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Would the sequence of partial sums from indefinitely adding $1.0 \times 10^{-n}$ to zero, correspond to the set of positive real numbers?

BLAZE
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Mark
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  • Is $n$ fixed or varying here? – πr8 Jan 05 '16 at 19:54
  • If $n$ increases at least once every $10$ times, then yes. I'd express it a bit differently, more like $$\sum_{n=0}^\infty a_n\cdot 10^{-n}$$ where $a_n\in{0,1,2,3,4,5,6,7,8,9}$ for all $n$. – Darth Geek Jan 05 '16 at 20:02
  • How would you express $e$ and $\pi$ with such a formula? – JB King Jan 05 '16 at 20:05
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    @DarthGeek The term "partial sum" indicates finite sums, so the answer is no - in particular, all partial sums are rational numbers, so you can't get $\sqrt{2}$ or $e$ or $\pi$. But the question is vague - what is meant by a sequence of partial sums? – Thomas Andrews Jan 05 '16 at 20:09
  • @ThomasAndrews But the question is not about partial sums but about sequences of partial sums. The sequence $1,2,3,3.1,3.11,3.12,3.13,3.14,3.141,\ldots$ is a sequence of such partial sums that can be associated with $\pi$. Then again, the wording is a bit odd and looks like there's not enough information about the behaviour of $n$. I'd recommend OP to change the wording to an unambiguous one, – Darth Geek Jan 05 '16 at 20:12
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    I'd recommend clarifying this question, Mark. It is entirely unclear what you mean by "the" sequence of partial sums. What do you really mean by that? – Thomas Andrews Jan 05 '16 at 20:13
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    But the question says "would the sequence ... correspond to the set of positive real numbers." Not "all sequences ..." but "the sequence." It is very badly worded question. @DarthGeek – Thomas Andrews Jan 05 '16 at 20:14
  • Can I suggest this previous question http://math.stackexchange.com/questions/1022227/real-numbers-as-decimals and links including exposition by Tim Gowers https://www.dpmms.cam.ac.uk/~wtg10/decimals.html – Mark Bennet Jan 05 '16 at 21:42
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    It seems that this account is also yours. It would probably be advisable that you merge the accounts, so you can participate in this thread as the OP. – Daniel Fischer Mar 26 '16 at 22:49

2 Answers2

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No, not all real numbers can be expressed as $\sum_{i=1}^k10^{-n}$. This is only the set of real numbers with a finite decimal expansion. For instance, the number $\frac{1}3=.333333\ldots$ is not a finite sum of terms of the form $10^{-n}$ since we have the bounds $$\frac{10^n-1}{3}\cdot 10^{-n}<\frac{1}3<\frac{10^n+2}{3}\cdot 10^{-n}$$ where the terms $\frac{10^n-1}3$ and $\frac{10^n+2}3$ can be seen to be consecutive integers, meaning $\frac{1}3$ is always between two of the partial sums. Another way to write this inequality is: $$0.\underbrace{33\ldots33}_{n-2\text{ times}}3<\frac{1}3<0.\underbrace{33\ldots33}_{n-2\text{ times}}4.$$

That said, numbers with finite decimal expansions are "dense" in the reals, which means that they can approximate any other number arbitrarily well. That is, if we chose some real number $x$ and some value $\varepsilon>0$ we can find a suitable number with finite decimal expansion which differs from $x$ by no more than $\varepsilon$. For instance, if we want a number that approximates $x=\pi$ to within $\varepsilon=10^{-4}$, we could see that the number $3.1415$ satisfies that and can be written of the desired form.

Milo Brandt
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    But nontheless, you can associate the sequence $1, 1.3, 1.33, 1.333,\ldots $ to an element of $\mathbb{R}$, namelly, to $1/3$. – Darth Geek Jan 05 '16 at 20:07
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    Yeah, the question asked about sequence of partial sums, which is an odd use of terminology. Basically, the question needs clarification about what is meant. – Thomas Andrews Jan 05 '16 at 20:11
  • @ThomasAndrews Ah. The question had seemed quite clear to me, but I had not thought of that very different interpretation of the question (I assumed they were equating the union of such sequences with the reals, but it seems you're suggesting they may have meant to equate the set of such sequences with the reals). I guess I'll leave this answer here unless the OP clarifies things. – Milo Brandt Jan 05 '16 at 20:13
  • Hasn't every number got infinite digits, 1 = 1.0... = 1.0' and wouldn't the sequence of partial sums be uncountably infinite when n is infinity? R0 = {0.0 x 10^-n}, R1 = {0.0 x 10^-n + 1.0 x 10^-n}, ... – Mark Jan 05 '16 at 22:05
  • @Mark I mean that, though $1$ an infinite decimal expansion $1.000\ldots$, it also has finite decimal expansions (e.g. $1=1.0=1.00=1.000$). The point is that some number lack a finite decimal expansion. Also, at least in standard analysis, $n$ can't be infinity (since the only reasonable definition of $10^{-\infty}$ is $0$ in $\mathbb R$) and a sequence of partial sums is bound to be countable. – Milo Brandt Jan 05 '16 at 22:19
  • @Mark A sequence of partial sums would have a countable number of terms. The kind of "sequence" you seem to be trying to create is not a mathematical object. – David K Jan 05 '16 at 22:42
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The vital missing pieces of information in the original question statement are the assumptions implied by this comment:

Hasn't every number got infinite digits, 1 = 1.0... = 1.0' and wouldn't the sequence of partial sums be uncountably infinite when n is infinity? R0 = {0.0 x 10^-n}, R1 = {0.0 x 10^-n + 1.0 x 10^-n}, ...

Yes, every number can be said to have infinitely many digits. The question is, what is $10^{-n}$ if $n$ is infinite? Whatever it is, it is not the smallest possible positive real number, because there is no such thing.

A reasonable interpretation would be that by "$10^{-n}$ where $n$ is infinite" you actually mean the limit of $10^{-n}$ as $n$ goes to infinity. This is evaluated as follows:

$$ \lim_{n\to\infty} 10^{-n} = 0.$$

This does not produce the sequence you want, because you can add $0$ to $0$ forever and it will always remain $0$.

In any case, there is no $n$ such that you can get "the next" real number by adding $10^{-n}$ to any real number, because there is never a unique "next" real number. Suppose $R_0$ is a real number and $R_1$ is "the next" real number greater than $R_0$. Then $x = (R_0 + R_1)/2$ is also a real number, but $R_0 < x < R_1$, so $R_1$ is not "the next" real number greater than $R_0$.

The best you can do by constructing any increasing sequence of partial sums is to generate a countable subset of the real numbers with uncountably many "missing" real numbers between each pair of numbers in your sequence.

David K
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  • The identifier representing any Von Neumann ordinal is arbitrary, so would there be a real number corresponding to every ordinal given that for consecutive identifiers, O1 and O2, representing successive ordinals, there is an identifier x = (O1 + O2)/2 with O1 < x < 02? – Mark Jan 08 '16 at 21:22
  • @Mark If you mean you want to map each Von Neumann ordinal to a unique real number, you can do that. Just like any other way of enumerating a countable sequence of real numbers, this gives you a way to talk about the "next" real number in that sequence. But no matter how you construct your sequence, it will "skip" infinitely many real numbers which will never show up in the sequence at all. That's what Cantor proved more than a century ago. Hence your sequence will not correspond to the positive real numbers. – David K Jan 08 '16 at 21:36
  • If an identifier, x', can be found that corresponds, x' := x, to any possible real number, x, then would the identifier sequence qualify for real number model, R', status even though it isn't the real real numbers? – Mark Jan 09 '16 at 20:31
  • @Mark We already have such an identifier for each real number $x$; that's part of how we construct any model of the real numbers. For example, a Dedekind cut of $\mathbb Q$ can identify a real number. But you will inevitably fail to model the real numbers if you insist upon drawing your identifiers only from a countable set. – David K Jan 09 '16 at 20:40
  • Von Neumann may have shown Cantor that a continuous set can consist of discrete numbers, so any sequence representing the set must be uncountable. – Mark Jan 10 '16 at 18:59
  • @Mark Are you referring to the "continuity" assumption of the Von Neumann-Morgenstern utility theorem? That terminology may be inspired by other uses of the word "continuity" but that doesn't mean you can apply it here. – David K Jan 10 '16 at 19:27
  • If the ordinals are essentially a single entity, class, because every ordinal can be ' traced' to one element, the empty set, then are the ordinals uncountable even though any sequence modelling them is not? – Mark Jan 16 '16 at 06:53