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 $R$.

  • $\begingroup$ When you map $[0, 1)$ to $0$ and $\{1\}$ to $1$, isn't that a non-trivial path? $\endgroup$ – user87690 Dec 22 '17 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 '17 at 16:57

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.