Prove that $f:\mathbb{R}\times\mathbb{Z}\rightarrow\mathbb{R}$, $f(x,y)=\frac{2}{\pi}\arctan x+2y$ is injective

Question

Recently in my discrete mathematics class we defined an injective function $f:A\hookrightarrow B$ in the following way:

A function $f:A\rightarrow B$ is injective if for all $a,b\in A$, $a\neq b\rightarrow f(a)\neq f(b)$.

or equivalently, $f(a)=f(b)\rightarrow a=b$. I attempted to extend this definition to multivariable functions:

A function $f:A_1\times A_2\times\dots\times A_m\rightarrow B_1\times B_2\times\dots\times B_n$ is injective if for all tuples $(x_1,x_2,\dots,x_m),(y_1,y_2,\dots,y_m)\in A_1\times A_2\times\dots\times A_m$, $f(x_1,x_2,\dots,x_m)=f(y_1,y_2,\dots,y_m)\rightarrow (x_1,x_2,\dots,x_m)=(y_1,y_2,\dots,y_m).$

Using this definition, I attempted to construct some examples of injective multivariable functions. I haven't yet come up with any functions $f:\mathbb{R}^2\hookrightarrow\mathbb{R}$, but I believe that $f:\mathbb{R}\times\mathbb{Z}\hookrightarrow\mathbb{R}$, $f(x,y)=\frac{2}{\pi}\arctan x+2y$ is an injective function based on the above definition. My reasoning is that since $y=\arctan(x)$ is injective and bounded below by $y=-\frac{\pi}{2}$ and above by $y=\frac{\pi}{2}$, the $\frac{2}{\pi}\arctan(x)$ part of $f$ only can change the value of $f$ by any amount between $-1$ and $1$. The $2y$ part, then, changes the "level" of the function, so when $y=0$, $f$ takes on values between $-1$ and $1$; when $y=1$, $f$ takes on values between $1$ and $3$; etc.

My difficulty is in proving that such a function is actually injective. The method we've used in class, i.e. manipulating the equation $f(a)=f(b)$ to show that $a=b$, doesn't easily work in this case, since it requires going from single values to tuples of values. I'm looking for a method to show that this function is injective (or for a counterexample if my example doesn't work! But preferably for a method to solve these kinds of problems). My goal is eventually to extend this simpler problem to construct a function $f:\mathbb{R}^2\hookrightarrow\mathbb{R}$, but I wanted to start with one that was easier to visualize first. With this in mind, is there a good method to prove that this function is injective?

Answer 1

Assume $\frac{2}{\pi}\arctan(x) + 2y = \frac{2}{\pi}\arctan(x')+2y'$.

Then $2(y-y') = \frac{2}{\pi}(\arctan(x')-\arctan(x))$.

What do you know about the left hand side? How many options are there for the right hand side then?

Answer 2

Let $g(x,y)=f(x,y)/2$. Proving $f$ is injective is the same as proving $g$ is injective.

You can note that $$ -\frac{\pi}{2}<\arctan x<\frac{\pi}{2} $$ so that $$ -\frac{1}{2}<\frac{1}{\pi}\arctan x<\frac{1}{2} $$ and therefore $$ y-\frac{1}{2}<g(x,y)<y+\frac{1}{2} $$ Hence, $y_1<y_2$ implies $$ y_1-\frac{1}{2}<g(x_1,y_1)<y_1+\frac{1}{2} \le y_2-\frac{1}{2}<g(x_2,y_2)<y_2+\frac{1}{2} $$ and, in particular, $g(x_1,y_1)\ne g(x_2,y_2)$, for every $x_1$ and $x_2$.

Thus $g(x_1,y_2)=g(x_2,y_2)$ implies $y_1=y_2$ and you can finish up.