# Weird generalization of disconnected topological spaces

I'm learning about connected topological spaces, and i came up with a related question i'm not able to answer. To make things simpler, i'm going to give some definitions. Let $$X$$ be a topological space. For a positive integer $$n \in \mathbb{N}$$, $$n \geq 2$$ we say that $$X$$ is $$n$$-separated if there exist non empty and pairwise disjoint open subsets $$U_1, ..., U_n \subset X$$ such that $$X = U_1 \cup \cdot \cdot \cdot \cup U_n$$, furthermore, we say that $$X$$ is countably separated if there exists a sequence $$\{ U_n \}_{n=1}^\infty$$ of non empty pairwise disjoint open subsets of $$X$$ such that $$X = \cup_{i=1}^\infty U_i$$. It's obvious that if a topological space is countably separated then it is $$n$$-separated for all $$n \geq 2$$. Now i'm trying to understand if the converse is true. The only thing i was able to notice is that if a topological space is $$n$$-separated for all $$n \geq 2$$, then it has infinitely many connected components. Indeed, suppose by contradiction that $$C_1, ..., C_m \subseteq X$$ are all the connected components of $$X$$, for some $$m \in \mathbb{N}$$, $$m \geq 1$$. By definition there are non empty and pairwise disjoint open subsets $$U_1, ..., U_{m+1} \subseteq X$$ such that $$X = \cup_{i=1}^{m+1} U_i$$. Notice that each $$C_i$$ is contained in one of the $$U_j 's$$ (Suppose by contradiction that $$C_i$$ intersects more than one of the $$U_j$$'s. Without loss of generality we can reorder the $$U_j$$'s and suppose that there is an $$1 < r \leq m+1$$ such that $$C_i$$ intersects $$U_1, ..., U_r$$ and doesn't intersect $$U_j$$ for $$j > r$$. This means that $$C_i = C_i \cap (U_1 \cup U_2 \cup \cdot \cdot \cdot \cup U_r) = (C_i \cap U_1) \cup (C_i \cap (U_2 \cup \cdot \cdot \cdot \cup U_r))$$, in contradiction with the fact that $$C_i$$ is connected) and therefore, because $$m + 1 > m$$, this implies that at least one of the $$U_i$$'s is empty. Contradiction.

Nice question - indeed, the converse need not hold in general!

Consider the topology on $$\mathbb{N}$$ (which for me contains $$0$$) generated by:

• all singletons $$\{n\}$$ with $$n>0$$, and

• all cofinite sets.

Basically, a set is open in this topology iff it either doesn't contain $$0$$ or it is cofinite. (Incidentally, this is the one-point compactification of the discrete topology on $$\mathbb{N}_{>0}$$.)

This is clearly $$n$$-disconnected for each $$n$$, but there is no infinite collection of disjoint open sets whose union contains $$0$$ at all.

• wait, isn't the sequence $\{ A_n \}_{n=1}^\infty$ given by $A_n = \{ n \}$ for each $n \in \mathbb{N}$ with $n \geq 1$ an infinite collection of disjoint open sets? Commented Sep 1, 2021 at 15:59
• @RickDoesMath Whoops, missed a clause. Fixed! Commented Sep 1, 2021 at 16:05