# Does the subset $(0,1)$ of $\Bbb{R}$ have the same cardinality as $\Bbb{R}$?

I want to check if the interval of real numbers $(0,1)$ has the same cardinality as $\Bbb{R}$. All that I need to prove is that I can get a bijective function in which domain is $\Bbb{R}$ and my image is the subset $(0,1)$. I have the function $$y=\frac{1}{2} + \frac{1}{2}\frac{x}{|x|+1}$$ I can formally prove that my function is bijective, but I don't know how to formally prove that this function is surjective. My question is: How can I prove $y$ is a surjection? And if I do so, is there still something else that my proof is lacking?

• Prove it by solving your equation for $x$. If the result is a real number for any given $y$, you are done. – MPW Feb 4 '15 at 22:28
• If you have formally shown $y$ is a bijection then by definition it is a surjection. – graydad Feb 4 '15 at 22:33
• @graydad: I assumed OP really meant he has shown it to be 1-1. – MPW Feb 4 '15 at 22:34
• Hint for the surjectivity: It is continuous, the limit at $+\infty$ is $1$, the limit at $-\infty$ is $0$. – Clement C. Feb 4 '15 at 22:36
• @MPW ah that would make sense. – graydad Feb 4 '15 at 22:37

To show $y$ is a surjection, let $a \in (0,1)$. We want to show there is some $x \in \Bbb{R}$ such that $y(x)=a$. To do this, set $a = \frac{1}{2}+\frac{x}{2(|x|+1)}$. Then $$2a = 1+\frac{x}{|x|+1} \\ \implies 2a(|x|+1)=|x|+1+x$$ Because of the absolute value found in the denominator of $y$, we will need to make cases.

Case 1: If $x \geq 0$ then $|x|=x$ so $$2a(|x|+1)=|x|+1+x \\ \implies 2a(x+1) = 2x+1 \\ \implies 2ax+2a-2x=1 \\ \implies x(2a-2)=1-2a \\ \implies x = \frac{1-2a}{2a-2} \\ = \frac{1}{2}\cdot\frac{2a-1}{1-a}$$ Now notice that $\frac{1}{2}\cdot\frac{2a-1}{1-a} \geq 0$ so long as $a \in [1/2,1)$. This is important to note since we got this equation by assuming $x \geq 0$. I will leave it up to you to complete the second case. You'll get a different equation when $x<0$ since $|x| = -x$, and you will need to verify that your equation behaves correctly for all $a \in (0,1/2)$. You can then conclude that no matter which $a \in (0,1/2) \cup [1/2,1) = (0,1)$ you begin with, you have a way of tracing it back to some $x \in \Bbb{R}$.

Your function can be written $$f(x)=\begin{cases} \dfrac{1}{2}\left(1+\dfrac{x}{1+x}\right) & \text{if x\ge0}\\[4px] \dfrac{1}{2}\left(1+\dfrac{x}{1-x}\right) & \text{if x<0} \end{cases}$$ For $x\ge0$, we have $f(x)\ge 1/2$, whereas, for $x<0$, we have $f(x)<1/2$.

Thus we can't have $f(x_1)=f(x_2)$ if $x_1<0$ and $x_2\ge0$ (or conversely). So, for injectivity, it suffices to show that

1. if $x_1<0$, $x_2<0$ and $f(x_1)=f(x_2)$, then $x_1=x_2$;
2. if $x_1\ge0$, $x_2\ge0$ and $f(x_1)=f(x_2)$, then $x_1=x_2$.

The same idea can be used for showing surjectivity. If $1/2\le y<1$, we want to find $x\ge0$ such that $f(x)=y$. This becomes $$2y=\frac{2x+1}{1+x}$$ or $$x=\frac{2y-1}{2(1-y)}$$ and this solution is good, for the right hand side is $\ge0$.

If $0<y<1/2$, we need $$2y=\frac{1}{1-x}$$ or $$x=\frac{2y-1}{2y}$$ and this solution is good, because the right hand side is $<0$.

An easier map that shows the same thing is $$f(x)=\frac{1}{2}+\frac{1}{\pi}\arctan x$$