# Partition of the real line.

I want to show that there exist sets $$A_x \ \forall x\in \mathbb{R}$$ s.t: $$A_x\cap A_y =\emptyset , \forall x\ne y$$, $$\cup_{x\in \mathbb{R}} A_x = \mathbb{R}$$ and $$\forall x\in \mathbb{R} : \ |A_x|=\aleph$$.

I thought of intervals in $$\mathbb{R}$$ such as $$(0,x)$$ but this doesn't cut it, since the first criterion of disjoint intervals doesn't hold.

I don't see how to define this. Any help?

Edit: for those who don't know $$\aleph$$ is the cardinality of $$\mathbb{R}$$ also known as the continuum.

• When you wote $\aleph$, did you mean $\aleph_0$? – José Carlos Santos Dec 28 '19 at 12:25
• @JoséCarlosSantos no, I meant what I wrote! – MathematicalPhysicist Dec 28 '19 at 12:27
• Then what does $\lvert A_x\rvert=\aleph$ mean? – José Carlos Santos Dec 28 '19 at 12:29
• I believe $\aleph$ means $2^{\aleph_0}$. I saw this notation before, though it is rare. – Mark Dec 28 '19 at 12:29
• @MathematicalPhysicist I believe it's more normal to use $\mathfrak{c}$ for the cardinality of the continuum, by the way. – Patrick Stevens Dec 28 '19 at 12:49

Let $$f:(0,1)\to\Bbb R$$ be a bijection. We will do the required activity with $$(0,1)$$ and then use our bijection to get it done for $$\Bbb R$$.
Take $$x\in(0,1)$$. Let its infinite binary expansion be $$0.x_1x_2x_3\cdots$$. Let $$B_x$$ be the set of all $$y\in(0,1)$$ such that the $$(2n-1)th$$ term in the infinite binary expansion of $$y$$ is $$x_n$$. Then $$\{B_x|x\in(0,1)\}$$ is an uncountable partition of $$(0,1)$$.
Now, let $$A_x=f(B_{f(x)})$$ for all $$x\in\Bbb R$$. We are done!
• do you mean $x_n$ instead of $x_i$? since as you wrote it, I don't see the connection between the indices $i$ and $2n-1$. – MathematicalPhysicist Dec 28 '19 at 13:03