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I need to show a partition of $\Bbb{R}$, call it $P$, s.t.:

  • $|P| = \aleph$
  • For all $X\in P$ we have $|X| = \aleph $

My aproch was to define an equivilance relation $S$ of $\Bbb{R}$ s.t. for $x,y\in \Bbb{R}$ whe have $xSy$ if and only if $x-y\notin \Bbb{Q}$ or $x=y$. Hence, for $X\in \Bbb{R}/\Bbb{Q}$ we can take $a\in X$ and so $X=\{a+r|r\in \Bbb{R}-\Bbb{Q}\}$ and $|X| = |\Bbb{R}-\Bbb{Q}| = \aleph$. Now the challenging part is showing that $|\Bbb{R}/\Bbb{Q}|=\aleph$, clearly $|\Bbb{R}/\Bbb{Q}|\geq\aleph_{0}$ since for two rational numbers $a,b\in \Bbb{Q}$ we have $a-b\in \Bbb{Q}$ (since $\Bbb{Q}$ is a field) and hence each rational number has its uniqe equivalence class. Also, $\bigcup_{q\in \Bbb{Q}}[q]_{S}\neq\Bbb{R}$ since, for example, $1\notin \bigcup_{q\in \Bbb{Q}}[q]_{S}$.

And here I'm pretty much stuck so any help would be appreciated, I need to show such a partition exists but I decided to try and find one, So if you can help me with that, I would be very thankful.

yotam maoz
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  • $S$ is certainly not an equivalence relation because we have $x \not S x$. Another problem is that your classes are not distinct, as the items related to $\sqrt 2$ and those related to $1+\sqrt 2$ are the same. It is not a partition. – Ross Millikan Dec 26 '20 at 20:03
  • I fixed the issue with $x\not{S}x$, but $\Bbb{R}/\Bbb{Q}$ is a partition, just $[\sqrt{2}]{S}=[1+\sqrt{2}]{S}$ – yotam maoz Dec 26 '20 at 20:08
  • But $[\sqrt 3]_S=[\sqrt 2]_S$ because the difference is irrational. Using the rationals either the sets or the number of sets will be countable. – Ross Millikan Dec 26 '20 at 20:13
  • Note that $rS0$ for all irrationals $r$, so for any 2 irrationals, $r_0S0Sr_1$, if $S$ was equivalence relation then the difference between any 2 irrationals would be irrational, but $e\not Se+1$ – ℋolo Dec 27 '20 at 00:43
  • I've never seen $\aleph$ used to mean $\mathfrak c$ before. Evidently some people have seen it, and others never have. There's a discussion here. https://math.stackexchange.com/questions/9475/symbol-for-the-cardinality-of-the-continuum – user4894 Dec 27 '20 at 02:25
  • @user4894: This is a conversation that takes place once a month in the comments of this site. Unfortunately for me, I am usually one of the sides in it, and it's getting real tired by now. – Asaf Karagila Dec 27 '20 at 13:08

2 Answers2

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Do you need to find such partition? Or just to show it exists?

If you only need to show it exists, note that $\aleph=2^{\aleph_0}=2^{\aleph_0+\aleph_0}=2^{\aleph_0}×2^{\aleph_0}=\aleph\times\aleph$, let $F:ℝ^2→ℝ$ be bijection, define $F_i(x)=F(i,x)$, and let $P=\{F_i''ℝ\mid i\inℝ\}$.

If you need to find such partition, you just need to find $F$, and now that you have $F$ the above construction will give you the desired partition

ℋolo
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We can give an explicit construction. It is easier to work on $[0,1)$. Express each real by its binary representation, using the terminating version where there is ambiguity. Our sets in $P$ will be identified with the infinite bit strings by taking the odd position bits of each real to decide on the class it belongs in. The even position bits give continuum many elements in each set, so we have partitioned $[0,1)$ into continuum many sets of size continuum.

Ross Millikan
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