Are all sets 'regions'?

Question

Specifically, we define the following terms :

• Neighborhood : For any point $z_0 \in \mathbb{C}$, its $\epsilon$-neighborhood, $N(z_0, \epsilon) = \{ z \in \mathbb{C} \mid \vert z - z_0 \vert \lt \epsilon \}$.

• Interior Point : For any subset $S$ of $\mathbb{C}$, a point $z_0 \in S$ is said to be an interior point of $S$, if there exists a neighborhood $N(z_0, \epsilon)$ such that $z \in N(z_0, \epsilon) \implies z \in S$.

• Boundary Point : For a subset $S$ of $\mathbb{C}$, a point $z_0 \in \mathbb{C}$ is said to be a boundary point of $S$ if every neighborhood $N(z_0, \epsilon)$ of $z_0$ contains points from both $S$ and the complement $S^C$ of $S$.

• Open set : A set $S \subseteq \mathbb{C}$, is said to be an open set if all points of $S$ are its interior points, ie. if $\forall z_0 \in S, \exists$, a neighborhood $N(z_0, \epsilon)$ such that $\exists z_1 \in N(z_0, \epsilon), z_1 \in S$ and $\exists z_2 \in N(z_0, \epsilon), z_2 \in S^C$.

• Region : A region is an open set along with some, none or all of its boundary points.

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This question asks weather all sets can be decomposed by removing their boundary points to give an open set and some, none or all boundary points (ie. a region). The accepted answer, argues that a non-empty set lacking any interior points cannot be decomposed in this way and is therfore not a region. In particular it cites $\{ z \in \mathbb{C} \mid Re(z) = Im(z), Re(z) >0\}$ or $\{(x,y) \mid x=y, x>0 \}$ as an example of such a set.

Here, we see that the empty set is vacuously open.

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I reasoned as follows -

1. If a set $S \in \mathbb{C}$ contains any internal points, we can decompose it as the open set of those internal points, plus the boundary points.

2. If it contains no boundary points, than the empty set fulfills the role of the open set. The matter now hinges on the question of weather the empty set can be considered to have any boundary points.

3. Any neighborhood shall necessarily have points from the universal set, $\mathbb{C}$. The empty set contains no points ;), can the neighborhood be considered to have points from it?

I am tempted to conclude that since there are no points to be contained, the neighborhood automatically includes the 'nothingness' from the empty set, and all sets are therefore regions.

On the other hand, it also seems correct to think that there is no point from $\{ \}$ in the neighborhood because there are simply no points to be had in the first place. This is bolstered by the fact that we do not consider 'empty neighborhoods' in our arguments, though it may be argued that this condition is separate from the issue and is an explicitly excluded factor.

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My question is -

Is it correct to say that the all points are boundary points of the empty set and all sets are regions? Can the empty set be considered to have boundary points? Why? What are the criteria for deciding such points?

Answer 1

The empty set $\emptyset$ does not have any boundary points. In order that $z_0 \in \mathbb C$ is a boundary point of $\emptyset$ it must have the property that every neighborhood $N(z_0, \epsilon)$ of $z_0$ contains points from both $\emptyset$ and the complement $\emptyset^C$ of $\emptyset$. But $\emptyset$ does not contain any point.

With your definition of a region in fact all sets are regions. However, it seems to me that this definiton is very unusual. For the standard definition and a little discussion of this concept see https://en.wikipedia.org/wiki/Region_(mathematics).