In any ordered metric space, let a set $S$ be sortable if it can be placed into 1-1 correspondence with $\mathbb{N}$ such that $x > y \iff n(x) > n(y)$ for all $x,y \in S$. I conjecture the following about sortable sets, and request help proving (or disproving!) them.
Conjecture 1: A set is sortable if and only if it is homeomorphic to $\mathbb{N}$. I conjecture this because the sortable sets I encountered, such as subsets of $\mathbb{N}, \{kn\}, \{n+k\},$ and $\{1/n\}$ are all homeomorphic to $\mathbb{N}$, whereas the unsortable sets, such as $\mathbb{Z}, \mathbb{Q},$ and $\{1/n\}\cup\{0\}$, are not.
Conjecture 2: If a set is countable but not sortable, is it somehow composed of sortable sets? $\mathbb{Z}$ is the union of two sortable sets, and while $\mathbb{Q}$ is not the union of any collection of sortable sets, it is the product of two sortable sets [*]. Is every countable set a union or product of a finite collection of sortable sets?
Proposition 3: An infinite sortable set is not compact. Proof: Consider the collection $O$ of open balls around each $s \in S$, each of radius less than $\min(d(n^{-1}(n(s)+1), s), d(n^{-1}(n(s)-1), s)$. Each $s \in S$ is in precisely one $o \in O$. Thus, $O$ is an open cover of $S$ with no finite subcover.
Besides the above, are there other noteworthy properties of sortable sets? Is there any discussion of sortable sets, or a similar concept, in the literature?
[*] By "$\mathbb{Q}$ is the product of two sortable sets," I mean that any rational number corresponds to two independently chosen natural numbers, the numerator and the denominator. However, I see now that this is not an injection, because, for example $\frac 1 2 = \frac 2 4$, whereas $(1,2) \neq (3,4).$ Perhaps I need to refine my statement: Can $\mathbb{Q}$ be somehow composed from finite sortable sets?
Best Answer
@SRobertJames, I add another answer because the explication is too long for the comments.
When you say "goes to infinity in two different ways," it can mean two things, neither of which is a topological property.
First, it can mean that if I take an $x \in \mathbb{Z}$, there is an infinite number of numbers higher than $x$ and an infinite number of numbers lower than $x$. This notion arises from your order and has nothing to do with the discrete topology of $\mathbb{Z}$.
For example, let's build another order $<'$ on $Z$ for which $\mathbb{Z}$ doesn't go to infinity. Take the set $X = [0, 1] \cap \mathbb{Q}$. $X$ is countable so you can define a function $f$ from $\mathbb{Z}$ to $X$. Now for any $x, y \in \mathbb{Z}$ define $x <' y \Leftrightarrow f(x) < f(y)$. Here the ordered set $(\mathbb{Z}, <')$ has a maximum and a minimum element, respectively $f^{-1}(0)$ and $f^{-1}(1)$, so there is no "going to infinity" with this order. So this is a property of $\mathbb{Z}$ seen as an ordered set with the canonical order, not as a topological space.
Similarly, if "goes to infinity in two different ways" is interpreted as "The two sequences $u_n = n$ and $v_n = -n$ are such that $d(0, u_n)$ and $d(0, v_n)$ diverges when $n$ goes to infinity", then this notion arises from the structure of $\mathbb{Z}$ as a metric space, not as a topological one. You need to have a distance to define your statement. If you take the discrete distance (i.e. the distance $d'$ such that $d'(x,y)$ is $0$ if $x=y$ and is $1$ otherwise) on $\mathbb{Z}$, then $\mathbb{Z}$ has a diameter of $1$ and nothing "goes to infinity" anymore. Once again, the "goes to infinity" parts here arises from the canonical metric structure of $\mathbb{Z}$, not from $\mathbb{Z}$ itself or its discrete topology.
Topology is interested in continuity, in how points are connected; there is no notion of how points are ordered or how far they are. And so, for topology, as long as your points aren't connected, it is precisely as if they live in totally different spaces. They are not topologically relatable to one another. This is why you don't have this notion of "goes to infinity" when dealing with isolated points. $\mathbb{Z}$ (with the discrete topology) is topologically equivalent to a collection of isolated points indexed by $\mathbb{Z}$, i.e. $\bigsqcup_{\mathbb{Z}} \star$ where $\star$ designate a singleton. When you write it that way, you see that you lost your metric and your order. And "pairing" isolated points equivoquely between two spaces exactly describes a bijection. This is how you can intuitively understand why when you have two spaces with the discrete topology, having them homeomorph and having them in bijection is exactly the same. Because for the discrete topology, spaces are literally seen as clouds of isolated points indexed by your original set.
I suspect that the notion you are interested in is not the one of homeomorphism but of monotonic homeomorphism. Here, you try in some sense to merge the topological structure of your space and the order structure of your space. In doing so, the natural "isomorphism" you seem to have in mind are the isomorphisms that preserve both the topological structure (homeomorphism) and ordered structure (monotonic function) of your space. And on this point, you are absolutely right; there exists no monotonic homeomorphism between $\mathbb{Z}$ and $\mathbb{N}$ because there exists no monotonic function between $\mathbb{Z}$ and $\mathbb{N}$. You can formally state this in category theory by saying that $\mathbb{Z}$ and $\mathbb{N}$ with the discrete topology are isomorphic in the category of topological spaces $\mathbf{Top}$, but that $\mathbb{Z}$ and $\mathbb{N}$ with their canonical order are not isomorphic in the category of ordered set $\mathbf{OrdSet}$. If you are interested in merging different structures on spaces, I highly suggest looking at category theory. It is a very powerful framework to think about those things.