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[Preface: please excuse my mathematical naïveté...]
I've learned (via: https://ncatlab.org/nlab/show/specialization+topology) that there is a way to turn any topological space $(X, \tau)$ into a pre-ordered set (Proset).

Interpreted as a proset, the binary $\leq$ relation (or lack thereof) between all the points of a space makes a lot more intuitive sense than defining a topological space as being built from open sets (which are just subsets that obey some arbitrary-looking axioms). The binary relation of a proset immediately makes clear the notion of adjacency/nearness/closeness between points and the notion of connectedness, whereas the notion of open sets feels like just blindly following rules.

[@HenningMakholm pointed out this is wrong:]
Moreover, once we consider a metric space, it is easily recognizable how a topological space generalizes a metric space. If we consider a topological space as a proset with just a binary $\leq$ relation between points then there is no notion of distance, however, if we define a metric on a space you automatically get a $\leq$ relation between points. E.g. for a metric space $(M, d)$ (where $M$ is a set and $d$ the metric function), then for any $a, b \in M$ we can choose an arbitrary point $c \in M$ to be our "origin" and then $a \leq b$ if $d(a,c) \leq d(b,c)$.

So my question is (assuming I haven't made totally fallacious statements thus far), given that prosets make topological notions far more intuitive and easy to reason about, why is topology developed with the mystifying (at least initially until you acquire an intuition) definition of open sets rather than prosets?

Henno Brandsma
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Brandon Brown
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  • For the plane, it doesn't seem like this notion would distinguish between different points that have the same distance to the center. – hmakholm left over Monica Jul 23 '17 at 20:27
  • For example, consider ${(x,y)\mid x>0, x^2+y^2<1}$ versus ${(x,y)\mid x\ge 0, x^2+y^2<1}\setminus{(0,0)}$. There's an order isomorphism of the entire plane taking one of these sets to the other, but one is open and the other is not. – hmakholm left over Monica Jul 23 '17 at 20:29
  • @HenningMakholm oh right, you're right. I can't believe I missed that. Hmm, I'll have to think/research more to see if a metric induces a proset. – Brandon Brown Jul 23 '17 at 20:33
  • This doesn't directly answer your question, but there is a (somewhat extreme) generalization of metric spaces where every topological space can be endowed a general metric that in some sense recovers the idea that a set is open if there is an open ball around each point. this is not light reading, but you can see some at Ittay Weiss's note on the subject, linked from this MSE answer. – Mark S. Jul 23 '17 at 20:45
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    The notion of open sets, and the axioms for a topology aren't that arbitrary once you look at the history of how they were designed and if you really try to understand why they are this way. Maybe a better way to introduce them would be through neighbourhoods (which gives an equivalent construction): then the axioms for a topology become really clear – Maxime Ramzi Jul 23 '17 at 20:49
  • There are ways of representing " nice" topological spaces algebraically, as sets. – gary Jul 23 '17 at 21:30
  • Topology was originally called Analysis Situs (Analysis of Position) and was concerned exclusively (or nearly so) with metric spaces....... Consider that $f:X\to Y$ is continuous iff whenever $f(x)\in V \subset Y$ where $V$ is open in $Y,$ there is an open $U\subset X$ with $x \in U$ such that $f$ maps $U$ into $V.$ This is the topological generalization of the classical "$\epsilon$-$\delta$ " def'n of continuity of a real function. – DanielWainfleet Jul 24 '17 at 04:21

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I think you've misunderstood the article. What it describes is a way to assign to any proset $P$ a topology - its specialization, $Spe(T)$. This goes the wrong way: it doesn't build a proset from a topological space.

Thinking about the inverse of this operation gives a way to assign a poset to a topological space, given by setting $a\le b$ if every open set containing $b$ also contains $a$; but this loses lots of information (for instance, any two $T_1$ topologies of the same cardinality yield the same proset, so this won't distinguish between e.g. $\mathbb{R}^2$ or $\mathbb{R}^3$). So this way of converting a space into a poset is a really bad one, from the point of view of understanding the original space; rather, we should think of this poset as capturing part of the information of the space, but not in general very much information.

The poset construction you've described for a metric space also loses information, although it also captures some information.

There is another way to "turn a space into a poset." To any space $X$, we may assign a poset $F(X)$ whose elements are the open sets in $X$, and where "$\le$" is "$\subseteq$." Note that elements of the poset are sets of points, not individual points, so we aren't really turning the space into a poset in the way I think you want. A point in the space $X$ then yields a filter in $X$: given $p\in X$, let $F(p)=\{U\in F(X): p\in U\}$.

This still loses some information, but not too much, and leads to the subject of pointless topology.

If you want to turn spaces into posets, here are some test problems for you:

  • Consider the discrete, trivial, cofinite, and cocountable topologies on $\mathbb{R}$ respectively. What are the prosets you want to assign to these? Are they different?

  • Consider the lower limit topology on $\mathbb{R}$. What prosets do you want to assign to this? Is it different from the one you want to assign to $\mathbb{R}$ with the usual topology?

I think that if you think about things like this, you'll quickly realize that topological structure carries far more information than is easy to fold into a proset whose elements must correspond to the points in the original space.

Indeed this is true in a precise sense: for infinite $\kappa$, there are only $2^\kappa$-many prosets of size $\kappa$ (up to isomorphism), but there are $2^{2^\kappa}$-many topological spaces with a set of points of size $\kappa$ (up to homeomorphism). So this means that we can't turn points in a space into elements in a proset without losing tons of information.

Noah Schweber
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    Note that $F(X)$ preserves all information about $X$ if $X$ is $T_1$: then the points of $X$ can be recovered from the order because "open set that has exactly one proper open superset" are exactly the sets of the form $X\setminus{x}$. – hmakholm left over Monica Jul 24 '17 at 01:38
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Viewing a space as a partial order of the points can only be done
for low level non Hausdorf spaces. On the other hand any topology
can be viewed as complete lattice of open sets with the subset order.
For a collection C of open sets, sup C is the union of the sets in C
and inf C is the interior of the intersection of the sets in C.

William Elliot
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There is an intriguing area that relates to your question namely that of finite Alexandroff spaces and their connection with posets. The most accessible sources for this are Jonathan Barmak's Algebraic topology of finite topological spaces and applications, SLN volume 2032,(2011),

and J. P. May., 2003, Finite spaces and simplicial complexes., URL www. math.uchicago.edu/~may/MISCMaster.html, notes for REU.

Another area worth following up is the theoryof Locales and Heyting Algebras which occur in the theory of tooses but also in a lot of the interactions of topology with theoretical computer science. Look at Steve Vicker's little book: Topology via logic. Volume 5 of Cambridge Tracts in Theoretical Computer Science. Cambridge University Press, Cambridge (1989).

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Tim Porter
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