# Using Connectedness to Define Topology

I have had this idea to define some kind of “topology” on the integers for graphical purposes. However, i do not know if this is approchable from the standard topological framework.

The idea is to define “connectedness” by stating what subsets of the integers are connected. Let $C$ be a collection of subsets in the integers that are stated to be connected. For every integer $i$ there exist a connected subset of the integers, and that is $\{i-1,i,i+1\}$

Is $C$ together with the integers is a topology? I am planing to use this concept for the topological study of computer graphics or pixelated visuals. I guess a better question would be Can declaring what subsets are connected be enough to define a unique topology?

• Do you want $C$ to be the only connected subsets of $\mathbb{Z}$? It means that if any set is not included in the collection $C$, it is not connected. – ChoF Apr 19 '18 at 22:36
• @ChoF yes. The question gives a construction as to specifically which sets are connected. – user512716 Apr 19 '18 at 22:53

This is possible to do if you give up the requirement that your topology is T$_1$, and it has already been done, with quite a few papers. Google khalimsky line or digital topology, I will include some reference and comments a bit later.

The general idea is that you declare each odd integer to be an open set (as in the usual topology) but the even integers are not open (though as usual each closed). Each even integer $2n$ has a minimal neighborhood consisting of itself together with the two neighboring odd integers, so called Khalimsky line. This makes the set of all integers connected, and so called COTS (connected ordered topological space) have been studied. The square of the Khalimsky line is the Khalimsky plane, and there is even a Jordan curve theorem for the Khalimsky plane.

Here is just one (early) paper in this area (you could find many more, for that matter, not just google, but a search on MSE website returns many results for digital topology).

Kong, T. Yung; Kopperman, Ralph; Meyer, Paul R.
A topological approach to digital topology.
Amer. Math. Monthly 98 (1991), no. 10, 901–917. It can be downloaded at http://www.jstor.org/stable/2324147

Also, Digital Topology, Azriel Rosenfeld
The American Mathematical Monthly
vol. 86, No. 8 (Oct., 1979), pp. 621-630 https://www.jstor.org/stable/2321290

Ralph Kopperman,
The Khalimsky Line as a Foundation for Digital Topology

Erik Melin, Connectedness and continuity in digital spaces with the Khalimsky topology
https://pdfs.semanticscholar.org/4e0f/3df375796f6c4dd7ab02980fbff402feef35.pdf http://uu.diva-portal.org/smash/get/diva2:306620/FULLTEXT01.pdf

• Note that under this topology every interval of integers $\{a, a+1, a+2, \ldots, a+n\}$ is connected, including $\{i, i+1\}$. This may not necessarily have been the asker's intention, since they specifically named intervals of three adjacent integers. – Dustan Levenstein Apr 19 '18 at 22:58

No. Consider $X=\{a,b\}$ with the topology $\{\emptyset, a, X\}$. All nontrivial sets are connected, but we can define also the discrete topology on $X$.

• @Mirko There was nothing about separation axioms, but I understand your point. However in the present version of the question the answer is negative. – Przemysław Scherwentke Apr 19 '18 at 22:40
• The discrete topology doesn't have the same connectedness structure as the one you gave; you could take the complement topology as an example. A more dramatic example of two different topologies on the same set with the same connectedness structure would be $\mathbb Q$ with the induced vs discrete topology. – Dustan Levenstein Apr 19 '18 at 22:47
• Also, I'm not too familiar with the Khalimsky plane, but I think Mirko is addressing the first question, "Is C together with the integers is a topology?", while you are addressing the second question, "Can declaring what subsets are connected be enough to define a unique topology?" – Dustan Levenstein Apr 19 '18 at 22:50
• @DustanLevenstein You are perfectly right. – Przemysław Scherwentke Apr 19 '18 at 22:52

There is no topology on $\mathbb{Z}$ whose connected sets are precisely $C \cup \{\mathbb{Z}\}$.

To see this, suppose we had such a topology, and consider the set $S = \{1,2,3,4,5,6\}$. This set is not connected since it's not in $C$, thus there are two disjoint open sets $A$ and $B$ such that $(A \cap S) \cup (B \cap S) = S$. One of these sets must contain all of $\{1,2,3\}$, otherwise $\{1,2,3\}$ would not be connected. Similarly, the other set must contain all of $\{4,5,6\}$, which means that without loss of generality $A\cap S = \{1,2,3\}$ and $B\cap S = \{4,5,6\}$. But this means that $A$ and $B$ are disjoint open sets that cover $\{2,3,4\}$, contradicting the fact that $\{2,3,4\}$ is connected.

• Outline of another contradiction: $\{ 1, 2, 3 \}$ and $\{ 3, 4, 5 \}$ would both be connected sets and they have nonempty intersection; therefore, their union $\{ 1, 2, 3, 4, 5 \}$ would also have to be connected. – Daniel Schepler Apr 19 '18 at 23:01