real analysis - Is there a bijection from $(-infty,infty)$ to $[0,1]$?

Previous questions established, for example, that a continuous bijection $f:(0,1)\to [0,1]$ does not exist, but the proof relied on continuity. Clearly, with continuity similar proofs can be invoked to show there is no bijection $f:(-\infty,\infty)\to[0,1]$.

Similarly, one can show there is no order-preserving bijection: Suppose there were, then there is an $x\in(-\infty,\infty)$ such that $f(x)=1$. But there exists some $y\in(-\infty,\infty)$ such that $y>x$, and for a bijection this requires $f(y)>f(x)=1$, which is a contradiction.

But what if the bijection need not be continuous? Is there a bijection, or can you prove there isn't?

Answer

Without continuity there is a bijection.

Let $x_1=\frac{1}{2}, x_2=\frac{1}{3},.., x_n=\frac{1}{n+1},...$. Define $f(x_1)=1, f(x_2)=0, f(x_3)=x_1,..., f(x_n)=x_{n-2},..$ and $f(x)=x$ for all $x$ not in the sequence.

Then $f$ is a bijection from $(0,1)$ to $[0,1]$.

Now use the fact that $\frac{1}{2}+\frac{1}{\pi}\arctan(x)$ is a bijection from $(-\infty, \infty)$ to $(0,1)$.