A bijection $f:\mathcal{A} \to \mathcal{B}$, where $\mathcal{A} \subset \mathbb{R}^n, \mathcal{B} \subset \mathbb{R}^k$, and $k

Question

The motivation for this post is the question of whether or not there exists a class of functions which are bijections between vector spaces of differing dimensions. Although it is simple enough to show that, over the reals at least, there are no such continuous functions, it seems that removing the constraint of continuity makes such functions possible. For instance, consider the function $f: \mathbb{R}^2 \to \mathbb{R}$, where if $x=\sum_{z \in \mathbb{Z}} x_z 2^z$ and $y=\sum_{z \in \mathbb{Z}} y_z 2^z$ then $$f(x,y)=\sum_{z \in \mathbb{Z}} x_z 2^{2z} +\sum_{z \in \mathbb{Z}} y_z 2^{2z+1}.$$ I.e., if the binary expansion of the real number $x$ is $$x=\ldots 0\, 0\, 0\, x_n\, x_{n-1}\, x_{n-2}\, \ldots,$$ where $n$ is the largest integer such that $x_n \neq 0$ and the binary expansion of the real number $y$ is $$y= \ldots 0 \, 0 \, 0 \, y_{k} \, y_{k-1} \, y_{k-2} \ldots, $$ where $k$ is the largest integer such that $y_k \neq 0$ (assume WLOG that $k>n$), then $$f(x,y)=\ldots 0 \, 0 \, 0 \, y_n \, 0 \, y_{n-1} \, 0 \, \ldots \, y_{n} \, x_{n} \, y_{n-1} \, x_{n-1} \, y_{n-2} \, x_{n-2} \, \ldots.$$ The question is: is this function a bijection? In theory, there could (should?) exist a bijection $g: \mathbb{R}^2 \to \mathbb{R}$ as the cardinality of $\mathbb{R}^2$ and $\mathbb{R}$ are the same (correct me if I'm wrong). A possible counterargument I've formulated to the injectivity of this function would be the pair of points $(x,y)=(.\overline{111} \, ,.0\overline{111})=(1,.1)$ and $(x,y)=(1,0)$. Is this function even well defined? Obviously, this function is incorrigibly discontinuous (and possesses a number of bizarre characteristics but that's talk for another post).

If this counter example convinces you that this function is not well defined, can you think of a way to define an equivalence class on the elements of the domain (e.g., if the binary expansion of two given elements in the domain converge to the same thing then we define the images of the two points to be the same) so that the function still constitutes a bijection? Additionally, can anyone else see any other flaws in the reasoning here or produce another 'flavor' of such functions? Thanks in advance!

Answer 1

The naive digit intermingling you describe does not produce a bijection, for reasons related to some numbers having multiple representations, like you sketch.

There are various approaches to fixing this. One, as suggested in comments, is to produce a non-surjective injection and then appeal to the Schröder-Bernstein theorem together with the obvious injection in the other direction.

Another (quite clever) workaround for producing a bijection directly is:

1. Never use a representation of a number that ends in repeated 0s.

2. Divide the representation of $x$ and $y$ into groups of digits, where each group consists of zero or more 0s followed by exactly one nonzero digit.

3. Concatenate the digit groups taking one group from $x$ alternately with one group from $y$.

Answer 2

This is not a bijection. Consider number $\frac{1}{3}=0.010101...$. Your function can't return it. If you want a bijection you can use $\mathbb{R}\leftrightarrow 2^{\mathbb{N}}$. It is described here. I think that there is a simple way for finding bijection between finite sets and finite sets $\cup$ complements of finite sets. Existence of your function $f$ is dependent on the sets A and B. If A is countable and B not, bijection does not exist. If they are both uncountable you should take into account continuum hypothesis.