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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?

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    $\begingroup$ Do you mean "generate" in the sense of subgroups? $\endgroup$
    – Arthur
    Commented Mar 13, 2016 at 22:40
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    $\begingroup$ 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". $\endgroup$
    – feralin
    Commented Mar 13, 2016 at 22:41
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    $\begingroup$ We can build a counterexample with the axiom of choice. I don't know of any simple answer, though. $\endgroup$ Commented Mar 13, 2016 at 22:43
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    $\begingroup$ @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 :) $\endgroup$
    – feralin
    Commented Mar 14, 2016 at 6:57
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    $\begingroup$ @EricTowers The set of rationals isn't uncountable, so it isn't a suitable counterexample to "all uncountable sets generate the reals". $\endgroup$ Commented Mar 14, 2016 at 16:05

4 Answers 4

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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!

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    $\begingroup$ 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? $\endgroup$
    – feralin
    Commented Mar 13, 2016 at 23:07
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    $\begingroup$ @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. $\endgroup$ Commented Mar 13, 2016 at 23:10
  • $\begingroup$ 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 math.stackexchange.com/questions/1696315/…. $\endgroup$ Commented Mar 13, 2016 at 23:11
  • $\begingroup$ 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? $\endgroup$ Commented Mar 14, 2016 at 1:15
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    $\begingroup$ @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. $\endgroup$ Commented Mar 15, 2016 at 17:46
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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 $\mathbb{R}$.

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    $\begingroup$ +1. Note that the axiom of choice is used here to conclude that $\mathbb{R}$ has a basis as a $\mathbb{Q}$-vector space. $\endgroup$ Commented Mar 13, 2016 at 22:52
  • $\begingroup$ 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 :) $\endgroup$ Commented Mar 13, 2016 at 22:56
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    $\begingroup$ @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. $\endgroup$
    – feralin
    Commented Mar 13, 2016 at 23:04
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    $\begingroup$ In spirit, you are correct, but you seem to be referring to $B_0$ as a basis, which it is not. $\endgroup$ Commented Mar 13, 2016 at 23:06
  • $\begingroup$ You're right. I was just being sloppy with terminology, sorry! $\endgroup$
    – feralin
    Commented Mar 13, 2016 at 23:09
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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. $0.(01)^\omega \notin S$.

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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 $0<t<1$.

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