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I tried to construct a total order on the set $2^{\mathbb{R}}$. I failed. I'm wondering, can it be done, constructively? Here's what I tried.

I'm sometimes going to treat elements of $2^{\mathbb{R}}$ as functions of type $\mathbb{R} \rightarrow \{0, 1\}$ and sometimes treat them as subsets of $\mathbb{R}$. Suppose $f, g \in 2^{\mathbb{R}}$ with $f \neq g$. Since $f \neq g$ then $\exists x\;f(x) \neq g(x)$. Define the set $X$ as follows:

$$X = \{x : f(x) \neq g(x)\}$$

Since $\{0, 1\}$ is clearly totally ordered, we can further split $X$ up into sets $F$ and $G$ where $f$ and $g$ are larger, respectively i.e.

$$F = \{x : f(x) > g(x)\}$$ $$G = \{x : g(x) > f(x)\}$$ $$X = F \cup G$$

So I tried to proceed as follows. If $F$ is empty then $G$ isn't so we'll say $g > f$. Conversely if $G$ is empty we'll say $f > g$. If both are non empty, then take the infimum of both sets and compare to decide which is bigger out of the two: whoever has the smaller infimum is the larger one. However, that's where I ran into an issue. E.g. we might end up with:

$$f = \{2^{-n} : n \in \mathbb{N}\}$$ $$g = \{3^{-n} : n \in \mathbb{N}\}$$

Where I'm just using a set-based definition of $f$ and $g$ rather than a function-based definition, as it's easier here. Then since $f$ and $g$ are disjoint, $F$ and $G$ are the very same sets, respectively:

$$F = \{2^{-n} : n \in \mathbb{N}\}$$ $$G = \{3^{-n} : n \in \mathbb{N}\}$$

Both infimums are $0$ i.e.

$$\inf{F} = \inf{G} = 0$$

So I'm stuck here, I can't find a way to constructively differentiate $f$ and $g$ in the case where both infimums are $0$. The construction above I believe does give rise to a partial order with elements being comparable whenever either $F$ or $G$ is empty or when the infimums differ. I did a quick check and that definition seems to satisfy transitivity. But clearly it's not total, it doesn't cover all possible $f, g$.

Notes

  • There are some questions on order theory like this one that seem to indicate that there exist some totally ordered sets with cardinality greater than that of $\mathbb{R}$. But I'm not sure if these are just proofs of existence rather than an explicit construction of an order.
  • I think the above construction can actually be simplified to saying that $f < g$ if there exists some point $x$ with $f(x) < g(x)$ and for all points $y < x$, $f(x) \leq g(x)$.
  • I think you can also simplify the above construction by taking the convention of the infimum of an empty set being positive infinity $+\infty$.
Colm Bhandal
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    I recall reading somewhere on this/these site/s that it is consistent with ZF that the quotient group $\Bbb R/\Bbb Q$ cannot be totally ordered (as a set). Since $\Bbb R/\Bbb Q\subseteq\mathcal P(\Bbb R)$, it is consistent with ZF that $\mathcal P(\Bbb R)\simeq 2^{\Bbb R}$ can't be totally ordered. –  Oct 07 '20 at 08:30
  • @Gae.S. That's interesting. I can see how the conclusion of that entailment would imply the non-existence of a construction that I ask for, but I don't follow how the entailment holds & I'd need to read up on why the premise holds – Colm Bhandal Oct 07 '20 at 10:47
  • @Gae.S. Actually I do see how the entailment holds. – Colm Bhandal Oct 07 '20 at 10:58
  • Good, point. I was thinking about a set-up where you are working in ZF, possibly with a weaker deductive system, but now I see that, if you are still working with a weaker deductive system, but with axioms encoding ZFC, then the argument I had in mind leaves open the possibility that whether or not the explicit statement you are trying to devise describes a total order on $2^{\Bbb R}$ is simply undecidable in ordinary ZF. –  Oct 07 '20 at 11:35
  • I have to admit I'm a bit out of my depth, but I think it's fair to say that my question is ambiguous because I don't specify what I mean by constructive. Any formal definition of constructive I'm sure would involve explicitly restricting ourselves to an axiomatic framework. However if it's provably undecidable in ZF then I guess that rules out any straightforward constructions like the one I attempted – Colm Bhandal Oct 07 '20 at 13:49

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