# Topology with only countable unions

Is there a special name for a set family $F \subseteq 2^X$ which has the axioms:

1. $\emptyset \in F$ and $X \in F$.
2. If $S, S' \in F$ then $S \cap S' \in F$.
3. (Countable Union). If $f:\mathbb{N} \to F$ is a sequence, then $(\bigcup_{i} f(i)) \in F$

I know that if $T = (F, X)$ is a topology then this holds. Is there a set family where these weaker axioms hold, but an important theorem in topology fails?

I am trying to understand why the field of Topology require arbitrary unions.

• – Mr. Chip Oct 22 '17 at 14:40
• The requirement for arbitrary unions is not because one thinks of being closed under the union operation, but because one thinks of analyzing an open set in terms of 'smaller' open sets. There are other, equivalent formulations of topology that make no reference at all to any infinitary construction, such as giving axioms that the closure operation must satisfy; e.g. as described in the middle section of this answer. – Hurkyl Oct 22 '17 at 14:52
• An example of such a space would be if we let $X$ be any uncountable set and $F$ consist of $X$ together with all countable subsets of $X$. Remains the question why we do not want this to be a topological space ... – Hagen von Eitzen Oct 22 '17 at 15:07
• Countable topology. – William Elliot Oct 22 '17 at 19:07
• If you think in terms of "neighborhoods" around points instead, then the arbitrary union axiom becomes almost sine qua non (at least from my point of view). Other axioms can also be understood in this interpretation too, except perhaps the empty set axiom. Clearly the universal set $X$ is a neighborhood of every point in $X$. Infinite intersection is not permitted because infinite intersection can easily lead to the singleton set becoming a neighborhood, which trivializes the whole notion. You may choose to think of the empty set as a neighborhood around nothing. – Project Book Oct 22 '17 at 21:38