How can we distinguish the two zeroes of these two double-line $T_1$-spaces?

Question

Motivation

If we have topological spaces which are not $T_1$, we can sort of measure the “failure to be $T_1$” by the specialization preorder $x\leq y$ iff $\operatorname{cl}(x)\subseteq\operatorname{cl}(y)$, which I will not go into now. Once the space satisfies $T_1$, this preorder is of course discrete, hence cannot provide us with further means of “distinguishing the behaviour of points”, whatever that means.

Getting „one layer up“ in the (strict) chain of implications $T_2 \implies$ compact sets are closed ($KC$) $\implies$ convergent sequences have a unique limit ($US$) $\implies T_1$[1], I tried to compare the following two simple

Examples

Let $A, B$ have as base set $\mathbb R$ disjointly accompanied by $\{x^\prime\mid x\in \mathbb R\}=: \mathbb R^\prime$ – I.e, we start with two copies of $\mathbb R$, and denote their elements by $x$ and $x\prime$, and the subsets by $\mathbb R$, $\mathbb R^\prime$, respectively – and the topologies generated by intervals in $\mathbb R_{\neq 0}$ and $\mathbb R^\prime_{\neq 0}$ together with

• for $A$: Sets of the form $$ (-a,0)\cup (-a^\prime,0^\prime)\cup\{0\}\cup(0,a)\cup(0^\prime,a^\prime) $$ and $$ (-a,0)\cup (-a^\prime,0^\prime)\cup\{0^\prime\}\cup(0,a)\cup(0^\prime,a^\prime) $$
• for $B$: Sets of the form $$ (-a,0)\cup\{0\}\cup(0,a) $$ and $$ (-a,0)\cup (-a^\prime,0^\prime)\cup\{0^\prime\}\cup(0,a)\cup(0^\prime,a^\prime) $$

It is easy to verify that $A,B$ are $T_1$ spaces. By considering the sequence ${(1/n)}_n$, which converges against $0$ and $0^\prime$ in both spaces, hence we have two examples of spaces in the “last interval” of the chain (i.e., $T_1\wedge \neg US$).

Now, the spaces are slightly different:

• in $A$, there is a local homeomorphism between $0$ and $0^\prime$, this is not the case in $B$
• intuitively, the neighborhoods of $0$ are “closer” to $0$ than the neighborhoods of $0^\prime$ in $B$, while they are “indistinguishable” in $A$
• An indicator for this is that $(1/n^\prime)_n\subseteq \mathbb R^\prime$ converges to both zeroes in $A$, but only to $0^\prime$ in $B$.
• A naive attempt to order points based on sequence convergence fails however: If we define a partial order $x\leq y$ iff all sequences converging to $y$ also converge to $x$, we will see that $0,0^\prime$ are incomparable in both $A$ and $B$ by just considering the constant sequences, respectively (this actually follows from $T_1$ alone).

Thus arises the following

Question

Is there a natural$^1$ partial order on the elements of $T_1$ spaces in which elements $x,y$ are incomparable if and only if any sequence converges to at most one of $x,y$?

Remark 1. I chose the last condition carefully based on the observation that

1. $US$ spaces should induce a discrete order and
2. In our spaces, the only comparable points would be $0,0^\prime$.

Remark 2. I tried to expand on the observation that in $B$, the closure of $(0,a]$ includes both zeroes whereas the closure of $(0^\prime,a^\prime]$ includes only $0^\prime$, which is a difference in behaviour to $A$. However, after one day of fiddling around, this didn't lead anywhere.

$^1$ by natural, I just mean „definable entirely in topological terms, not depending on our example or on different axioms“

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[1] Wilansky, Albert. "Between T 1 and T 2." The American Mathematical Monthly 74.3 (1967): 261-266.

Answer 1

First, let me remark that question as stated is highly unnatural because of your restriction to sequences, which see very little of the topology of some spaces. So, instead of sequences I will consider nets.

With that said, there is an obvious fix to your "naive" definition: just exclude the constant nets (and nets which have constant nets as subnets). Note that this is natural and not ad hoc in the context of $T_1$ spaces, since the $T_1$ condition is all about constant nets (it is equivalent to saying that every constant net has only one limit).

So, let's define $x\leq y$ if every net which converges to $y$ and is not frequently equal to $y$ also converges to $x$. Equivalently, $x\leq y$ if every neighborhood of $x$ contains a deleted neighborhood of $y$ (that is, a set of the form $U\setminus\{y\}$ where $U$ is a neighborhood of $y$). It is easy to check that this relation is transitive on $T_1$ spaces (though it is not antisymmetric, so it is just a preorder).

This, I think, captures the idea you are trying to define. Note, though, that it does not have the exact property you ask for: you can have $x$ and $y$ which are incomparable but such that some net (or even some sequence) converges to both of them. For instance, consider the line with doubled origin except that the neighborhoods of the first origin $0$ are of the form $\{0\}\cup(-a,0)\cup((0,a)\cap\mathbb{Q})$ and the neighborhoods of the second origin $0'$ are of the form $\{0'\}\cup((-a,0)\cap\mathbb{Q})\cup(0,a)$. Then $0$ and $0'$ are incomparable (neither one's neighborhoods always contain deleted neighborhoods of the other), but the sequence $(1/n)$ converges to both of them.

Note moreover that by the symmetry of this example, if $0$ and $0'$ were to be comparable for some natural preorder, then we would have to have both $0\leq 0'$ and $0'\leq 0$ (in particular, this must hold for any preorder $\leq$ which is preserved by homeomorphisms). But now consider what happens if we add a third origin $0''$ whose neighborhoods have the form $\{0''\}\cup((0,a)\setminus\mathbb{Q})$ (and, if we want to maintain the symmetry between $0$ and $0'$, we could also add a fourth origin with the corresponding neighborhoods on the negative side). There are sequences which converge to both $0''$ and $0'$ (take irrationals approaching $0$ from above) and so $0''$ must be comparable to $0'$. Since both $0'\leq 0$ and $0\leq 0'$, this means $0''$ must also be comparable to $0$. But there is no net which converges to both $0''$ and $0$ (they have disjoint neighborhoods). So, this proves there cannot be any natural preorder which satisfies your condition.