Prove that an uncountable space with the countable-closed topology satisfies the countable chain condition.

Prove that an uncountable space with the countable-closed topology satisfies the countable chain condition.

Let $(X,T)$ be an uncountable space with the countable-closed topology.

Then all nonempty open sets are of the form $X \backslash C_i$ where $C_i$ is a countable subset of $X$.

Let $\{X \backslash C_i\}$ be a disjoint family of open subsets.

Then any $(X \backslash C_1) \cap (X \backslash C_2) = \emptyset$ iff $X \backslash (C_1 \cup C_2) = \emptyset$ iff $X = C_1 \cup C_2$.

But the union of two countable sets is never an uncountable set, so how can this satisfy the countable chain condition?

• What is the countable chain condition? Is that the condition that every collection of disjoint nonempty open sets is countable? If your space does not even have two disjoint nonempty open sets, doesn't that imply that it satisfies the countable chain condition? A set with just one element is countable, isn't it? – bof Jun 28 '16 at 23:57
• On the 3rd line of your question, by "all open sets are of the form" I think you meant "all nonempty open sets . . .", right? – bof Jun 28 '16 at 23:57
• So the only disjoint collection of open sets is $X$ itself, which is a family of $1$ set? And yes, nonempty open sets. – Oliver G Jun 29 '16 at 0:21
• For each countable set $C\subset X,$ the one-element set $\{X\setminus C\}$ is a collection of disjoint open sets. If the empty set is allowed to be a member of the collection, then also $\{X\setminus C,\emptyset\}$ and $\{\emptyset\}$ are such collections. But, as you noticed, no disjoint collection of open sets can have more than three members, and no disjoint collection of nonempty open sets can have more than two members. – bof Jun 29 '16 at 0:26

Recall that a topological space $X$ has the countable chain condition if every family of pairwise disjoint open sets is countable (or, equivalently, there are no uncountable families of pairwise disjoint open sets).

A space is called hyperconnected (or irreducible) if every two nonempty open sets meet.

Trivially, any hyperconnected space has the countable chain condition: any family of pairwise disjoint sets has at most two elements, the empty set, and one nonempty open set.

What you have noticed is that the countable closed topology on an uncountable set is hyperconnected (and therefore has the countable chain condition).

To be more general, given a cardinal number $\kappa \geq \aleph_0$, and a set $X$ of cardinality at least $\kappa$, by the co-${<}\kappa$-topology on $X$ we mean the topology where the open sets are $\varnothing$ and every set whose complement has cardinality ${<}\kappa$. (So the countable closed topology on an uncountable set is the co-${<}\aleph_1$ topology, and the co-finite topology on an infinite set is the co-${<}\aleph_0$ topology.) Such spaces are always hyperconnected:

• If $U,V$ are nonempty open sets, then $X \setminus U$ and $X \setminus V$ have cardinality ${<}\kappa$. If $U \cap V = \varnothing$, then $X = X \setminus ( U \cap V ) = ( X \setminus U ) \cup ( X \setminus V )$, but by rules of cardinal arithmetic since $| X \setminus U |, | X \setminus V | < \kappa$, then $| ( X \setminus U ) \cup ( X \setminus V ) | < \kappa$, which contradicts that $|X| \geq \kappa$.

(This is basically your observation cast in a more general setting.)