Let $X$ be uncountable and $A$ be a countable subset of $X$. Then $X$ and $X \setminus A$ are equinumerous.
I have previously proved that Suppose that $X$ is infinite and that $A$ is a finite subset of $X$. Then $X$ and $X\setminus A$ are equinumerous here. I have a feeling that the result can be extended by this theorem, so I tried to give a shot.
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!
My attempt:
Lemma 1: If $A \sim \Bbb N$ and $B \sim \Bbb N$, then $A\cup B \sim \Bbb N$.
Lemma 2: If $X$ is uncountable, then there exists $B\subsetneq X$ such that $B \sim \Bbb N$.
There are only two cases.
$A$ is finite (I presented a proof here)
$A$ is infinite
Then $A$ is countably infinite and consequently $A\sim\Bbb N$.
Since $X$ is uncountable and $A$ is countable, then $X\setminus A$ is uncountable. By Lemma 2, there exists $B\subsetneq X\setminus A$ such that $B \sim \Bbb N$.
Let $f_1:(X\setminus A)\setminus B \to (X\setminus A)\setminus B$ be the identity map on $(X\setminus A)\setminus B$, then $f_1$ is a bijection.
Since $A\cup B\sim \Bbb N$ by Lemma 1 and $A \sim \Bbb N$, there exists a bijection $f_2:B \to A\cup B$.
We define $f:X\setminus A \to X$ by $f(x)=f_1(x)$ for all $x\in (X\setminus A)\setminus B$ and $f(x)=f_2(x)$ for all $x\in B$. Thus $f$ is a bijection.
Hence $X\setminus A \sim X$.
