Constructing one to one function

Question

Let X and Y be two disjoint sets . Suppose further that X ~ R ( set of real numbers ) and that Y ~ N . Show that Z, which is the union of X and Y forms a one to one correspondence with R .

Any suggestions

Answer 1

Given bijections $f:X\to \Bbb R$ and $g:Y\to\Bbb N$, we could build $h:X\cup Y\to\Bbb R$ as follows:

$$h(x):=\begin{cases} 2f(x)+1 & \text{if $x\in X$ and $f(x)\in\Bbb N$} \\ \phantom2f(x) & \text{if $x\in X$ and $f(x)\not\in\Bbb N$} \\ 2g(x) & \text{if $x\in Y$} \end{cases}.$$

Explanation. $X$ already fits perfectly into $\Bbb R$. But we can make place for countably many other elements by shifting all integers to the odd integers $x\mapsto 2x+1$, leaving free places at the even integers. Now we map all the elements of $Y$ onto the even integers which were unused otherwise.

Answer 2

Suppose $f: X \rightarrow \mathbb{R}$ and $g: Y \rightarrow \mathbb{N}$ are bijective. Define $k : \mathbb{N} \cup -\mathbb{N} \rightarrow -\mathbb{N}$ as $$k(n) = \begin{cases} - 2 |n| + 1, & \text{if } n > 0, \\ - 2 |n| & \text{if } n < 0.\end{cases}$$

Define $h: X \cup Y \rightarrow \mathbb{R}$ as follows: $$h(z) = \begin{cases} g(z), & \text{if } z \in Y,\\ k(f(z)), & \text{if } z \in f^{-1}(\mathbb{N} \cup - \mathbb{N}),\\ f(z) & \text{otherwise.}\end{cases}$$