Question: what (if any) results in classical homotopy theory rely on the separability of the real interval?
For a more expanded, fleshed out, and (in my opinion) interesting version of the context given below, please check out the following essay: https://machineappreciation.wordpress.com/2022/06/03/to-what-extent-does-classical-homotopy-theory-rely-on-the-real-numbers/
Context:
Let me give some more context to this question. In classical homotopy theory (no model categories), the definition of two maps being homotopic is really simple. A homotopy from $f:X \rightarrow Y$ to $g:X \rightarrow Y$ is a continuous map $H:X \times I \rightarrow Y$ such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for all $x$. Here $I$ denotes the real closed interval $[0,1] \subseteq \mathbb{R}$ This homotopy is often thought of as some “continuous process” that takes us from $f$ to $g$.
There’s something fundamentally dissatisfying to me about this definition. In particular, I usually think of the real numbers as an “analytic” object, not a topological one. Thinking of $\mathbb{R}$ strictly as an order, it is immediately clear that we could take any linear order with endpoints $\langle L, \leq \rangle$, turn $L$ into a topological space with the order topology, and define an analogous notion of $L$-homotopy, where the real interval $I$ is replaced with $L$. Since $L$ is a linear order, $L$-homotopy maintains this “continuous process” intuition.
I know that there is a lens through with the real closed interval is the terminal coalgebra in a category of bipointed spaces endowed with a wedge product, but I’m also interested in this generalization.
One key property of “ordinary” homotopy is that it is an equivalence relation. If we want $L$-homotopy to be an equivalence relation, we need to impose some conditions on the order. To get symmetry, we need to impose that $L$ is “reversible”, in the sense that $L$ is order-isomorphic to $L^*$, the reversed order on $L$. To get transitivity, we need to impose that $L$ is “divisible”, in the sense that there is a point $\ell \in L$ such that $\{x \in L : x \leq \ell \} \cong \{ x \in L : x \geq \ell \} \cong L$ as orders. WE can think of this as a sort of wedge product.
Finally, $L$ needs to be dense and Dedekind complete as an order. If $L$ is not dense and Dedekind complete, $L$ will not be connected as a topological space. Thus we would get a trivial homotopy theory where all maps are $L$-homotopic.
So all in all, we know that $L$ needs to be a reversible, divisible, Dedekind complete dense linear order with endpoints. It is a well known fact that the real closed interval $[0,1] \subseteq \mathbb{R}$ is the unique separable Dedekind complete dense linear with endpoints. $[0,1]$ also happens to be divisible and reversible. The essence of my question is what do we lose by giving up separability?
Here are two examples of different intervals we could use that meet the needed conditions.
- Take the hyper-real numbers $^*\mathbb{R}$. These are not Dedekind complete, but we could take their Dedekind completion to get a new order (at the expense of losing the algebraic properties) $\overline{^*\mathbb{R}}$. Take the closed interval $[0,1] \subseteq \overline{^*\mathbb{R}}$.
- Take the closed unit square $[0,1] \times [0,1] \subseteq \mathbb{R}^2$. Put the lexicographical order on it.
What do we lose when we do homotopy with respect to these intervals? That is, what results do we lose when we throw away separability? A satisfying answer to this question, to me, would look like “The X-Theorem relies on the separability of the real closed interval”.