Is there any uncountably infinite set that does not generate the reals?

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?

Solution 1:

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!

Solution 2:

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 $\mathbb{R}$.