A category of spaces in which $\mathbb{R}_{>0}$ is not isomorphic to $\mathbb{R}$ $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$I am writing up notes on totally positivity in flag varieties. I often have the following situation: I have a complex variety $X$ defined over $\RR$ and a real semi-algebraic subset $M$ of $X(\RR)$. I have $M \cong \RR_{>0}^n$ in the following sense: There is some affine patch $T$ in $X$, isomorphic to $(\CC^{\times})^n$, such that $M$ is the orthant $\RR_{>0}^n$ in $T(\RR)$.
The simplest example is to take $X = \mathbb{P}^1$ and $M$ to be the set of points $[x:y]$ with $x$, $y \in \RR_{>0}$. A more interesting example is to consider $M=\{ (x,y) : xy>1,\ x,y > 0 \}$ in $\RR^2$. Note that $M$ is contained in the open set $T = \{ x(xy-1) \neq 0 \}$, and that the map $(x,y) \mapsto (x, xy-1)$ is an isomorphism between $T$ and $(\CC^{\times})^2$, with $M$ corresponding to $\RR_{>0}^2$. This latter example points out that there is no uniqueness to $T$; the torus $\{ y(xy-1) \neq 0 \}$ could play the same role.
So, when I am writing these notes, I want to say $M \cong \RR_{>0}^n$. And now the nitpicking reader who lives in my head says "in what sense is $M$ isomorphic to $\RR_{>0}^n$ and not also isomorphic to $\RR^n$?".
I had hoped the answer was "they are not isomorphic as semi-algebraic sets." But, if I understand the definitions correctly (see eg these notes), $\RR$ and $\RR_{>0}$ are isomorphic as semi-algebraic sets: The map $x \mapsto x+\sqrt{x^2+1}$ is an isomorphism $\RR \longrightarrow \RR_{>0}$ in the category of semi-algebraic sets. So my question is :

What is a reasonable category of spaces -- ideally one which already appears in the literature -- for which $\RR^n$ and $\RR_{>0}^n$ are not isomorphic?

 A: The category of uniform spaces should fit the bill: $\mathbf R^n$ and $\mathbf R_{>0}^n$ are homeomorphic, but not uniformly continuously. Here $\mathbf R_{>0}$ has the uniform structure inherited from the real line, not the one induced by the group structure. (I see now that Noah Schweber made the same suggestion in a comment.)
That said, I kind of disagree with the premise of the question. I think you can justify writing $M \cong \mathbf R_{>0}^n$ rather than $M \cong \mathbf R^n$ on account of the first isomorphism being more canonical than the second one. Indeed the first isomorphism is fixed once you've made the choice of an affine patch of $X$ with an isomorphism to $\mathbb G_m^n$, whereas the second isomorphism depends in addition on the noncanonical choice of a homeomorphism  between $\mathbf R^n$ and $\mathbf R_{>0}^n$.
A: Let $X$ be a metric space.  For every $x\in X$ and every $r\in {\mathbb R}$ let $B(r,x)$ be the punctured  ball of radius $r$ around $X$.
Now fix $x$ and let $r_1\neq r_2\in {\mathbb R}$.  Consider the following condition:
No connected component of $B(r_1,x)$ is equal to any connected component of $B(r_2,x)$.
In ${\mathbb R}$, the above condition holds for every $x,r_1,r_2$.  In ${\mathbb R}_{> 0}$, it doesn't.  So maybe you want a category that keeps track of the structure of the lattice of the connected components of the various $B(r,x)$.
Unfortunately, this won't work as it stands in dimensions higher than 1.
