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I have understood, I think, the construction of the real numbers as the set of equivalence classes of Cauchy sequences. That is, if $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences, then we say they are equivalent if $\lim_{n\to \infty} \lvert a_n - b_n\rvert = 0$. The set of equivalence classes is then a field under the "obvious" operations. This is what I believe is the completion of $\mathbb{Q}$ with respect to the absolute value.

How does one go from this to the fact that any real number can be described as "infinite decimals"? That is, given one of the equivalence classes $[\{a_n\}]$ how do we get that

$$ [\{a_n\}] = \sum_{i=-n}^\infty a_i10^{-i} $$ ?

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    $\begingroup$ Note that every decimal expansion is a Cauchy sequence, for example $3, 3.1, 3.14, 3.141, 3.1415, 3.14159, \ldots$. $\endgroup$ – Connor Harris Oct 8 at 20:53
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    $\begingroup$ @ConnorHarris True, but I believe OP is asking something more like "given a Cauchy sequence, how can you prove there is a decimal expansion to describe it? $\endgroup$ – 79037662 Oct 8 at 20:57
  • $\begingroup$ @79037662: Yes, that is what I want to know. But also how I can find it. $\endgroup$ – John Doe Oct 9 at 12:13
  • $\begingroup$ We probably can prove that there is no convergence class that does not contain at least one of the sequences described by Connor Harris. Some even contain two, as $1.99999999999999....=2.000000000000...$ $\endgroup$ – David Oct 11 at 12:38
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    $\begingroup$ The responses so far just go to show that your question is not an easy one to answer! $\endgroup$ – TonyK Oct 11 at 22:50
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Your question is "Why can any number $x$ be described as "infinite decimals"".

Do you know about the floor function on the reals, aka the archimedian property ? If you don't, I can explain it here.

Consider the sequence $(b_n)_{n\geq 1}$ defined by $$b_n=\frac{\lfloor 10^{n}x \rfloor}{10^{n}} \tag{1}$$

Then by construction, we have $$ 10^{n}x-1 \lt b_n \leq 10^{n}x \tag{2} $$ It follows that $$ 10b_n-1 \leq 10(10^n x)-1= 10^{n+1}x-1 \lt b_{n+1} \leq 10^{n+1}x =10(10^n x) \lt 10(b_n+1) \tag{3} $$

So the integer $d_n=b_{n+1}-10b_n$ is strictly between $-1$ and $10$, in other words it is a digit.

Next, consider the sequence $(s_n)$ defined by $s_n=\frac{b_n}{10^n}$. By construction, we have $|s_n-x| \leq \frac{1}{10^n}$ so the sequence $(s_n)$ converges to $x$. But $$s_n=b_0+\frac{d_1}{10}+\frac{d_2}{10^2}+\ldots+\frac{d_n}{10^n}=b_0.d_1d_2\ldots d_n \tag{4}$$

This justifies viewing $x$ as an "infinite decimal"

$$ x=b_0.d_1d_2\ldots \tag{5} $$

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    $\begingroup$ But you are using $x$ as if it were a fully-formed element of $\Bbb R$, which is not at all in the spirit of Cauchy sequences. Can you define $b_n$ in terms of the sequence $(a_i)$? I think that is what this question is really about. $\endgroup$ – TonyK Oct 11 at 22:48
  • $\begingroup$ @TonyK 1) If you wish you can replace everywhere $x$ in this answer with $\lim_{n\to\infty}{a_n}$. This way everything is in terms of the sequence $(a_n)$. 2) What do you mean by "fully-formed element of $\mathbb R$" and "spirit of Cauchy sequences" $\endgroup$ – Ewan Delanoy Oct 12 at 4:17
  • $\begingroup$ Here, Cauchy sequences are being used to define the reals. So to use a "fully-formed" real number (whether you call it $x$ or $\lim_{n\to\infty}a_n$) is assuming as understood what the OP is trying to get to grips with. $\endgroup$ – TonyK Oct 12 at 8:44
  • $\begingroup$ @TonyK The OP says he already knows that the real numbers are a field and he already knows that the real numbers are the completion of $\mathbb Q$. I did not use anything else except the floor function, offering more explanations if needed. $\endgroup$ – Ewan Delanoy Oct 12 at 8:59
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Here we map each Cauchy equivalence class $0 \lt [(a_n)] \lt 1$ to its decimal expansion.

Recall the definition of decimal fractions, said another way, the finite decimal expansions. The class $[(a_n)]$ might be represented by a constant sequence with all the terms equal to a decimal fraction. We can then unambiguously associate to this class a terminating decimal expansion.

To handle the remaining Cauchy equivalence classes, we need to prove the following:

Lemma 1: Let $0 \lt [(a_n)] \lt 1$ be given and suppose it can't be represented by a finite decimal expansion. Then for every $k \ge 1$ there exist one and only one decimal fraction

$\tag 1 \frac{h}{10^k} \text{ with } 0 \le h \lt 10^k$

such that for every representative $(b_m)_{\,m \ge 0} \in [(a_n)]$ there exist $N \in \Bbb N$ such that

$\tag 2 \text{For every } n \ge N, \; \frac{h}{10^k} \lt b_n \lt \frac{h+1}{10^k}$

Proof Sketch: Use the triangle inequality over $\Bbb Q$ and the definition of a Cauchy sequence.

Proposition 2: The mapping $k \mapsto h$ gives the first $k$ digits of the infinite decimal expansion to the right of the decimal point, by placing the $k$ digits of the $\text{base-}10$ expansion of the integer $h$ in accordance with the prescription of our positional numeral system
(it may be necessary to 'pad' with zeroes right after the decimal point).


We can directly build the decimal expansion for a real 'Cauchy' number $x$ using the above theory. If $x \lt 0$ we build it for $-x$ and then put the minus sign back.

You can choose any Cauchy sequence representing $x \gt 0$ and you will get the same decimal expansion. Using very simple arguments you 'strip out' the integer part (expansion left of the decimal point) which might actually be $x$. In any case, the problem is reduced to building the expansion for a Cauchy sequence of rationals $(a_n)$ where there exist $N$ such that for $n \ge N$, $\;0 \lt a_n \lt 1$
and $(a_n)$ is not a decimal fraction.

Let $k = 1$. For some $N$ all the terms $a_n$ for $n \ge N$ can be guaranteed to lie in one of the $10$ sub-open-interval of length $\frac{1}{10}$. This '$h$' starts the process of building, from left to right the digits after the decimal point, the decimal estimates for $x$.

You then keep repeating, at each $k$ step dividing each new 'target' into $10$ sub-open-intervals of equal length $10^{-k}$ to get the next decimal digit in $\{0,1,2,\dots,9\}$.

When you get up to the $k^{th}$ build, $d = 0.d_1 d_2 \dots d_k$, we have

$\tag 3 x \gt d \text{ and } x - d \lt \frac{1}{10^k}$

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    $\begingroup$ This looks good to me. Your crucial idea is to treat decimal fractions separately, so there is no problem with $0.4999\ldots$ and such like numbers. $\endgroup$ – TonyK Oct 15 at 13:13
  • $\begingroup$ @TonyK Yes, I avoided all mention of the $.9999....$ stuff, leaving it as a subtle 'background thing' for the astute reader to observe. $\endgroup$ – CopyPasteIt Oct 15 at 13:16
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To extend the great answer of Ewan Delanoy:

You define the real numbers as a structure fulfilling pregiven properties (the axioms). Now, you need to show that all models of the structure (all "matching" constructions) are isomorphic (with respect to constants, relations and operations) and that the construction as infinite decimals is a model of your structure. This is what you want to ask in the language of mathematical logic.

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Well given a Cauchy sequence $\{a_n\}$ we can construct another Cauchy sequence $\{b_n\}$ that is equivalent to the first and also $$b_{n+1}=b_n+\frac{c_n}{10^{n+1}}$$ where $c_n\in\{0,...,9\}$

i.e.: $\{b_n\}$ can be interpretated as a decimal expansion.

Let's do so: suppose without loss of generality that the sequence $\{a_n\}$ is non-decreasing. All sequences have a monotone subsequence, so we can always take a monotone subsequence of $\{a_n\}$. If it is non-increasing the construction will be analogous, changing the way of some inequalities. For a fixed $n$ we have some $N(n)$ such that if $l,k\geq N$ then $$|a_l-a_k|<\frac{1}{2\cdot 10^{n+1}}$$ define $b_n$ as beign the number of the form $\frac{M}{10^n}$ which minimizes the difference $a_N-b_n\geq 0$. As the naturals are well ordered there is always such a number.

Note that $a_N-b_n-10^{-n}< 0$ (otherwise $b_n$ wouldn't minimize the difference above), so $|a_N-b_n|< 10^{-n}$, this way is easy to see that the sequence $b_n$ is Cauchy and equivalent to $\{a_n\}$. Also the sequence $b_n$ is non-decrescing.

Now it's only left to prove that $b_{n+1}-b_n$ is a number of the form $\frac{c}{10^{n+1}} $, with $c\in\{0,...,9\}$. By our definition we have: $$|b_{n+1}-b_n|=|b_{n+1}-a_{N(n+1)}+a_{N(n+1)}-b_n|\leq|b_{n+1}-a_{N(n+1)}|+|a_{N(n+1)}-b_n|\leq$$ $$\leq\frac{1}{2\cdot 10^{n+1}}+\frac{1}{2\cdot 10^{n+1}}=\frac{1}{10^{n+1}} $$ thus as $b_n$ non-decrescing, $b_{n+1}$ can only be of the form $b_n+\frac{c_n}{10^{n+1}}$.

Note: intuitivelly we are taking aproximations of the form of decimals of the limit of $\{a_n\}$ from the left.

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  • $\begingroup$ This doesn't work. The problem is, even though $b_n\le a_N$, we can't be sure that $b_n$ is not greater than the limit of the $a_i$. You can't necessarily deduce a decimal digit without having to look arbitrarily far ahead; your $N(n)$ isn't far enough (and neither would any function that depends only on a finite number of the $a_i$). To put it another way: your $(b_n)$ is not necessarily a non-decreasing sequence. $\endgroup$ – TonyK Oct 11 at 22:44
  • $\begingroup$ I see, you are right, there is no way of being sure where the limit will be just by making the terms of the sequence close... But is there no way of making a sequence of $b_n$ without computing the limit? $\endgroup$ – Arararararagi-kun Oct 11 at 22:53
  • $\begingroup$ In general, no. $\endgroup$ – TonyK Oct 11 at 23:47
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Well, I have been thinking about my first answer and I think it's kind of confusing... I assume you want a natural way of associating a cauchy sequence with a decimal expansion, without calculating the real number related to it.

We can think of which intervals the final segment of the sequence is and put lowers bounds using a decimal expansion, which comes quite natural.

From some point on, the sequence is contained in an interval of the form $[n,n+1]$, unless it converges to an integer, which has a trivial decimal expansion.

Let's suppose that $a_n\in[0,1]$ and give an expansion in binary, which is similar to the decimal and whose construction is clearer.

We start dividing the interval in two: $I_0=[0,1/2],I_0'=[1/2,1]$. If the first part has infinite points of the sequence we set $b_1=0$, otherwise we set $b_1=1$. It is clear that $b_1\leq a_n$ for all $n>N$, for some $N$.

Now we repeat the process, dividing each segment into two and choosing the extremity of lowest with infinite points to be $b_n$.

We can also suppose that there are no cases where 2 intervals has infinite points, because if they both have the limit must be the common endpoint and it has a finite binary expansion.

It is clear that the numbers $b_n$ form a decimal expansion.

The sequence of $b_n$ is equivalent to $a_n$: the final segment of $\{a_n\}$ lies within the same interval as $b_n$, whose length is $2^{-n}$. So we must have: $$|b_n-a_k|<2^{-n}$$ If $k$ is sufficiently big.

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