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

elementary-set-theory

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!

Best Answer

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 $A\cap B=\emptyset$.

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