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https://en.wikipedia.org/wiki/Bounded_set defines boundedness of a set as

A subset S of a metric space (M, d) is bounded if there exists r > 0 such that for all s and t in S, we have d(s, t) < r.

after saying that

The word 'bounded' makes no sense in a general topological space without a corresponding metric.

However, I do not see why the triangular inequality is required for $d$. Can I define boundedness the same way if $d$ is a premetric satisfiying

  • d(x, y) ≥ 0
  • d(x, x) = 0
  • d(x, y) = d(y, x)

?

Make42
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  • take promenade https://arxiv.org/pdf/1610.07594.pdf – janmarqz Aug 29 '20 at 14:12
  • @janmarqz: Sorry, I am not familiar with the expression "take promenade". Does it mean "take a look"? Do you mean that the PDF contains examples of bounded sets that are subsets of premetric spaces? – Make42 Aug 29 '20 at 14:14
  • @janmarqz: Also, besides removing the triangular equation, a premetric also removes the need to have "d(x,y) = 0 if and only if x = y" and replaces it with "d(x,x)=0". The "distance space" in the PDF seems to only remove the triangular equation. – Make42 Aug 29 '20 at 14:17
  • yes... there, in the article that space are dubbed distance spaces – janmarqz Aug 29 '20 at 14:17
  • well, it is one of many "variations"... let's keep on reading – janmarqz Aug 29 '20 at 14:19
  • @janmarqz: mmh... expect for Definition 3.1 and 3.3 I could not find anything more relating to my question. Did I miss something? – Make42 Aug 29 '20 at 14:24
  • I dunno, these are my 1st reading on the subject – janmarqz Aug 29 '20 at 14:26

1 Answers1

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No, it does not really make sense for a premetric, at least not with the usual definition of a boundedness notion, in which you want the union of two bounded sets to be bounded.

It does, however, make sense once you do have triangle inequality, e.g. pseudometrics. It also makes sense, for example, once you fix an equivalence class of pseudometrics up to quasi-isometry.

All of this does not invalidate the claim that it does not make sense to talk about bounded sets in a topological space: the point is not that there cannot be a notion of boundedness in a space which has no metric (or even is not metrisable at all), but rather that there is no unique way to assign a notion of boundedness to a pure topological space (i.e. a topological space with no extra structure).

For example, if you look at $\mathbf R$, then you have the standard notion of boundedness that comes from the Euclidean metric, but you also choose a metric which is bounded, making all sets bounded, and the topology won't tell you which ones to choose.

tomasz
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  • Can you show, i.e. add a proof, that a premetric space is not "stable under finite unions"? In your third paragraph you discuss topological spaces, but a premetric space is not quite that pure. We do have the premetric afterall. An example of such a premetric space would be $(S, d)$, where $S$ is the set of all possible sequences of length $N\in\mathbb N$ and $d$ is the dynamic time warping distance, which does not fulfill the triangular equation. – Make42 Sep 05 '20 at 10:34
  • @Make42: The "bounded sets" in a premetric space are not necessarily closed under union. For example, if you consider $\mathbf N$ with premetric given by $d(a,b)=0$ when $a,b>0$ and $d(0,a)=d(a,0)=\lvert a\rvert$ for arbitrary $a$, you have a counterexample. Yes, a premetric space is not a pure topological space, but the sentence you cite is about pure topological spaces. – tomasz Sep 05 '20 at 12:21
  • You are right, the quote was for general topological space - I missed that somehow. Can you dumb down with a practical example why your examples is not closed under union - I am kind of at a loss. – Make42 Sep 05 '20 at 12:49
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    $\mathbf N$ is obviously not bounded, but ${0}$ and $\mathbf N\setminus {0}$ both have diameter $0$. – tomasz Sep 05 '20 at 14:48
  • I had another look at the topic. I am actual only interested in a single subset and not a space like the Bornological space, so there would not be a second set available for a union. Also, the "inclusion" condition is not necessary. Is this now well defined for a subset in a premetric space? – Make42 Oct 17 '20 at 16:21