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:
Copyright © 2018, Konrad Burnik