Continuous extension of Euclidean spaces?

Question

I am wondering if it is possible to "continuously" increase the dimension of Euclidean spaces — in other words, would it be possible to define Euclidean spaces of non-integer dimensions with nice topological properties?

I have thought about the way to generalize Euclidean space with nonnegative real dimension, and here are some axioms that I have set.

A sequence $\mathcal{R}$ of generalized topological spaces is given by the following data and properties:

• For each nonnegative real $d \geqslant 0$, there corresponds a topological space $\mathcal{R}(d)$.
• If $d \geqslant 0$ is an integer, then $\mathcal{R}(d)$ is homeomorphic to $\mathbb{R}^d$.
• If $d, e \geqslant 0$ satisfies $d \neq e$, then $\mathcal{R}(d)$ and $\mathcal{R}(e)$ are not homeomorphic to each other.
• For each pair of nonnegative reals $d \geqslant e \geqslant 0$, there corresponds an embedding (i.e. a continuous injection) $\rho_{ed} : \mathcal{R}(e) \rightarrow \mathcal{R}(d)$.
• If $d \geqslant 0$, then $\rho_{dd}$ is an identity function on $\mathcal{R}(D)$.
• If $d \geqslant e \geqslant f \geqslant 0$, then $\rho_{ed} \circ \rho_{fe} = \rho_{fd}$.

Sequences of generalized Euclidean spaces, however, might not be set-theoretically unique, so we can define isomorphisms between such sequences. Two sequences $\mathcal{R}_1$ and $\mathcal{R}_2$ of generalized Euclidean spaces are said to isomorphic if:

• There exists a proper mapping $\varphi : \mathbb{R}_{\geqslant 0} \rightarrow \mathbb{R}_{\geqslant 0}$.
• For all $d \geqslant 0$, $\mathcal{R}_1(d)$ and $\mathcal{R}_2(\varphi(d))$ are homeomorphic to each other.

Now I wonder if such sequence of generalized Euclidean spaces exists, and if it is unique up to isomorphism provided that it exists.

Any feedback on either existence/uniqueness problem or general background of the question would be highly appreciated.

Answer 1

Perhaps not an answer; too long for a comment.

Mathematicians usually look for generalizations when they have a problem to solve rather than a definition that seems as if it might be generalizable, so my first question would be "why do you want to do this, other than the fact that it seems interesting?"

You could start by thinking about fractals, which are geometric objects with well defined fractal dimension that need not be integral.

A google search for nested fractal found lots of links; perhaps some of them will be fruitful. You could also try fractal embedding or other synonyms that might address your fourth bullet.

Answer 2

$\newcommand{\Reals}{\mathbf{R}}$Here's a sketched construction of an increasing family of subsets of $\Reals^{\infty}$ (the space of eventually-zero sequences of reals) satisfying the stated axioms, modulo the existence of a set-valued function $P:[0, 1) \to \mathcal{P}(\Reals)$ with the properties that if $0 \leq s < t < 1$, then $P(s) \subset P(t)$ is a proper subset not homeomorphic to $P(t)$. (A prospective $P$ is to let $K$ denote the Cantor ternary set, use the axiom of choice to fix a bijection $p:(0, 1) \to K$, and to take $P(0) = \varnothing$ and $P(d) = p(0, d)$ if $0 < d < 1$.)

For each positive integer $k$, identify $\Reals^{k}$ with the set of sequences in $\Reals^{\infty}$ whose first $k$ components are arbitrary and all remaining components are zero.

Fix an increasing bijection $h:[0, 1) \to [0, \infty)$, such as $h(x) = x/(1 - x)$.

If $k \leq d < k+1$, define \begin{align*} \Reals(d) &= \bigl[\Reals^{k} \times (-h(d - k), h(d - k))\bigr] \cup \bigl[P(d - k) \times \Reals^{k}\bigr] \\ &= \{(x_{j})_{j=1}^{\infty} \in \Reals^{k+1} : |x_{k+1}| < h(d - k) \text{ or } x_{1} \in P(d - k)\}. \end{align*} The idea is to "thicken" $\Reals^{k}$ monotonically by expanding along the $(k + 1)$th coordinate, while "tagging" the result with the cylinder $P(d - k) \times \Reals^{k}$ to get mutually non-homeomorphic sets.

When $d = k$, the open interval $(-h(d - k), h(d - k))$ should be interpreted as the singleton $\{0\}$, so that $\Reals(k) = \Reals^{k}$ when $k$ is an integer. The embeddings $\rho_{ed}$ are set-theoretic inclusions.

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Assuming this construction (or something like it) works, there's no hope of "uniqueness up to isomorphism". I haven't carefully checked details, however, and invite others to bolster or modify this answer accordingly. (It's likely there are better ways to "tag" the thickening, i.e., procedures that generate increasing families of sets that are clearly pairwise non-homeomorphic.)