Bijection between an infinite set and its union of a countably infinite set

Solution 1:

Take a countably infinite sequence from $A$ which includes $S\cap A$, denoted by $(a_n)_{n\geq 1}$. And denote also $S\cup \{a_n, n = 1,2,\cdots\} = \{b_n, n = 1,2,\cdots\}$, since the union of two countable sets are still countable.

Then one explicit bijection between $A$ and $A\cup S$ is the following $f$:

$f(x) = x$ if $x \not\in \{a_n, n = 1,2,\cdots\}$, $f(a_{k}) = b_k$

Clearly it's a surjection. The fact that it's an injection is implied by the way we pick $(a_n)_{n\geq 1}$.

There can't be a bijection between $\mathbb{N}$ and $\mathbb{N}\cup \mathbb{R}$

Solution 2:

Let $S=\{s_n\}_{n\in\mathbb N}$. As $A$ is infinite, then it has an infinite countable subset $\{a_n\}_{n\in\mathbb N}$.

Define now the following bejection $f:A\to A\cup S$:

$$ f(x)=\left\{\begin{array}{llll} x & \text{if} & x\in A\smallsetminus \{a_n\}_{n\in\mathbb N},\\ a_{n} & \text{if} & x=a_{2n-1},\\ s_{n} & \text{if} & x=a_{2n}. \end{array} \right. $$

There is no bijection between $\mathbb N$ and $\mathbb R\cup\mathbb N$, as there is no bijection between $\mathbb N$ and $(0,1)$. This is done using a standard (Cantor's) diagonal argument.