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Let $S$ be a set with continuously many elements, and $\_<\_:\_$ be a three-placed relation such that for any $x,y,z$ in $S$, $x<y:z$ iff $x$ is closer to $z$ than $y$ is. Also, by definition, for any $x,y$ and a given $z$ in $S$, $\neg x<y:z \wedge \neg y<x:z$ if and only if $x \approx y:z$, or $x$ and $y$ are equally far from $z$. However, there is no distance definable between any two elements in S. S does not form a metric space.

For any given $a$ in $S$, the following holds for any $x,y,z$ in $S$:

(1) $\neg x<x:a$

(2) If $x<y:a$, then $\neg y<x:a$

(3) If $x<y:a$ and $y<z:a$, then $x<z:a$

In this case, what does it mean for a subset of S to be open? Would the following work?

A subset $S^*$ of S is an open set iff $\forall x _{\in S^*}(\forall y_{\in S^*}((x = y) \vee(\exists z(y<z:x))))$.

Asaf Karagila
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Lory
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  • It seems that depending on relation $ \circ<\circ :\circ $, the whole space $S$ may or may not be open due to your definition, which is not good. You would have to include "unboundendness condition" to ensure that $S$ is open: $(\forall x, y \in S)(\exists z \in S)x<z:y$ – Esgeriath Jan 08 '23 at 18:41
  • Do not just ask. Show your attempts to prove that "it works", i.e. that it (=?) satisfies the definition of a topology. – Anne Bauval Jan 08 '23 at 18:50
  • This definition seems kinda strange, when $z$ does not relate to $S^$ at all, perhaps you meant $\ldots(\exists z\in S^)\ldots$? – Esgeriath Jan 08 '23 at 18:53
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    Your three-placed relation is equivalent to: for each $a\in S,$ some total preorder on $S.$ It leaves me perplexed. – Anne Bauval Jan 08 '23 at 19:02
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    I think you probably want some more conditions on your definition. For example, according to the current definition it is possible to have some point $y$ that is closer to $x$ than $x$ is. – N. Virgo Jan 09 '23 at 04:12
  • @Esgeriath Thanks! I want S to be open, so the unboundedness condition you mentioned seems necessary. Also, you're right that z would be in S* as well. – Lory Jan 09 '23 at 18:25
  • @Anne Bauval Can you elaborate on the aspect of the relation that's perplexing? – Lory Jan 09 '23 at 18:27
  • @N.Virgo Thanks, that seems right! – Lory Jan 09 '23 at 18:27
  • I was simply doubtful about the possibility to build something like Frousse did some minutes later. – Anne Bauval Jan 09 '23 at 18:56

1 Answers1

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Edit $1$. This assumes that for any fixed $p$, $\_\approx\_:p$ is an equivalence relation (which makes $\_<\_:p$ a strict total preorder).

Edit $2$. Slightly strengthening the definition so to allow the relation to give e.g. the trivial topologies.

Now, clearly a metric induces a relation as you defined (let's call it a proximity relation). For a metric space $(M,d)$ and for $x,y,z\in M$, let $x<_d y:z$ iff $d(z,x)<d(z,y)$. Let's view this as the prototypical example.

When a proximity relation is given by a metric, you probably want the metric topology to coincide with that of the relation. Here's a natural way of doing so:

Let $(X,<)$ be a set equipped with a proximity relation. For $p,x\in X$, define $B_x(p)=\{y\in X,\ y<x:p\}$ (this is like open balls for a metric space). Also let $p\in X$ be degenerate if $\forall x\in X, x\approx p:p$.

Then a set $S\subset X$ is open iff for every $p\in S$, $x\approx p:p\Rightarrow x\in S$ and when $p$ is non-degenerate, $\exists x\in X, \ x>p:p$ such that $B_x(p)\subset S$.

You can verify that the collection of open sets indeed forms a topology and that it coincides with the metric topology if the proximity relation is induced by a metric.

Frousse
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  • According to the OP's definition we would not necessarily have $p\in B_x(p)$. (I think that's probably a mistake on the OP's part.) – N. Virgo Jan 09 '23 at 04:17
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    @N.Virgo We're only looking at $B_x(p)$ when $p$ is in $S$, so we don't need to test for it. – Frousse Jan 09 '23 at 05:40
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    But it does get ugly if we want some standard non-metric topologies to arise from the relation. I edited accordingly. – Frousse Jan 09 '23 at 05:51
  • Thanks! It seems to me that there is a typo in "a set $S\subset X$ is open iff for every $p\in S$, $x\approx p:p\Rightarrow x\in S$ and when $p$ is non-degenerate, $\exists x>p:p, B_x(p)\subset S$." Is that correct? What do you mean by "$\exists x>p:p$"? Do you mean $\exists x x>p:p$? – Lory Jan 09 '23 at 19:23
  • @Anti-Tachyon I read that bit as "there is an $x$ further than $p$ wrt $p$". But I'll make it slightly less informal. – Frousse Jan 09 '23 at 20:48