How can a bijection from $\mathbb R^n \to \mathbb R$ exist when a non square matrix is not invertible?

Question

I have found references online that say there exists a bijection between $\mathbb{R}^n$ and $\mathbb{R}$. I have also found references that say that if a bijection exists between sets, then an inverse exists. However, I know that any matrix transformation, $A$, between $\mathbb{R}^n$ and $\mathbb{R}$ is in $\mathbb{R}^{n\times 1}$ which is not square and thus does not have an inverse.

How is this not a contradiction on the various claims? Thank you for the help in clearing this up.

Answer 1

It might help you to know what sort of bijection it actually is. Here's an example, if you think of $\mathbb{R}^n$ as just a list of n numbers, then take the n numbers and interleave their digits to form a single number. For example in $\mathbb{R}^3$:

(3.45, 27.9, 0.045) -> 20370.490504005

Now it's fairly clearly that this is a bijection, and fairly clear you can invert it. But you can't represent it with a matrix, because it's not at all a linear mapping.

Answer 2

There is no contradiction. Every bijection from $\mathbb{R}^n$ onto $\mathbb R$ has an inverse, whereas (if $n>1$) there is no linear bijection from $\mathbb{R}^n$ onto $\mathbb R$.