Simply stated, I've been trying for a long time to either find in the literature, or derive myself, a notion of path in Cech closure spaces, that specialises to paths in a topological space, and to graph-like paths in so-called "quasi-discrete closure spaces".

Let me recall the definitions:

A closure space is a pair $(X,C)$ where $C : \mathcal P (X) \to \mathcal P (X)$ is a function satisfying $C(\emptyset) = \emptyset$, $A \subseteq C(A)$, $C(A \cup B) = C(A) \cup C(B)$.

A continuous function $f$ is a function between two spaces such that $f(C(A)) \subseteq C(f(A))$

A topological space is (via the Kuratowski definition) a closure space with the additional axiom $C(C(A)) = C(A)$ (idempotence of closure).

Any reflexive relation $R$ generates a closure space by $C(A) = \{y \in A | \exists x \in A . x R y\}$. That's called a "quasi-discrete closure space".

Topological paths are defined as continuous functions from the unit interval.

Let me now make two examples.

Example 1: $\mathbb R^2$. Topological paths work fine (indeed!).

Example 2: the closure space on $\mathbb N$ generated by the successor relation. It's a nice closure space, but topological paths exist that do not "follow the edges" of non-symmetric relations, due to "directionality" of $R$; topology (e.g. the unit interval) is intrinsically symmetric; relations are not. For an example of this consider the set $\{a,b\}$ and the relation $R = \{ (a,b) \}$. This generates a quasi-discrete closure space. Consider the function $f : [0,1] \to \{a,b\}$ with $f(0) = b$ and $f((0,1])) = a$. This function is continuous but not a graph-like path in $R$.

Further clarifications (due to comments)

I understand that [0,1]-paths in topological spaces can't be directional. That's absolutely the case and for good reasons. But then, is there a more general construction that becomes the "natural" notion of path in closure spaces, and in topological spaces, it is not directional, since this is very natural in topological spaces?

Let's say it more formally: perhaps there's a universal construction in the category of closure spaces, of which topological spaces are a full subcategory, that captures the notion of paths in such a way that in directional graph structures like quasi-discrete closure spaces, paths are directional, and in topological spaces, paths are in one-to-one-correspondence to classical, topological paths, a.k.a. $[0,1]$-morphisms?

  • $\begingroup$ When you map $[0, 1)$ to $0$ and $\{1\}$ to $1$, isn't that a non-trivial path? $\endgroup$
    – user87690
    Dec 22, 2017 at 11:07
  • $\begingroup$ I have to slightly edit my question since I realise that, trying to be concise, I merged two different issues with different types of paths. 1) consider the topological paths; take the quasi-discrete closure space generated by graph a->b. Then you can map [0,1) to a and 1 to b. Great! But you can also map 0 to b and (0,1] to a. This is a continuous function but not a graph path. 2) If you would try to use the natural numbers with closure generated by the successor relationas as path indexes, then "no nontrivial paths exist" e.g in euclidean spaces, due to cardinality. $\endgroup$
    – vincenzoml
    Dec 22, 2017 at 16:57

2 Answers 2


I don't understand precisely what you mean by a 'graph-like path'. You appear to be adding some kind of structure to $[0,1]$ (like an order) to define a 'graph-path' from the interval to $R$, but the ordering of the set is unimportant for questions of continuity between topological and closure spaces, unless that ordering is somehow reflected in the closure structure, as it is for your set $R$.

The most natural definition of a path in a closure space $(X,c)$ is what you've already suggested, simply a map from the topological interval to a closure space, i.e. $\gamma:(I,\tau) \to (X,c)$, where $\tau$ is the ordinary topological closure on the interval $I = [0,1]$. What your example in the comment illustrates is simply (the very desirable property) that $[0,1)$ and $(0,1]$ are homeomorphic via a homeomorphism that extends to the entire interval (as such a homeomorphism has to).

If you want to add some directionality to your paths, then you should do it by modifying the closure structure on $I$. One possibility would be to define a closure structure $u$ on $I$ by:

$u(A) = \{t \in I \mid \exists a \in A : a \leq t\}$

Now the function $f: I \to R$, $f((0,1]) = a$, $f(0)= b$, is no longer continuous as a function from $(I,u) \to (R,c_R)$, but the function $g:I \to R$, $g([0,1)) = a$, $g(1)= b$ is continuous.

However, the identity is not continuous as a map from $(I,u) \to (I,\tau)$, nor from $(I,\tau) \to (I,u)$. Since $(I,u)$ is itself a topological space, and the identity is clearly continuous as a map from $(I,u) \to (I,u)$ or $(I,\tau) \to (I,\tau)$, maps from $(I,u)$ cannot simply be thought of as generalizations of topological paths. Note, too, that $[0,1)$ is no longer homeomorphic to $(0,1]$ as subspaces of $(I,u)$.

If you want a notion of path which is a single map from $[0,1] \to X$, and such that the directionality of $X$ is encoded in the closure structure, then directional paths and classical topological paths have to be different, for exactly the reason which comes up in your example.

Suppose, on the other hand, that you had a topological path $f:([0,1],\tau) \to (X,c)$ such that $f$ was continuous but $g(t) = f(1-t)$ was not. Then there is a set $A \in [0,1]$ such that $g(\tau(A)) \not\subset c(g(A))$. However, this implies that, for the set

$B = \{1-t \mid t \in A\}$,

we have $f(\tau(B)) \not\subset c(f(B))$, and therefore $f$ is not continuous, a contradiction. (Alternately, $g = f \circ \alpha$, where $\alpha:(I,\tau) \to (I,\tau)$ is $\alpha(t) = (1-t)$. Therefore, $g$ must be continuous, a contradiction.)

You could perhaps get around this by considering equivalence classes of continuous directional paths (i.e. from $(I,u) \to (X,c)$, or $I$ with another closure structure) with the same image. You would still have to show that, in the topological case, there is a function in every equivalence class that is also continuous for $(I,\tau) \to (X,c)$, and that, for every topological path $(I,\tau) \to (X,c)$, there is a directional path $(I,u) \to (X,c)$ with the same image. Given the example of the identity $(I,u) \to (I,\tau)$ not being continuous, however, I'm doubtful that this will work.

  • 1
    $\begingroup$ Your $(I,u)$ itself is a topological space, i. e. $u(u(A))=u(A)$ holds for any $A\subseteq I$. $\endgroup$ Mar 17, 2019 at 22:11
  • $\begingroup$ Indeed. I've edited the end of the post to reflect this. Thanks. $\endgroup$
    – APR
    Mar 17, 2019 at 23:32
  • $\begingroup$ Thanks! But I'm not getting your point; you're suggesting that "directional paths" and "classical topological paths" must be two different things, defined differently. Which works, but I do think there must be a way to define a single notion that specialises to topological paths on topological spaces "because topological spaces are not directional", and to directional paths on those closure spaces derived from a relation; and moreover, when the relation in question is symmetric, also those "directional paths" are in practice, not directional (because there is no "direction" in the spaces). $\endgroup$
    – vincenzoml
    Mar 22, 2019 at 15:18
  • $\begingroup$ By a "graph like path" in a relation R I mean a natural-numbers indexed sequence of points p, such that for all i, R(p(i),p(i+1)). $\endgroup$
    – vincenzoml
    Mar 22, 2019 at 15:24
  • $\begingroup$ I've edited the answer to show that topological paths can't also be directional at the same time. $\endgroup$
    – APR
    Mar 23, 2019 at 16:18

I believe my other answer shows that defining a path as a map from the topological closed interval won't work, but now define a path to be a map $\gamma:((0,1],\tau) \to (X,c)$, where, again, $\tau$ is the standard topological closure. Restricted to topological spaces, this definition contains all of the closed paths, although it clearly also contains others, but you gain the directionality you want in your graph-like spaces.

Note that in both of these answers, an asymetry is built into the domain. In the first answer this was done in the closure structure, and here it's in the space itself. This is necessary, since in any possible notion of 'directional path', if, indeed, a path is going to be a map between spaces, the map needs to preserve a structure equivalent to 'direction' from the domain to the range. Therefore, the domain needs to be asymetric, and there are exactly two ways to do this in a closure space: either build the asymetry into the closure structure or build it into the space.


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