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At first I was looking for a way to have a function definition which is one-to-one from $\mathbb{R}^n$ to $\mathbb{R}$. Then I have seen this question and it seems according to the rank-nullity theorem, this is not possible.

How about the vector spaces over $\mathbb{Q}$ ? Is it possible to define a one-to-one function from $\mathbb{Q}^n$ to $\mathbb{Q}$ such that

$$ f : \mathbb{Q}^n \rightarrow \mathbb{Q} $$

If this is true, do you have an example of a function that satisfy it?

steve cleng
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    Rank-nullity is about linear maps, not maps in general. – Vercassivelaunos Mar 24 '22 at 14:11
  • As pointed above, the link you're referring to talks about linear mappings. If you drop the linear constraint the answer is yes. It is a standard proof that there exists bijections between $\mathbb{N}\times\mathbb{N}$, and $\mathbb{N}$, and $\mathbb{Q}$. From there you can find your mapping. About the case $\mathbb{R}^n$, have a look at: https://math.stackexchange.com/questions/3129820/bijection-f-mathbbr-times-mathbbr-to-mathbbr?rq=1 or also https://math.stackexchange.com/questions/4131432/proof-that-mathbbr2-cong-mathbbr?rq=1 – Leo Mar 24 '22 at 14:29

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$\mathbb{R}^n$ and $\mathbb{R}$ have the same cardinality, so by definition, a bijection exists between the two. The same is true for $\mathbb{Q}^n$ and $\mathbb{Q}.$ In fact, for any infinite set $S,$ $S^n$ and $S$ have the same cardinality, and so a bijection exists between the two.

The rank-nullity theorem is about linear maps, not functions in general. Every $n$-dimensional vector space over the scalar field $\mathbb{R}$ is isomorphic to $\mathbb{R}^n.$ The rank-nullity theorem is essentially the statement that although $\mathbb{R}^n$ and $\mathbb{R}$ are isomorphic as sets, they are not isomorphic as vector spaces.

Angel
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