_n$ is an upper bound of $S$ for all $n$.
In this case we will show that $[\alpha] = [\xi]$.
Since $[c_n,d_n]$ is a constructive interval sequence , we know that $\displaystyle \lim_{n \to \infty} d_n - c_n = 0$.
Therefore, for $n>M$, we can ensure $d_n - c_n < \frac{1}{N}$ for some natural number $N$.
Now take $[a_n,b_n] \in [\xi]$. We know that $b_n - a_n < \frac{1}{n}$ and $b_n$ is an upper bound of $S$, $a_n$ is some left end point of some constructive interval sequence of some $[\alpha] \in S$.
For $n>M$, we know that $c_n + \frac{1}{N}$ is an upper bound of $S$.
We also know that for $n>N$, $a_n + \frac{1}{N}$ is an upper bound of $S$ by definition.
Take some $n> max(N,M)$.
Suppose $|c_n - a_n| > \frac{1}{N}$.
then $c_n - a_n > \frac{1}{N}$.
This implies that $c_n > a_n + \frac{1}{N}$
But, $[a_n + \frac{1}{N}]$ is an upper bound of $S$, and so the constructive interval sequence $[m_i,a_n + \frac{1}{N}]$ where $\displaystyle \lim_{i \to \infty} (a_P + \frac{1}{N}) - m_i = 0$, certainly belongs to $[a_P + \frac{1}{N}]$.
We discussed earlier that any left end point of any constructive interval sequence of some $[\alpha]$ in $S$ may not exceed a right end-point of a constructive interval sequence of an upper bound of $S$.
Thus we reach a contradiction.
Similarly, we reach a contradiction if $a_n - c_n > \frac{1}{N}$ for some $n> max(N,M)$.
Thus for all $n> max(N,M)$ , $|c_n - a_n| \leq \frac{1}{N}$.
Since $N$ was any natural number, we have shown that $[\xi] = [\alpha]$.
Case 2 - For some natural $n$, $d_n$ is not an upper bound of $S$.
This means that for all $n>M$ for some natural $M$, neither $c_n$ nor $d_n$ are upper bounds of $S$.
Now, as $b_n$ (a right endpoint of $[a_n,b_n] \in [\xi]$) is always an upper bound of $S$ by definition, we have $b_n - d_n > 0$ for all $n>M$.
By lemma 3, an equivalent definition for $[\alpha] < [\beta]$ was that if $\{I_n\} = [a_n,b_n] \in [\alpha]$ and if $\{K_n\} = [c_n,d_n] \in [\beta]$, for all $n>L$ for some natural $L$, $d_n - b_n > \epsilon$ for some positive rational $\epsilon$.
Since we know that in our case, $b_n - d_n > 0$ for all $n>M$, it is not possible for $[\alpha] > [\xi]$.
Thus by the law of trichotomy, $[\alpha] \leq [\xi]$.
$\square$
Thus we have shown that $[\xi]$ is an upper bound of $S$. All that is left to do now is to show that it is indeed the least upper bound of $S$.
Suppose, for the sake of contradiction, for some upper bound $[\beta]$ of $S$, $[\beta] < [\xi]$.
Let $\{K_n\} = [c_n,d_n] \in [\beta]$. Then for all $n>M$, $a_n - c_n > \epsilon > 0$.
However, as $[c_n,d_n]$ is a constructive interval sequence, for all $n>T$ where $T$ is some natural number, we can ensure that $d_n - c_n < \epsilon$
Thus, for $n> max(T,M)$ , $d_n < a_n$.
But $d_n$ is a right end point of a constructive interval sequence of an upper bound of $S$, and $a_n$ is a left endpoint of some $[\alpha] \in S$. Therefore, $a_n \leq d_n$ for all $n$.
This yields a contradiction.
Therefore $[\beta] < [\xi]$ is false, so by the law of trichotomy, $[\xi] \leq [\beta]$ for any upper bound $[\beta]$ of $S$.
Thus $[\xi]$ is indeed the least upper bound of $S$, and $\mathbb{R}$ under this construction is complete.
$\blacksquare$
For the sake of completeness (no pun intended), I will verify that all the field axioms hold and that the previously defined operations of $+$ and $*$ are well-defined in my next post. (Hopefully by the end of today or tomorrow). We will also stop denoting real numbers as '$[ ]$' and refer to them as normal.
Let's prove the Monotone Convergence Theorem:
$Theorem \ (Monotone \ Convergence) - $ Let $\{a_n\}$ be a sequence of non-strictly increasing real numbers, that is, $a_1 \leq a_2 \leq a_3 ...$.
Furthermore, suppose there exists a real number $b$ such that $a_n \leq b$ for all $n$.
Then $\displaystyle \lim_{n \to \infty} a_n$ exists.
$Proof \ (Monotone \ Convergence \ Theorem) - $ Consider the set non-empty subset of the real numbers, $\{a_n | n \in \mathbb{N} \}$. Clearly, this set is bounded above by $b$.
Therefore, by the least upper bound property of the real numbers, there must exist a number $\xi$ that is the least upper bound of the aforementioned set.
We will show that $lim_{n \to \infty} a_n = \xi$.
Consider $|a_n - \xi| = \xi - a_n$.
For the sake of contradiction, suppose there does not exist a natural number $m$, such that $\xi - a_m < \frac{1}{n}$ where $n$ is some arbitrary natural number.
Then for all $m$, $\xi - a_m \geq \frac{1}{n}$ and $a_m \leq \xi - \frac{1}{n}$ for all $m$.
This is a contradiction, as then $\xi - \frac{1}{n} < \xi$ is a smaller upper bound than $\xi$ of the aforementioned set.
Thus, there must exist a natural number $m$ such that $\xi - a_m < \frac{1}{n}$, which implies, as $a_n$ is non-strictly increasing, that $\xi - a_k < \frac{1}{n}$ for all $ k> m$.
As $n$ was arbitrary, we are finished.
$\blacksquare$
In the next post we will also consider the case where $\{a_n\}$ is a non-strictly decreasing sequence of real numbers.
Thanks for reading.
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