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Let us consider Euclidean space $\mathbb{R}^n$. We say it is $n$-dimensional because each vector in it is an $n$-tuple $(x_1,...,x_n)$. However, it is possible to represent this exact same space using only a single number by creating a bijection $\mathbb{R}^n\to\mathbb{R}$ (e.g. for $n=3$). Therefore, by the conventional definition of dimension, we must conclude that $\mathbb{R}^n$ is in fact one-dimensional.

How do we resolve this? What is a rigorous definition of dimension?

This question was inspired by this one.

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lemon
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  • Dimension of a vector space is defined in terms of linear maps. The claim that dimension is an invariant amounts to saying there is no bijective linear map between $\mathbb R^n$ and $\mathbb R^m$ when $n \neq m$. So in particular, any bijection between $\mathbb R^n$ and $\mathbb R$ cannot be linear. – Dustan Levenstein Jul 22 '14 at 22:48
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    A bijection between $\mathbb{R}^n$ and $\mathbb{R}$ can, for $n \neq 1$, not be linear. Thus the spaces are not isomorphic as vector spaces. – Daniel Fischer Jul 22 '14 at 22:48
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    "by the conventional definition of dimension" - what definition do you mean? – Marcin Łoś Jul 22 '14 at 22:49
  • Great! @DanielFischer and Dustan: is there a proof that such a bijection cannot be linear? – lemon Jul 22 '14 at 22:51
  • @MarcinŁoś that the dimension is the least number of coordinates required to represent any point within the space. – lemon Jul 22 '14 at 22:52
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    @ozo this is a general algebraic fact, since there are $n$ linearly independent vectors in $\Bbb R^n$, any linear bijection must map a basis to another basis, i.e the target set needs to be of the same size, but that would produce a bijection between the finite sets ${1,2,3,\ldots, n}$ and ${1}$ which is impossible when $n>1$. – Adam Hughes Jul 22 '14 at 22:57
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    The rank formula, for example. That shows that the kernel of every linear $T\colon \mathbb{R}^n \to \mathbb{R}$ has dimension $\geqslant n-1$, so for $n > 1$, it cannot be injective. For $n = 0$, there is not even a non-linear bijection. – Daniel Fischer Jul 22 '14 at 22:57
  • I'm not 100% sure that [tag:dimension-theory] fits here. You're asking about the dimension of a vector space. – Asaf Karagila Jul 22 '14 at 23:02

2 Answers2

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You can resolve this when you notice that any such bijection does not preserve the vector space structure of $\Bbb R^3$. In simpler words, it is not a linear transformation.

(Assuming the axiom of choice, you can find such bijection which preserves the additive structure, and therefore the structure of a vector space over $\Bbb Q$, but never as a vector space over $\Bbb R$.)

The point is that cardinality "shakes off" disregards the structure and only considers the underlying set. But dimension requires structure, so you can't just define the dimension of a set as a vector space or a topological space, without specifying the topology or the vector space structure as well.

What is true is that $\Bbb R$ can be endowed with a $3$-dimensional real vector space structure. But this structure is incompatible with the usual structure of $\Bbb R$ as an ordered field.

Asaf Karagila
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  • Thank you. You hit the nail on the head by highlighting the fact that cardinality shakes off structure. I was imagining the two spaces as being identical. – lemon Jul 22 '14 at 23:05
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For vector spaces, the definition of dimension is the number of vectors in a basis for the vector space.

A basis for a vector space V over a field $\mathbb{F}$ is a set of vectors such that every vector in V is a unique linear combination of the basis vectors. That is, $\{ v_1, ..., v_k \}$ is a basis when every vector $v \in V$ can be written uniquely as $v = a_1 v_1 + \dots + a_k v_k$ for scalars $a_1$, \dots, $a_k \in \mathbb{F}$.

It is possible to prove all bases for the same vector space have the same number of vectors, and this number is defined to be the dimension.

For $\mathbb{R}^n$, a basis is $(1,0,\dots,0)$, $(0,1,\dots,0)$, ..., $(0,0,\dots,1)$ so it has dimension $n$.

DavidButlerUofA
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