In Part 1 I posed the following question:
"Does there exist a metric space such that it has an infinite metric base with the following properties:
- is not dense;
- for all finite metric bases of we have ?"
Let be defined as
for all . Then is a metric on , denoted by and is a metric space.
After a brief chat with professor Zvonko Iljazović, he proposed the metric space and the integer lattice as the candidate for such a set. Here I will explain why this set is a good candidate.
Let . Then is a countable set, which is not dense in which can be seen by taking for example and then . By [1] the space does not have a finite metric base, therefore itself can not contain a finite metric base. What is left to prove is that is a metric base for . The rest of this blog post is concerned with proving this fact.
First, note that the following result was mentioned briefly in our CCA 2018 presentation.
Proposition 1. Let such that either , or , Let . Then is a metric base for .
Interestingly enough, the article [1] already gives a complete characterisation of finite metric bases for squares in the digital plane for a couple of metrics including . My conjecture is that this also gives a complete characterisation of finite metric bases for however, this is a topic for another blog post. For now, it seems that the result stated in Proposition 1 coincides with some parts of results from [1] and as such will be sufficient for our needs.
We will also need the following claim, which is straightforward to establish.
Proposition 2. Let and be metric spaces. If is a metric base for and is an onto mapping such that there exists with the property
for all , then is a metric base for .
By using Proposition 2 we can now easily find a metric base for an arbitrary square where by using the known result of Proposition 1 and finding a suitable map .
Proposition 3. Let be a metric base for . Let where be defined as
for all . Then is a metric base for .
Note that by setting it is easy to verify that defined in Proposition 3 satisfies the property of from Proposition 2.
Finally, we now have everything we need to prove that is indeed a metric base for .
Proposition 4. is a metric base for .
Proof.
Let . Let and assume that
for all .
Note that we can always choose such that and . Namely, we can take for example such that and set . Now set and .
From Proposition 1, it follows that any three corners of the square are a metric base for . By Proposition 3 there is a map such that is a metric base for . Since , then for all implies .
Therefore, is a metric base for .
Q.E.D.
References
- Robert A. Melter and Ioan Tomescu "Metric Bases in Digital Geometry", Computer Vision, Graphics, and Image Processing Volume 25, Issue 1, January 1984, Pages 113-121 DOI: https://doi.org/10.1016/0734-189X(84)90051-3
- Konrad Burnik, Zvonko Iljazović, "Effective compactness and uniqueness of maximal computability structures", CCA 2018 presentation slides:
Effective_compactness_and_uniqueness_of
Copyright © 2018, Konrad Burnik