Can we divided the irrationals into a countable disjoint union of subsets, none of which has a rational limit point?

Question

If we partite $\mathbb{R}\backslash\mathbb{Q}$ as $\cup_{i\in\mathbb{N}}A_i=\mathbb{R}\backslash\mathbb{Q}$, $A_i\cap A_j=\emptyset$ if $i\ne j$, can it hold that $A_i$ has no rational limit point?

Actually, we have irrational perfect sets, but they don't seem to be able to contain much of $\mathbb{R}\backslash\mathbb{Q}$. Or if we can express the irrationals as a disjoint union of such sets, then we are done. But this too seems hopeless.

Furthurmore, does it hold that $\cup_{i\in\mathbb{N}}\bar{A_i}=\mathbb{R}$?

Actually, this is a strengthening of the first problem. The first one seeks to prove that whatever partition we make, the outcome cannot be too coarse. The second one further discusses whether it is still dense.

Camilo Arosemena-Serrato · Accepted Answer · 2013-11-13T18:43:37.770

5

No, it cannot hold.

Your problem asks for a family $\{A_n\}_{n\in\mathbb N}$ such that $A_i'\cap\mathbb Q=\emptyset$ for all $i\in\mathbb N$, but this means $\overline{A_i}\cap\mathbb Q=\emptyset$, as $\overline{A_i}=A_i\cup A_i'$ and $A_i\subseteq\mathbb R\setminus\mathbb Q$, but this is equivalent to $\mathbb Q\subseteq \overline{A_i}^c=int(A_i^c)$ for all $i$. Thus your problem, ignoring the disjointedness part, is equivalent to finding a family $\{A_i\}_{i\in\mathbb N}$ of sets of reals such that $\bigcap_{i\in\mathbb N} A_i=\mathbb Q$ and $\mathbb Q\subseteq int(A_i)$ for all $i$, but this cannot happen because of Baire's First Category Theorem:

Let $\mathbb Q=\{q_i:i\in\mathbb N\}$ be an enumeration of $\mathbb Q$. As $\mathbb Q\subseteq int(A_i)$ for each $i$, each $int(A_i)$ is open dense in $\mathbb R$, then clearly each $int(A_i)\setminus\{q_i\}$ is open dense in $\mathbb R$; as $int(A_i)$ either contains an open interval that contains $q_i$; in such case $q_i\in \overline {int(A_i)\setminus\{q_i\}}$, or $q_i\notin int(A_i)$. Therefore $\bigcap_{i\in\mathbb N}(int(A_i)\setminus\{q_i\})$ is dense in $\mathbb R$, by 1. However $\bigcap_{i\in\mathbb N} int(A_i)\subseteq\bigcap_{i\in\mathbb N} A_i=\mathbb Q$, which implies $\bigcap_{i\in\mathbb N}(int(A_i)\setminus\{q_i\})=\emptyset$. Contradiction.

Now, if you ask for $\bigcup_{i\in\mathbb N} \overline{A_i}=\mathbb R$, it's even easier, as this means $\bigcap_{i\in\mathbb N} int(A_i^c)=\emptyset,$ however, $\mathbb Q\subseteq int(A_i^c)$ for each $i$, in consequence each $int(A_i^c)$ is open dense, so you obtain a contradiction using 1.

edited Nov 13 '13 at 18:43

answered Nov 11 '13 at 05:34

Camilo Arosemena-Serrato

9,656

It's actually $\overline{int(A)}\cap\mathbb{Q}=\emptyset$. It implies $\overline{A}\cap\mathbb{Q}=\emptyset$ in this case though. – Anonymous Coward Nov 11 '13 at 06:25
@AnonymousCoward what do you mean? – Camilo Arosemena-Serrato Nov 11 '13 at 13:07
I mean we should exclude the isolated point (i.e. consider the derived set, not the closure). – Anonymous Coward Nov 11 '13 at 13:44
@AnonymousCoward But there is no need to, as $\overline {A_i}=A_i\cup A_i'$ and $A_i\subseteq \mathbb R\setminus\mathbb Q$, so $A_i'\cap\mathbb Q=\emptyset$ is equivalent to $\overline{A_i}\cap\mathbb Q=\emptyset$. – Camilo Arosemena-Serrato Nov 11 '13 at 14:03
@AnonymousCoward, have you checked my edit? – Camilo Arosemena-Serrato Nov 13 '13 at 23:54
I think you took it wrong. The second problem is discussed without assuming the sets having no rational limit point. But $\mathbb{Q}\subset\mathrm{int}(A_i^c)$ in your argument is a consequence of that. – Anonymous Coward Nov 14 '13 at 02:21
It is unbelievable that the assertion is always false. For the trivial partition $A_1=\mathbb{Q}$ and empty otherwise, clearly $\overline{A_1}=\mathbb{R}$. – Anonymous Coward Nov 14 '13 at 02:22
What are the $A_i$'s in the second question?, you weren't clear enough... – Camilo Arosemena-Serrato Nov 14 '13 at 13:24
I have now modified the problem to be clear. $A_i$ is always a partition of the irrationals. In the first question I ask whether at least one of them they have at least one rational limit point. In the second one I ask whether they always have every point as a limit point (of at least one of them), which is a considerable strengthening of the first one. – Anonymous Coward Nov 14 '13 at 16:10

Can we divided the irrationals into a countable disjoint union of subsets, none of which has a rational limit point?

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