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I came upon a question and I'm stuck on solving it.

Does there exist any continuous function $f:\Bbb R^2\to \Bbb R$ such that $$f(x_1,y_1)=f(x_2,y_2)\iff x_1=x_2\quad,\quad y_1=y_2$$?

What if we drop the continuity constraint? (though, the function is better be well-defined)

Originally, this question comes from a physical background. If two inertial frames measure the 4-tuples $(\Delta x,\Delta y,\Delta z,\Delta t)$ and $(\Delta x',\Delta y',\Delta z',\Delta t')$, we have $$\begin{cases}\Delta x^2+\Delta y^2+\Delta z^2-c^2\Delta t^2=\Delta x'^2+\Delta y'^2+\Delta z'^2-c^2\Delta t'^2 &,\quad\text{Special Relativity}\\ \Delta x^2+\Delta y^2+\Delta z^2=\Delta x'^2+\Delta y'^2+\Delta z'^2\\\text{and} \\\Delta t=\Delta t' &,\quad\text{Classical Physics} \end{cases}$$ I am OK with special relativity, but about the the second equivalence, I wonder if a function $f$ is found such that $$f(\Delta x^2+\Delta y^2+\Delta z^2,c^2\Delta t^2)=f(\Delta x'^2+\Delta y'^2+\Delta z'^2,c^2\Delta t'^2)\\\iff \\ \Delta x^2+\Delta y^2+\Delta z^2=\Delta x'^2+\Delta y'^2+\Delta z'^2\\\text{and} \\\Delta t=\Delta t' $$

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Mostafa Ayaz
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1 Answers1

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For $f$ continuous, let $f(a,b)>f(c,d)>f(e,g)$.

Choose any path from $(a,b)$ to $(e,g)$ avoiding $(c,d)$. The image of this path is a path which must include all points between $f(a,b)$ to $f(e,g)$.

Thus there is another point with the same image as $(c,d)$.

A neat (?) function $f$ if the continuity constraint is dropped.

For $x\in \Bbb R$ define $a(x)=\frac{e^x}{2e^x+1}$.

For $x\in \Bbb Q$ define $b(x)=0.5+\frac{x}{\sqrt 5}$. For $x\in \Bbb R- \Bbb Q$ define $b(x)=x$.

Using the continued fraction form for irrational numbers in $(0,1)$, define $c([0;x_1,x_2,...],[0;y_1,y_2,...])=[0;x_1,y_1,x_2,y_2,...]$.

We can then define the function $f$ from $\Bbb R^2\to \Bbb R$ by $$f(x,y)=c(ba(x),ba(y)).$$