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


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


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

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  • $\begingroup$ It looks correct, let me ask you, did you proved Schroeder Bernstein theorem? If yes then there is one liner proof using the fact that for infinite cardinality $|A\times A|=|A|$ $\endgroup$ – ℋolo Sep 14 '18 at 19:03
  • $\begingroup$ @Holo I has proved Schroeder Bernstein theorem many times :) Please elaborate more on your approach! Does it utilize ordinal numbers? Can you suggest some textbooks that contain materials about cardinal arithmetic? $\endgroup$ – LE Anh Dung Sep 14 '18 at 23:39
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    $\begingroup$ I posted the proof, I don't know textbooks so I can't help you there but I would suggest to study ordinal numbers, this is very useful for a lot of proofs $\endgroup$ – ℋolo Sep 15 '18 at 0:10
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(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 $|A|+|B|=|A|\cdot|B|=\max\{|A|,|B|\}$.

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  • $\begingroup$ No, they key inequality is that the product is the maximum. It hides the whole difficulty of the problem. $\endgroup$ – Andrés E. Caicedo Sep 15 '18 at 0:37
  • $\begingroup$ @AndrésE.Caicedo this is indeed important part, I said in the comments of the post that I'm usinf this fact I also know that OP is already in the middle of trying to solve that equality, what I meant is that this is the only inequality OP didn't saw before as far as I know(from OP's previous questions) $\endgroup$ – ℋolo Sep 15 '18 at 0:44
  • $\begingroup$ So $|A\times A|=|A|$ for an infinite set $A$ is a direct consequence of such inequality? $\endgroup$ – LE Anh Dung Sep 15 '18 at 2:55
  • $\begingroup$ @LeAnhDung no, note that we used that fact to find the last equality! For this you have to prove beforehand that $|A\times A|=|A|$ $\endgroup$ – ℋolo Sep 15 '18 at 2:57
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    $\begingroup$ Oh I see. Let me tackle that :) $\endgroup$ – LE Anh Dung Sep 15 '18 at 2:59

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