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Show that $\mathbb{R}$ is a disjoint union of $\mathfrak{c}$ sets of cardinal $\mathfrak{c}$, where $\mathfrak{c} = | \mathbb{R} | = 2^{\aleph_0}$.

I find this problem very interesting and very challenging at the same time. Since any set of sets with cardinality $\mathfrak{c}$ I can think about are not disjoint. There is probably a specific definition of this $\mathfrak{c}$ sets of cardinal $\mathfrak{c}$ which immediately solves the problem. However, I would like to understand how can I possibly think about this.

user642796
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Joaquin Liniado
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3 Answers3

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HINT: Show that $\Bbb R$ and $\Bbb{R^2}$ have the same cardinality. Now show that $\Bbb R^2$ is the union of $\frak c$ sets of size $\frak c$.

(Note that the proof itself generalizes to any infinite set $A$ such that $A$ and $A^2$ have the same cardinality; which is all of them if you assume the axiom of choice.)

Asaf Karagila
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  • That's a good idea. – Gregory Grant May 18 '15 at 14:41
  • I can show that $\mathbb{R}$ and $\mathbb{R}^2$ have the same cardinality and therefore write $\mathbb{R}^2=\bigcup_{x \in \mathbb{R}}(x,\mathbb{R})$ which is a union of $\mathfrak{c}$ sets of cardinal $\mathfrak{c}$. The implication that therefore $\mathbb{R}$ can be wrote like this is immediate? – Joaquin Liniado May 18 '15 at 14:50
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    @JoaquinLiniado Yes, think about the images of ${(x,y) \mid y\in\mathbb{R}}$ under a bijection $\mathbb{R}^2\rightarrow \mathbb{R}$. You know that the union will be all of $\mathbb{R}$, and because it is a bijection, you should be able to show that the images are all disjoint as well. – Hayden May 18 '15 at 14:53
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We can construct an uncountable collection of paiwise disjoint sets of real numbers using decimal expansions.

The collection of subsets $A$ of positive integers such that both $A$ and complement of $A$ are infinite is an uncountable collection.

For such a set $A$, denote by $S_A$ the set of all those real numbers whose decimal expansion has digit $0$ exactly at positions specified by $A$ and nowhere else.

$S_A$ is an uncountable set, because there are infinitely many positions for non-zero numbers where we can place other numbers arbitrarily (for example numbers using the digits 1 and 2 alone in those positions are uncountable as $A^C$ is an infinite set by choice.)

P Vanchinathan
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Another decimal-type solution.
Any real number $x$ can be written in "binary" form $x = n+\sum_{i=1}^\infty a_i 2^{-i}$, where $n \in \mathbb Z$, and $a_i \in \{0,1\}$. Let us choose non-terminating expansions, if there is more than one expansion.
For $0 \le t \le 1$, let $A_t$ be the set of all reals $x = n+\sum_{i=1} a_i 2^{-i}$ such that $$ \limsup_{N \to \infty}\frac{1}{N}\sum_{i=1}^N a_i = t $$ Then each $A_t$ has cardinal $\mathfrak c$ and $\mathbb R = \bigcup_{0 \le t \le 1} A_t$

GEdgar
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