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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.

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

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  • $\begingroup$ 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 $2n-1$. $\endgroup$ – MathematicalPhysicist Dec 28 '19 at 13:03
  • $\begingroup$ Yes, I mean that. Corrected the typo. $\endgroup$ – Martund Dec 28 '19 at 13:04

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