Let $B$ be a binary image, i.e. a matrix of binary values (0 or 1). Let $R$ be a simply connected region in the image, i.e. $R$ consists of a set of 1s (the background pixels are all 0s) and there are no holes in $R$ (a hole consists of 0 pixels surrounded by 1s).
The 8-neighborhood of a point $N_{8}(x)$ is defined as the set of all points within unit distance (horizontal/vertical/diagonal) of $x$. (the 8-neighborhood of a point is the set of all points that a king could move to from that point in a game of chess)
The 4-neighborhood of a point $N_{4}(x)$ is defined similarly, but the diagonals are now excluded (so only up, down, left and right).
The 4-perimeter of $R$ is defined as: $P_{4} = \{x \in R| N_{4}(x)- R \neq \varnothing \}$
The 8-perimeter of $R$ is defined as: $P_{8} = \{x \in R| N_{8}(x)- R \neq \varnothing \}$
A set $R$ is 8-connected if for any pair of points $x,y \in R$ there exists a path $x, x_{2}, x_{3, }... x_{n},y \in R$ such that $x_{i} \in N_{8}(x_{i-1})$ for all $i$ and $y \in N_{8}(x_{n})$. A similar definition for 4-connected follows.
How can I prove that the 4-perimeter of a simply connected region is 8-connected and that the 8-perimeter is 4-connected?