How to work out a bijection?

Question

Let $X$ be an infinite set. $Y$ is a finite subset of $X$. Prove there is a bijection $f:X-Y\rightarrow X$.
What I already solved is the case where $X$ is a countable set. We can assume $X=\{a_1,a_2,\cdots,a_n,\cdots\}$, and $Y=\{a_1,a_2,\cdots,a_k\}$.
Then, $f:X-Y\rightarrow X, a_{k+i}\mapsto a_i$ is a bijection.
However, if $X$ is not countable we can not assume $X=\{a_1,a_2,\cdots,a_n,\cdots\}$. In this case how can I work out the bijection?

Answer 1

Pick any countably infinite subset of $X$ that includes $Y$, apply your solution there, and let $f$ be the identity function elsewhere.

Answer 2

Let $X$ be an infinite set and $t\in X$. Then there exists a sequence $(x_n ) \in X^{\mathbb{N}}$ such that $x_1 =t $ and $x_j \neq x_i $ for $i\neq j .$ Now define $$f_{t,X} (x) =\begin{cases} x \mbox{ if } x\in X\setminus \bigcup_{j=1}^{\infty} \{x_j \}\\ x_{n+1} \mbox{ if } x=x_n \end{cases} $$ so $f_{t,X} :X\to X\setminus \{t\} $ is an bijection. Now let $Y$ be an finite subset of $X.$ Say $Y=\{ y_1 , y_2 ,..., y_k\}$ then then the superposition $ f =f_{y_k , X\setminus \{ y_1 , y_2 ,..., y_{k-1}\}}\circ ...\circ f_{y_2 , X\setminus\{y_1\}} \circ f_{y_1 , X}$ is an bijection between $X\setminus Y$ and $X.$