Some kind of "bijection" between $\mathbb R$ and $\mathbb N$

Question

The problem is: prove the existence of function $f: \mathbb{R}\times\mathbb{R} \rightarrow \mathbb{N}$ such that $f(x,y)=f(y,z)\implies x=y=z$. I was thinking about $f(x,y) = \left\{ \begin{array}{ll} 1 & \textrm{iff $x=y$}\\ something & \textrm{ iff $x\neq y$} \end{array} \right.$ but I guess we need to have bijection from $\mathbb{N}$ to $\mathbb{R}$ to do that, which don't exist because $|\mathbb{N}| < |\mathbb{R}|$. EDIT: let's make it clear. If we take some $c \neq g$, $f(c,c)$ can't be equal to $f(g,g)$.

Answer 1

There seems to be a misunderstanding: the original condition

$$f(x, y)=f(y, z)\implies x=y=z$$

does not imply the new (added in an edit) condition

$$f(x, x)=f(y, y)\implies x=y.$$

To see this, consider the following counterexample: let $f: \{1, 2\}\rightarrow \{1, 2, 3\}$ be given by

• $f(1, 1)=f(2, 2)=1$,

• $f(1,2)=2$, and

• $f(2, 1)=3$.

This $f$ satisfies the original condition, but not the new condition.

It's easy to show that there are no functions $\mathbb{R}^2\rightarrow \mathbb{N}$ satisfying the new condition. Perhaps surprisingly, if we restrict attention to the original condition, there are such functions! Moreover, we don't even need something silly like the axiom of choice to build them!

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Construction. Let $\mathcal{A}=\{A_r: r\in\mathbb{R}\}$ be a family of sets of natural numbers other than $0$ with the property that $$r\not=s\implies A_r\setminus A_s\not=\emptyset.$$

Now define $f(r, s)$ as follows:

• If $r\not=s$, then $f(r, s)$ is the least element of $A_r\setminus A_s$.

• If $r=s$, then $f(r, s)$ is $0$.

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Verification. Now suppose $f(x, y)=f(y, z)$ and it is not the case that $x=y=z$.

First, let's see that the numbers $x, y, z$ must be distinct. Suppose $x=y$. Then $f(x, y)=0$. But since $y\not=z$ (by assumption that "$x=y=z$" fails) then $f(y, z)\in A_y\not\ni0$, so $f(y, z)\not=f(x, y)$. Similarly, we can't have $y=z$.

So $x, y, z$ are distinct. But then $f(x, y)\in A_x\setminus A_y$, and $f(y, z)\in A_y\setminus A_z$. In particular, we have

• $f(x, y)\not\in A_y$, but

• $f(y, z)\in A_y$.

Oops.

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Caveat. Of course I'm skipping a crucial step here: showing that such a family $\mathcal{A}$ actually exists. Since there was some disagreement over whether such families actually exist, let me give a concrete one here:

Fix your favorite bijection $F$ between $\mathbb{N}$ and the set of finite binary strings $2^{<\omega}$; separately, fix your favorite bjection $G$ between $\mathbb{R}$ and the set of infinite binary strings $2^\omega$. Now, given a real $r$, we view each real $r$ as an infinite sequence, and let $A_r$ be the set of natural numbers standing for initial segments of that sequence: more formally, we let $$A_r=\{F^{-1}(\sigma): \sigma\prec G(r)\}.$$

The resulting family is in fact an almost disjoint family: the intersection $A_r\cap A_s$ for $r\not=s$ is always finite, but every $A_r$ is infinite!