On base-b expansion of nonnegative reals

[psie]

TL;DR Summary

It has probably been asked numerous times before, yet I couldn't find anything complete enough that rigorously proves that every nonnegative real number has a base-$b$ expansion.

In Folland's real analysis text, twice (so far) he drops a very similar statement about base-$b$ expansions of nonnegative reals.

The first time is when discussing the proof of $\operatorname{card}(\mathcal{P}(\mathbb N))=\mathfrak c$, where he says that every positive real number has a base-$2$ decimal expansion. The existence of such expansion is crucial for the proof. The other time is when discussing the Cantor set, where he says that every real number in $[0,1]$ has a base-$3$ decimal expansion which is unique unless the number is of the form $p3^{-k}$ for integers $p,k$. Some have said "Yes, but this is due to $0.\overline{9}=1$." But I don't think this explains, or rather proves, why the expansion is unique (it does not prove existence either).

I'm looking for lecture notes, books, anything you could imagine citing yourself if you'd want the reader to learn more about existence and uniqueness of base-$b$ expansions of nonnegative reals. If it's possible to give a treatment of this as reply to this post, I'd be happy to accept it.

 

[fresh_42]

You cannot achieve uniqueness, e.g. $1=0.\overline{9}.$ One way to define real numbers is through equivalence classes modulo zero-Cauchy sequences. So every real number is a representative of such a class. Is your question why there always exists a sequence $(a_n)_{n\in \mathbb{Z}} \subseteq \{0,1,\ldots,b-1\}$ with almost all members with negative indices are zero such that a given real number $r$ equals
$$
r=\sum_{k=0}^\infty a_{k-n}b^{-k+n}\;?
$$
The point with every real number is that we need a definition for real numbers. The equation above is already a rational Cauchy-sequence and the real number $r$ is in my definition simply the limit that we add to the rationals. I guess, things become a bit more technical if we use Dedekind cuts instead.

Say we use the definition by those equivalence classes. Then the question reduces to whether every rational number can be expressed this way. That's probably some boring work to do but not difficult.

 

[psie]

You cannot achieve uniqueness, e.g. $1=0.\overline{9}.$ One way to define real numbers is through equivalence classes modulo zero-Cauchy sequences. So every real number is a representative of such a class. Is your question why there always exists a sequence $(a_n)_{n\in \mathbb{Z}} \subseteq \{0,1,\ldots,b-1\}$ with almost all members with negative indices are zero such that a given real number $r$ equals $$r=\sum_{k=0}^\infty a_{k-n}b^{-k+n}\;?$$

Yeah, I think that's my question more or less. I wonder if

1. Why there exists a sequence $(a_n)_{n\in\mathbb Z}\subset\{0,1,\ldots,b-1\}$ with finitely many members being nonzero for nonpositive indices such that $r=\sum_{k=-\infty}^\infty a_kb^{-k}$? (I think if we do not allow $a_k=b-1$ for infinitely many positive indices, one can get a unique expansion.)
2. If we take the example with the Cantor set and $[0,1]$, why the expansion is unique (or why it isn't) for numbers not of the form (or of the form) $x=p3^{-k}$ where $p,k$ are integers such that $x\in[0,1]$?

The equation above is already a rational Cauchy-sequence and the real number $r$ is in my definition simply the limit that we add to the rationals.

I don't understand this sentence. Yes, the terms in the sum are rationals. Do you mean the terms form a rational Cauchy sequence? What do you mean by "the limit that we add to the rationals"?

Dedekind cuts would be interesting I'm looking at Rudin's book, and he uses Dedekind cuts to construct the reals, though I lean more towards construction via Cauchy sequences.

 

[fresh_42]

Yeah, I think that's my question more or less. I wonder if

1. Why there exists a sequence $(a_n)_{n\in\mathbb Z}\subset\{0,1,\ldots,b-1\}$ with finitely many members being nonzero for nonpositive indices such that $r=\sum_{k=-\infty}^\infty a_kb^{-k}$? (I think if we do not allow $a_k=b-1$ for infinitely many positive indices, one can get a unique expansion.)

We need to allow infinitely many digits. This means we have to cope with equivalences anyway. Avoiding infinitely many $b-1$ appears artificial to me. And it is not obvious that this solves the problem. What about the number $\pi \cdot 0.\overline{9}$? The moment we use equivalence classes we fail uniqueness: $a$ cannot be distinguished from $a+ \lim_{n \to \infty}b^{-n}.$

1. If we take the example with the Cantor set and $[0,1]$, why the expansion is unique (or why it isn't) for numbers not of the form (or of the form) $x=p3^{-k}$ where $p,k$ are integers such that $x\in[0,1]$?

I'm not sure I understand this question. What do Cantor sets have to do with it?

I don't understand this sentence. Yes, the terms in the sum are rationals. Do you mean the terms form a rational Cauchy sequence? What do you mean by "the limit that we add to the rationals"?

I meant that
$$
r=\lim_{m \to \infty}\underbrace{\left(\sum_{k=0}^m a_{k-n}b^{-k+n}\right)}_{=a_m}
$$
which is a Cauchy sequence $(a_m)_{m\in \mathbb{N}}\subseteq \mathbb{Q}$ of rational numbers and the given real number $r$ is its limit. "Add to the rationals" was sloppy for the construction of real numbers by equivalence classes of Cauchy sequences of rational numbers modulo those Cauchy sequences that have $0$ as their limits. The construction is basically to take $\mathbb{Q}$ and a topological completion via Cauchy sequences, i.e. expand $\mathbb{Q}$ by those limits. The details take a few lines, about five pages in my book: from an Archimedean ordered field via topological completion of the rational numbers to the real numbers.