Partition of the real line.

Question

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.

I believe $\aleph$ means $2^{\aleph_0}$. I saw this notation before, though it is rare. — Mark, Dec 28 '19 at 12:29
By the "continuum hypothesis, [tex]2^{ℵ_0}= R[/tex] so you want Ax= {x}. — user247327, Dec 28 '19 at 12:34
@user247327 what you suggest won't work since $|A_x|=| { x} | =1$ and not $\aleph$! — MathematicalPhysicist, Dec 28 '19 at 12:37
@MathematicalPhysicist Then I suggest you mention that $\aleph$ is the cardinality of the continuum in your question since I have never seen that notation before. — balddraz, Dec 28 '19 at 12:38
@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
@PatrickStevens to tell you the truth, I don't understand why it's more normal? the book I use uses this notation. But I guess since Cantor wasn't really normal... :-D — MathematicalPhysicist, Dec 28 '19 at 12:53
Why does the English word "lame" more normally mean "unable to walk" than "a razor on a stick used for slashing dough"? It just… is more normal. — Patrick Stevens, Dec 28 '19 at 12:54
How do you define "normal"? do you call the author who wrote the book I am using not so "normal"? anyway, I appreciate it if someone can help me with my question instead of declaring what is "more normal". — MathematicalPhysicist, Dec 28 '19 at 12:57
To some extent, this is a duplicate of https://math.stackexchange.com/questions/183361/examples-of-bijective-map-from-mathbbr3-rightarrow-mathbbr/. — Asaf Karagila, Dec 28 '19 at 13:16
Closely related: A partition of the unit interval into uncountably many dense uncountable subsets AND Uncountable collection AND Uncountable family of uncountable compact subsets of $\mathbb{R}$ AND Nice partition of $\mathbb{R}$ into uncountably many uncountable sets. — Dave L. Renfro, Dec 28 '19 at 14:15

Martund · Accepted Answer · 2019-12-28T13:03:45.687

2

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!

edited Dec 28 '19 at 13:03

answered Dec 28 '19 at 12:57

Martund

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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
Yes, I mean that. Corrected the typo. – Martund Dec 28 '19 at 13:04

Partition of the real line.

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