$\mathbb{R^+}$ is the disjoint union of two nonempty sets, each closed under addition.

Question

I saw Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition. and have a question related to the answer (I'm not sure if this is the right place to post it);

Why don't we just take $\mathcal{A}=\{A\subseteq \mathbb{R^+}\ :\text{A is closed under addition and all elements of A are irrational}\}$, $\mathcal{B}=\{B\subseteq \mathbb{R^+}\ :\text{B is closed under addition and all elements of B are rational}\}$ and order both of them with the partial order $\subseteq$ so that they will satisfy the chain condition? Then, they have maximal elements $\bigcup{\mathcal{A}}$ and $\bigcup{\mathcal{B}}$ respectively. So it remains to verify that $\bigcup{\mathcal{A}}$ is the set of all irrationals and $\bigcup{\mathcal{B}}$ is the set of all rationals. Is this approach correct?

Answer 1

I'll add my two cents to the existing answers and point out exactly what went wrong in your argument.

It is absolutely correct that $\bigcup \mathcal{A}$ is the set of all irrationals and $\bigcup \mathcal{B}$ is the set of all rationals. But, as was pointed out in the comments, the set of all irrational numbers is not closed under addition, i.e. $\left(\bigcup \mathcal{A}\right) \not \in \mathcal{A}$.

The thing is, you didn't use Zorn's lemma correctly. I mean, yes, Zorn's lemma is applicable to the partially ordered set $\mathcal{A}$, and there is at least one maximal element $X \in \mathcal{A}$. But $X$ doesn't have to be larger then every $A \in \mathcal{A}$. The only thing you know about $X$ is that there is no $A \in \mathcal{A}$ such that $X \subsetneq A$.

In case of the partially ordered set $\mathcal{B}$, it turns out that $\bigcup \mathcal{B}$ really is the only maximal element in $\mathcal{B}$, but this is not because of Zorn's lemma. Rather, it can be thought of as a lucky coincidence.

Answer 2

There is a reason why you shouldn't try to explicitly construct such a decomposition. If $\mathbb{R}^{+}$ is a disjoint union of $A$ and $B$, where both $A$ and $B$ are closed under addition then neither one of $A$ and $B$ is either Lebesgue measurable or has Baire property since if $X$ is either a non meager set of reals with Baire property or $X$ has positive Lebesgue measure then $X + X$ has non empty interior. It is well known that such sets cannot be provably defined within set theory (ZFC).

Answer 3

While the sum of rational numbers is always rational, the sum of irrational numbers is not always irrational.

If you want to split $\Bbb R^+$ completely you will have to mix the two, as Lord_Farin commented $\pi$ and $4-\pi$ are both irrational, but their sum is rational.