# Use Cantor-Schroder-Bernstein to prove |X1|=|X2|

If $X_1 = \left\{\text{all functions }f: \mathbb{ R}\rightarrow \mathbb{ R}\right\}$ and $X_2=\left\{\text{all functions }g:\mathbb{R}\rightarrow\mathbb{R}\text{ such that }g(0)=0\right\}$,

$a)$ Use Cantor-Schroder-Bernstein to prove |$X_1$|=|$X_2$|
$b)$ Find a concrete bijection between both sets

For part $a)$ I said that because $X_2 \subseteq X_1$, then $|X_2|\leq|X_1|$. So now I just need to find an injection between $X_1$ and $X_2$. I have tried a couple of functions but I always get that two functions in $X_1$ that differ only on their value for $x=0$ map to the same function in $X_2$, so it's not injective.

So I don't know how to procede from here. Any help would be greatly appreciated.

$a)$ Let $\phi:\mathbb{R}\rightarrow (0,2)$ be a bijection (for example, $\phi(x)=\frac{x}{1+|x|}+1$), and let $\Phi:X_1\rightarrow X_2$ be given by $$\Phi(f)(x)=\begin{cases}f(\phi^{-1}(x))&\text{, if }x\in(0,2)\\ 0&\text{otherwise}\end{cases}.$$ Let's show that $\Phi$ is injective: suppose that $f_1,f_2:\mathbb{R}\rightarrow\mathbb{R}$ satisfy $\Phi(f_1)=\Phi(f_2)$. That means that, for every $x\in\mathbb{R}$, $\Phi(f_1)(x)=\Phi(f_2)(x)$. Given $y\in\mathbb{R}$, we have $\phi(y)\in(0,2)$, thus $$f_1(y)=f_1(\phi^{-1}(\phi(y))=\Phi(f_1)(\phi(y))=\Phi(f_2)(\phi(y))=f_2(\phi^{-1}(\phi(y))=f_2(y),$$ so $f_1=f_2$. Therefore, $\Phi$ is injective. Since $X_2\subseteq X_1$, we obtain, by Cantor-Bernstein, $|X_2|=|X_1|$.
$b)$ Let $\psi:\mathbb{R}\rightarrow\mathbb{R}\setminus\left\{0\right\}$ be any bijection (for example, $\psi(x)=x$ for $x\not\in\mathbb{N}=\left\{0,1,2,\ldots\right\}$ and $\Psi(n)=n+1$ for $n\in\mathbb{N}$). Let $\Psi:X_1\rightarrow X_2$ be given by $$\Psi(f)(x)=\begin{cases} f(\psi^{-1}(x))&\mathrm{, if }x\neq 0\\ 0&\mathrm{otherwise}\end{cases}.$$ To show that $\Psi$ is bijective, we use arguments similar to the ones in part $a)$.
• I don't understand why the function in part $a)$ is injective, or why you used the interval $(0, 2)$. Could you please clarify? – ItsMe May 2 '14 at 0:25
• @user146502 I used the interval $(0,2)$ simply because it does not contain $0$ and because it is easy to find a bijection between $\mathbb{R}$ and it (specifically, $\phi$). For $\Phi$ being an injection, see the edited answer. – Luiz Cordeiro May 2 '14 at 0:36
• @LuizCordeiro how would you prove surjectivity for the part $b)$? – Lstoi May 2 '14 at 6:04
• @Lstoi Given a function $g\in X_2$, let $f:=g\circ \psi:\mathbb{R}\rightarrow\mathbb{R}$. Then $\Psi(f)=f\circ\psi^{-1}=g$ in $\mathbb{R}\setminus\left\{0\right\}$, and $\Psi(f)(0)=0=g(0)$, so $\Psi(f)=g$. – Luiz Cordeiro May 2 '14 at 22:58