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!

  • $\begingroup$ 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 $|A| = |B|$? $\endgroup$ – Daniel Mroz Sep 15 '18 at 14:48
  • $\begingroup$ @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? $\endgroup$ – MadnessFor MATH Sep 15 '18 at 14:57
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Mroz Sep 15 '18 at 15:17
  • $\begingroup$ You should be able to prove this without foundation. Obviously, a different proof will then be needed. $\endgroup$ – Andrés E. Caicedo Sep 15 '18 at 15:21
  • 1
    $\begingroup$ I saw it. It seems fine. $\endgroup$ – Andrés E. Caicedo Sep 15 '18 at 15:28

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 $A\cap B=\emptyset$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.