How do I prove this?
The problem contains the following "hint": Prove that X/Y contains a denumerable family of pairwise disjoint denumerable subsets.
I am not sure how that proves cardinal equivalence. My first instinct was to try to show that a bijection exists between X and X\Y, but that may be extremely difficult which would be why the hint was given.
Any help would be appreciate, and also carefully describe how the solution was found.
Thank you.
Here's an idea for when $Y =\{y\}$ is a singleton:
Let $Z \subset X$ be any denumerable subset, $Z = \{z_1, z_2, \cdots\}$, where we take $z_1 = y$. Define a map $\phi : X \to X \setminus Y$ by sending $x \mapsto x$ for all $x \in X \setminus Z$, and by sending $z_i \mapsto z_{i + 1}$. It's not hard to see that this is a bijection.
How do you extend this method to the situation where $Y$ is countable? Think about how you would do this for $|Y| = 2, 3$, etc., and extrapolate to $Y$ countable.
Let $g\colon \omega\to Y$ be a bijection. Find an injection $f\colon \omega\to X\setminus Y$ recursively as follows: Assume we have already found an injective map $f_n\colon n\to X\setminus Y$ (the case $n=0$ being trivial). Then $$ x\mapsto\begin{cases}f_n(x)&\text{if }x<n,\\g(x-n)&\text{if }x\ge n\end{cases}$$ defines an injection $\omega\to X$. By assumtion, this is not a bijection, hence there exists some $x_n\notin Y\cup \operatorname{im}f_n$. Define $f_{n+1}\colon n+1\to X\setminus Y$ per $f_{n+1}(n)=x_n$ and $f_{n+1}(x)=f_n(x)$ otherwise. Ultimately define $f(n)=f_{n+1}(n)$. Now you can define $h\colon X\to X\setminus Y$ per $$ h(x)=\begin{cases}f(2g^{-1}(x))&\text{if }x\in Y,\\ f(2f^{-1}(x)+1)&\text{if }x\in \operatorname{im}f,\\ x&\text{otherwise}.\end{cases}$$ Why is this a bijection?
We assume it has already been shown that $X$ has a denumerable subset $D$. Then there is a bijection $\beta$ from $\mathbb{N}$ to $D$. (For definiteness, $\mathbb{N}$ starts at $0$.)
Write $d_i$ instead of $\beta(i)$. Let $Y$ be the set of $d_i$ such that $i$ is even.
We define a mapping $\varphi$ from $X$ to $X\setminus Y$ as follows:
(1) If $x=d_i$ for some $i$, let $\varphi(x)=d_{2i+1}$;
(ii) If $x\not\in D$. let $\varphi(x)=x$.
It is not difficult to verify that $\varphi$ is a bijection from $X$ to $X\setminus Y$.
Remark: One can modify the argument to show that if $Y$ is a given denumerable subset of $X$, then there is a bijection from $X$ to $X\setminus Y$.
The idea is that we choose a denumerable subset $D$ of $X\setminus Y$. Let $D_0$ be the even index members of $D$, and $D_1$ the odd index members. Send $X\setminus Y$ to $(X\setminus Y)\setminus D_0$ as in the answer above. That makes room to send $Y$ to $D_0$.
