elementary set theory - Let $A,B$ be infinite sets such that $Acap B=emptyset$ and $|A|=|B|$. Then $|Acup B|=|A|$

Let $A,B$ be infinite sets such that $A\cap B=\emptyset$ and $|A|=|B|$. Then $|A\cup B|=|A|$.

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My Attempt:

Lemma: Any infinite set can be partitioned into a family of countably infinite sets. (I presented a proof here)

We denote $X$ and $Y$ are equinumerous by $X\sim Y$.

By Lemma, $A$ can by partitioned into a family $(A_i\mid i\in I)$ where $A_i \sim \Bbb N$.

Thus there exist a bijection $f_i:\Bbb N\to A_i$ for all $i\in I$. Hence every $a\in A_i$ is determined by $f_i(k)$ for a unique pair $(i,k)\in I\times\Bbb N$. It follows that $A\sim I\times\Bbb N$.

We have $|A|=|I\times\Bbb N|=|(I\times\Bbb N_1)\cup(I\times\Bbb N_2)|$ where $\Bbb N_1=\{n\in \Bbb N\mid n\text{ is even}\}$ and $\Bbb N_2=\{n\in \Bbb N\mid n\text{ is odd}\}$. It's clear that $B\sim A\sim I\times\Bbb N\sim I\times\Bbb N_1\sim I\times\Bbb N_2$.

We have:

1. $I\times\Bbb N_1\sim A$ and $I\times\Bbb N_2\sim B$




2. $(I\times\Bbb N_1) \cap (I\times\Bbb N_2)=\emptyset$




3. $A\cap B=\emptyset$

Thus there is a bijection from $A\cup B$ to $(I\times\Bbb N_1)\cup(I\times\Bbb N_2)$ and hence $A\cup B\sim (I\times\Bbb N_1)\cup(I\times\Bbb N_2)$

As a result, $|A|=|(I\times\Bbb N_1)\cup(I\times\Bbb N_2)|=|A\cup B|$.

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

Answer

(if $A,B$ disjoint we have $|A|+|B|=|A\cup B|$ if not $|A|+|B|=|(A\times\{0\})\cup (B\times\{1\})|$, we have $|A|=|A\times\{a\}|$so we can always assume the two are disjoint)

($|A|\cdot|B|=|A\times B|$)

Assuming $A$ and $B$ are not the empty set.

$$\max\{|A|,|B|\}\le|A|+|B|\le|A|\cdot|B|\le\max\{|A|,|B|\}\cdot\max\{|A|,|B|\}=\max\{|A|,|B|\}$$

The only inequality you need to be concern about is

$|A|+|B|\le|A|\cdot|B|$

Try to prove it yourself.

This shows that for $A,B$ such that $|A|\ne0\ne|B|$ and $|A|$ or $|B|$ are infinite then $|A|+|B|=|A|\cdot|B|=\max\{|A|,|B|\}$.