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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Sep 1, 2021 at 15:59
  • $\begingroup$ @RickDoesMath Whoops, missed a clause. Fixed! $\endgroup$ Commented Sep 1, 2021 at 16:05

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