I am reading a famous textbook (4th Edition) by Kolmogorov and Fomin (translated from Russian to Japanese).
There is the following exercise in this book:
Let M be an infinite set.
Let $A$ be a finite or countable set.
Prove that $M\sim M\cup A.$
My solution is the following.
Is my solution correct?
Is my solution a standard solution for this exercise?
I want to know a standard solution for this exercise and I want to improve my solution.
Thank you!
(1) In the case $A$ is a finite set.
Since $M$ is an infinite set, $M$ contains a countable set $B$.
Obviously, $B\sim B\cup A$.
So, $M=(M\setminus B)\cup B\sim (M\setminus B)\cup (B\cup A)=M\cup A.$
(2) In the case $A$ is a countable set.
If $A\setminus M$ is a finite set, then $M\sim M\cup (A\setminus M)=M\cup A$ by (1).
We consider the case $A\setminus M$ is a countable set.
Since $M$ is an infinite set, $M$ contains a countable set $B = \{b_1,b_2,b_3,\dots,b_n,\dots\}$.
Let $B_1:=\{b_1,b_3,b_5,\dots\}$.
Let $B_2:=\{b_2,b_4,b_6,\dots\}$.
Then, $B=B_1\cup B_2$ and $B_1$ and $B_2$ are countable sets.
So, $B\sim B_1$ and $B_2\sim A\setminus M$.
So, $M\setminus B_2=B_1\cup (M\setminus B)\sim B\cup (M\setminus B)=M$.
So, $M=(M\setminus B_2)\cup B_2\sim M\cup (A\setminus M)=M\cup A.$
Best Answer
I think your proof is correct. But it can be simplified.
In case (2), given that $B$ and $A$ are both countable, $B\sim B\cup A$. So $M=B\cup (M\setminus B)\sim (B\cup A)\cup(M\setminus B)=M\cup A$.