Digital topology

Remarks

Example: $S = \set{(0, 0), (1, 1)}$ is $8$-connected, since the two points are $8$-adjacent. Non-example: the same $S$ is not $4$-connected, because $\abs{0 - 1} + \abs{0 - 1} = 2$ and $S$ has no other points. Let $S = \set{(1, 0), (0, 1), (-1, 0), (0, -1)}$. With $8$-adjacency for $\Z^2 \setminus S$, the point $(0, 0)$ is $8$-adjacent to $(1, 1) \in \Z^2 \setminus S$, so the complement is $8$-connected and the "curve" $S$ encloses nothing. With $4$-adjacency for the complement, $(0, 0)$ is a $4$-component by itself, so $S$ separates it from the outside. For this reason one pairs $8$-adjacency for a set with $4$-adjacency for its complement, or the reverse. The Khalimsky topology on $\Z$ is a genuine topology: the sets $\set{2n + 1}$ and $\set{2n - 1, 2n, 2n + 1}$ for $n \in \Z$ form a base, and $\Z$ is connected in it.