Bijective function between $\mathbb{R}^n$ and $\mathbb{R}^m$ I suspected that it was not possible to define a bijection $f \colon \mathbb{R}^m \to \mathbb{R}^n$ where $m,n \in \mathbb{N}$ and $m>n$.

After coming across Why are the cardinality of $\mathbb{R^n}$ and $\mathbb{R}$ the same? , I now suspect that it is in fact possible to define a bijection $f \colon \mathbb{R}^m \to \mathbb{R}^n$.


My reasons for believing that such a bijection exists:

*

*This being established, we can say that there exists a bijection $g \colon \mathbb{R}^m \to \mathbb{R}$.

*By the same argument, there exists a bijection $h \colon \mathbb{R} \to \mathbb{R}^n$.

*Thus the function $h \circ g \colon \mathbb{R}^m \to \mathbb{R}^n$ is bijective.

Can I now conclude that $\mathbb{R}^m$ and $\mathbb{R}^n$ have the same cardinality. So, does the set of all $m$ tuples with real entries have the same number of elements as the set of all $n$ tuples with real entries, even when $m\neq n$?
 A: This should really be a comment but I don't have enough reputation at the moment.
In most cases your intuition is correct. You can have a bijection going from $ \mathbb R^n $ to $ \mathbb R^m $ when $ m\neq n $, but such an $ f $ exhibits pathological behavior.
Assume for example that $ f\colon \mathbb R^n\to \mathbb R^m $ and $ f^{-1}\colon \mathbb R^m\to \mathbb R^n $ are differentiable: by the chain rule, the differential of $ f $ calculated in $ 0 $ is linear isomorphism between $ \mathbb R^n $ and $ \mathbb R^m $, and this of course implies that $ m = n $.
(You can require something less than full differentiability and still have the same result, but the proof is more technical)
A: COMMENT.-Basically $\mathbb R^n$ and $\mathbb R^m$ have same cardinality so there is a bijection as desired. You know that $I=]0,1[$ is in bijection with $\mathbb R$. Since
$$x\in I\iff x=0.a_1a_2a_3\cdots a_n\cdots$$ try carefully something like
$$x\rightarrow(0.a_1a_3a_5\cdots, 0.a_2a_4a_6\cdots)$$ to show that $\mathbb R$ is in bijection with $\mathbb R^2$. Then go to distinct dimensions $n$ and $m$.
