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Let $A$ be a set. Define a set B such that $|A|=|B|$ and $A\cap B=\emptyset$.

My attempt:

Let $B=\{\{A\}\cup a\mid a\in A\}$.

It's clear that $|A|=|B|$. Next we prove $A\cap B=\emptyset$.

If $A\cap B\neq\emptyset$, then there exists $c$ such that $c\in A$ and $c\in B$. Since $c\in B$, then $A\in c$. Thus $A\in c\in A$, which contradicts Axiom of Regularity. Hence $A\cap B=\emptyset$.

Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

Akira
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  • The general idea is good, yes. If this is for a course in axiomatic set theory, you may want to give further justification for some statements. Why does $B$ exist? How do you know that $|A| = |B|$? – Daniel Mroz Sep 15 '18 at 14:48
  • @DanielMroz It seems that the comment of one user is not correct, so he deleted it. So my proofs is ok but lack some clarification? – Akira Sep 15 '18 at 14:57
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    If you’re uncertain of a proof, that’s probably a sign that some details are missing. Try working them out until you’re convinced. – Daniel Mroz Sep 15 '18 at 15:17
  • You should be able to prove this without foundation. Obviously, a different proof will then be needed. – Andrés E. Caicedo Sep 15 '18 at 15:21
  • @AndrésE.Caicedo I will give it a try :). I seem to find a simpler way to partition X into countably infinite sets at https://math.stackexchange.com/a/2917868/368425. You have offered useful comments in this post before. Could you please have a look at it? – Akira Sep 15 '18 at 15:27
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    I saw it. It seems fine. – Andrés E. Caicedo Sep 15 '18 at 15:28
  • Thank you @AndrésE.Caicedo :) – Akira Sep 15 '18 at 15:48
  • Hi @AndrésE.Caicedo! On the basis of your previous comment, i have presented a proof without appealing to Axiom of Regularity and posted it as an answer below. Could you please have a look at it? Thank you for your help! – Akira Sep 19 '18 at 01:26
  • Sorry I don't follow why you have enclosed $A$ itself in set brackets in you definition of $B$ in set builder notation, from my understanding this is effectively defining a singleton, for which the solitary element is the set $A$ itself. I may be wrong, I also get very confused with axiomatic questions, but that would be the first thing that I see that I am unclear of – Adam Ledger Sep 19 '18 at 01:30
  • @Adam Your understanding is correct: this is effectively defining a singleton. This creation of ${A}$ from $A$ follows Axiom of Pairing (https://www.wikiwand.com/en/Zermelo%E2%80%93Fraenkel_set_theory#/4._Axiom_of_pairing). – Akira Sep 19 '18 at 05:37
  • @LeAnhDung interesting well I will bookmark it and have another look at it later, but I will say it's very commendable that you are approaching these kinds of axiomatic questions that draw near to the paradoxical, I only commenced Abstract Algebra this year, and have touched on Russell's paradox, but it is a very slow process for me – Adam Ledger Sep 19 '18 at 06:14
  • @Adam my process is slow too :) Don't worry! You will get better over time. – Akira Sep 19 '18 at 06:17

1 Answers1

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We have $\exists a'\notin \bigcup A$. If not, Russell's paradox appears. Thus $a'\notin a$ for all $a\in A$.

Let $B=\{a\cup \{a'\} \mid a\in A\}$.

We define a mapping $f:A\to B$ by $f(a)=a\cup \{a'\}$ for all $a\in A$.

  1. $f$ is injective

Let $a_1,a_2\in A$ and $f(a_1)=f(a_2)$. Then $a_1\cup \{a'\}=a_2\cup \{a'\}$. Since $a'\notin a_1$ and $a'\notin a_2$, $a_1\cup \{a'\}=a_2\cup \{a'\} \iff a_1=a_2$. Hence $f$ is injective.

  1. $f$ is surjective

For any $b\in B$, there exists $a\in A$ such that $b=a\cup \{a'\}$. Thus $f(a)=b$. Hence $f$ is surjective.

As a result, $f$ is bijective and consequently $|A|=|B|$.

  1. $A\cap B=\emptyset$

For any $a\in A, a'\notin a$. For any $b\in B, a'\in b$. Thus $a\neq b$ for all $a\in A$ and $b\in B$. Hence $A\cap B=\emptyset$.

Akira
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