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Does there exist an uncountably infinite set $X \subseteq \mathbb R$ such that $\mathbb R \neq \left<X\right>$? I can't think of any, but I'm also having trouble trying to prove that no such subset exists.

For example: $\mathbb R$ is uncountable and obviously $\mathbb R = \left<\mathbb R\right>$. The Cantor set $C$ is uncountable, and we know that $C - C = [0, 1]$, so then since $\mathbb R = \left<[0, 1]\right>$ we know that $C$ also generates $\mathbb R$. Also the set of irrationals $\mathbb R \setminus \mathbb Q$ is uncountable, but we can generate all the rational numbers by fixing one irrational number $\alpha$ and then saying the any rational number $x$ shall be $(\alpha + x) - \alpha$, since both $\alpha + x$ and $\alpha$ are irrational.

So the examples that quickly come to mind all generate the reals. Is there a simple counterexample?

feralin
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    Do you mean "generate" in the sense of subgroups? – Arthur Mar 13 '16 at 22:40
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    Yeah. When I say "generates the reals" I mean "generates a subgroup of the reals which, in fact, ends up being all of the reals". – feralin Mar 13 '16 at 22:41
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    We can build a counterexample with the axiom of choice. I don't know of any simple answer, though. – Ben Grossmann Mar 13 '16 at 22:43
  • If you "generate in the sense of subgroups", aren't you limited to finite sums? (I.e., if you use the rationals and generate in the sense of subgroups, you only get the rationals; reaching the reals requires infinite sums.) For instance, a free group is limited to finite words. – Eric Towers Mar 14 '16 at 06:07
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    @EricTowers yes, generated subgroups are limited to finite sums, but I'm not sure what your point is. I hope I didn't write something blatantly stupid in my question :) – feralin Mar 14 '16 at 06:57
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    @EricTowers The set of rationals isn't uncountable, so it isn't a suitable counterexample to "all uncountable sets generate the reals". – Daniel Wagner Mar 14 '16 at 16:05
  • @DanielWagner : I don't claim that it is. I claim that "generate in the sense of subgroups" is not what I understand a Schauder basis to do. – Eric Towers Mar 15 '16 at 05:53
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    @EricTowers : ​ Did some comment before yours refer to Schauder bases? ​ ​ ​ ​ –  Mar 15 '16 at 09:54
  • @EricTowers I think you're misreading the OP; the question is whether there is a set of reals which (1) doesn't generate all of $\mathbb{R}$ by finite sums, but (2) is uncountable. Generally questions like this are easily settled by the axiom of choice, and sometimes (not in this case though) require it. – Noah Schweber Mar 16 '16 at 19:32

4 Answers4

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Well, the easiest way this can happen is if the continuum hypothesis fails - that is, if there is an uncountable set of reals $X$ such that $\vert X\vert<\vert\mathbb{R}\vert$. In this case it's easy to see that $\vert\langle X\rangle\vert=\vert X\vert<\vert \mathbb{R}\vert$, so the subgroup generated by $X$ is not all of $\mathbb{R}$.

Now, it is consistent with the usual axioms of set theory (ZFC) that the continuum hypothesis fails. However, it is also consistent that the continuum hypothesis holds; so this isn't really a solution. Can we do better?

Sure! Using the axiom of choice, we can show there is an uncountable set $X$ of reals such that the subgroup generated by $X$ doesn't contain $\pi$ (say). The way we do this is: let $\mathbb{P}$ be the set of all sets of reals $X$ which generate subgroups not containing $\pi$. Order $\mathbb{P}$ by inclusion. By Zorn's Lemma - a consequence of the axiom of choice (in fact, equivalent to it) - $\mathbb{P}$ has a maximal element, and it's not hard to show that such an element can't be countable.

But this still isn't great, because this $X$ is hard to describe - can we get an explicit example?

The answer, perhaps surprisingly, is yes! (Certainly it's surprising to me - in an early version of this answer, before I'd thought it through, I wrote that the answer to this subquestion is no.) See https://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets. Although we do need the axiom of choice to get a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space, we can get explicit uncountable linearly independent sets of reals in ZF alone. Then, given such a set, we can:

  • Examine the set given, and note that it doesn't generate all of $\mathbb{R}$. (I believe that ZF proves that no Borel set is a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space; certainly ZFC does.)

  • Or, just remove a single element, and then call the result our $X$. Con: marginally less "sweet." Pro: No complicated analysis needed!

Community
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Noah Schweber
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    Can you explain the statement "$\left|\left<X\right>\right| = \left|X\right|$" in the first paragraph? Is that similar to the statement that a countable set can only generate a countable subgroup? – feralin Mar 13 '16 at 23:07
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    @feralin Bingo. Remember that an element of $\langle X\rangle$ can be represented by a finite string of elements of $X$, together with a finite string of elements of $\mathbb{Z}$ (the corresponding coefficients). The number of such pairs of strings is just $X^{<\omega}\times \mathbb{Z}^{<\omega}$, which is just the size of $X$ again . . . assuming the axiom of choice, that is. Technically, it's possible to have a set of reals generate a strictly larger subgroup (e.g. a Dedekind-finite infinite set of reals will do this) but (a) it still won't be all of $\mathbb{R}$ and (b) that's weird. – Noah Schweber Mar 13 '16 at 23:10
  • Actually now that I think about it I'm not sure (a) is correct (although (b) certainly is); models where the axiom of choice fails are weird. I've asked a question about it http://math.stackexchange.com/questions/1696315/spanning-the-reals-with-a-small-set-choicelessly. – Noah Schweber Mar 13 '16 at 23:11
  • I'm struggling to understand what you're claiming to be right and what you're claiming to be wrong. Can you restructure the answer so that a newcomer to this question will be able to follow it easily? – goblin GONE Mar 14 '16 at 01:15
  • @goblin The italicized paragraph referred to a previous version of the answer, not what had appeared above. Better? – Noah Schweber Mar 14 '16 at 01:30
  • @NoahSchweber, thank you - yes, that's much clearer. – goblin GONE Mar 14 '16 at 01:32
  • It's consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, in which case every subset of $\mathbb{R}$ is Borel, so I'd be surprised if "ZF proves that no Borel set is a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space", although not for codable Borel sets. ​ ​ –  Mar 14 '16 at 16:23
  • @RickyDemer But remember that a basis has to be linearly independent! In models where the reals are a countable union of countable sets, getting large linearly independent sets is very hard. So that doesn't quite work. – Noah Schweber Mar 14 '16 at 16:27
  • That's why I'm not asserting non-provability. ​ ​ –  Mar 14 '16 at 16:29
  • "By Zorn's lemma ℙ has a maximal element, and it's not hard to show that such an element can't be countable." Could you sketch an argument for this? It seems to boil down to the original question... – alexis Mar 15 '16 at 11:44
  • @Ricky, I don't see why: By construction, the maximal element of ℙ does not even generate the reals-- let alone be equinumerous to them. – alexis Mar 15 '16 at 16:04
  • @alexis : ​ Oh, I was assuming Noah had defined $\mathbb{P}$ is the obvious way, rather than using $\pi$. ​ ​ ​ ​ –  Mar 15 '16 at 16:07
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    @alexis Suppose $X$ is a maximal element of $\mathbb{P}$ which is countable. Consider $Y=X\cup{\pi}$. $Y$ is again countable, so the $\mathbb{Q}$-subspace (not additive group!) generated by $Y$ is countable and hence not all of $\mathbb{R}$. Pick a real $s$ not in this subspace; then the subgroup generated by $X\cup{s}$ doesn't contain $\pi$, so $X$ wasn't maximal. – Noah Schweber Mar 15 '16 at 17:46
  • @NoahSchweber that makes sense, thanks! – alexis Mar 15 '16 at 19:43
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Note that $\mathbb{R}$ is a vector space over $\mathbb{Q}$; let $B$ be a basis. As $\operatorname{dim}_{\mathbb{Q}}\mathbb{R}$ is uncountable, $B$ is uncountable. Now let $b \in B$ and set $B_0 = B\setminus\{b\}$; note that $B_0$ is uncountable.

Suppose $b \in \langle B_0\rangle$, then there are $\alpha_1, \dots, \alpha_k \in \mathbb{Z}$ and $b_1, \dots, b_k \in B_0$ such that $b = \alpha_1 b_1 + \dots + \alpha_k b_k$. This is impossible as $\{b, b_1, \dots, b_k\}$ is linearly independent ($B$ is a basis).

Therefore $B_0$ is an uncountable set which does not generate $\mathbb{R}$.

Michael Albanese
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    +1. Note that the axiom of choice is used here to conclude that $\mathbb{R}$ has a basis as a $\mathbb{Q}$-vector space. – Noah Schweber Mar 13 '16 at 22:52
  • Absolutely right. In view of your extensive answer, I was going to edit mine to make the implicit use of the axiom of choice clear, but you beat me to it :) – Michael Albanese Mar 13 '16 at 22:56
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    @MichaelAlbanese am I understanding your answer correctly? If I take any basis of $\mathbb R$ over $\mathbb Q$, then remove one element (or, I suppose, even a countable number of them!), then that element is no longer generated by the basis, but the basis is still uncountable. – feralin Mar 13 '16 at 23:04
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    In spirit, you are correct, but you seem to be referring to $B_0$ as a basis, which it is not. – Michael Albanese Mar 13 '16 at 23:06
  • You're right. I was just being sloppy with terminology, sorry! – feralin Mar 13 '16 at 23:09
  • @MichaelAlbanese I wish I could accept both answers! They're both useful, but Schweber provided more insight and information. Thanks for your input! – feralin Mar 13 '16 at 23:12
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Here is an example of a rather direct construction:

Given some $x \in \mathbb{R}$, let $\lambda(x)(n)$ denote the number of consecutive identical digits in the decimal expansion of $x$ starting at position $2^n$. (We prefer the expansion ending in $0^\omega$ to the one ending in $9^\omega$ here).

Now let $$S := \{x \in \mathbb{R} \mid \exists k \in \mathbb{N} \ \forall n \in \mathbb{N} \quad \lambda(x)(n) \geq 2^n - k\}$$

$S$ is uncountable, as eg the choice of $k = 1$ already allows us to choose infinitely many digits of an element of $S$ independently. $S$ contains $0$ and is closed under substraction and addition, this only requries the choice of a larger $k$. Finally, $S$ is not $\mathbb{R}$, e.g. $0.(01)^\omega \notin S$.

Arno
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generate in the sense of fields
Let $X \subseteq \mathbb R$ be a set with Hausdorff dimension zero, and furthermore all Cartesian products $X \times X \times \dots \times X$ have Hausdorff dimension zero. Then the field $F$ generated by $X$ also has Hausdorff dimension zero (so $F$ is not all of $\mathbb R$). You can construct Cantor sets $X$ like this, which are uncountable.

plug
G. A. Edgar & Chris Miller, Borel subrings of the reals. Proc. Amer. Math. Soc. 131 (2003) 1121-1129
LINK
Borel sets that are subrings of $\mathbb R$ either have Hausdorff dimension zero as described, or else are all of $\mathbb R$.

Also: see the references there for subgroups of the reals (due to Erdös and Volkmann) with Hausdorff dimension $t$ for any $t$ with $0<t<1$.

GEdgar
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