# Does a family of non-void sets contain void set?

I'm confused with the definition of topology and topological space. Let me state the two definitions:

1. Given a set $X$. Then $\tau$ is a topology of $X$ if it satisfies 3 conditions: $X$ and $\varnothing$ belong to $\tau$; finite intersections of members of $\tau$ belong to $\tau$; arbitrary unions of members of $\tau$ belong to $\tau$. (this definition is based on Wikipedia)

2. Given a family of sets $\tau$ which satisfies 2 conditions: the intersection of any two members of $\tau$ is a member of $\tau$ and the union of each subfamily of $\tau$ is a member of $\tau$. Then $\tau$ is the topology of the set $X=\bigcup \left \{ U:U\in \tau \right \}$. (this definition is based on the book General Topology of John L. Kelley)

Confusion comes from the second definition. If $\tau$ is a family of non-void sets, then $\varnothing$ does not belong to $\tau$. Therefore we cannot ensure that $\varnothing$ is a member of $\tau$. However, Kelley also wrote in his book that $\varnothing$ is always a member of a topology because it is the union of the members of the void family. I do not understand this point, so it leads to my question in the title.

Thanks so much.

• I don't see anywhere in (2) where you require the sets to be non-empty. – Asaf Karagila Aug 11 '15 at 11:20
• @AsafKaragila because void set belongs to every topology, as I write below (2). – Tien Kha Pham Aug 11 '15 at 11:22
• But I don't see where in (2) it is said that $\varnothing\notin\tau$. You just conclude that later on yourself that $\tau$ is a collection of non-empty sets. And I'm trying to understand why you do that, so we can clarify the confusion better. – Asaf Karagila Aug 11 '15 at 11:23
• Based on (2), an arbitrary family of sets which satisfies the two conditions is a topology for some space. So I consider the family $\tau$ such that $\varnothing \not \in \tau$. For example, $\tau$ contains $\left\{1\right\}$ and $\left\{1,2\right\}$. Follow (2), $\tau$ is a topology for $X=\left\{1,2\right\}$, but it actually is not. – Tien Kha Pham Aug 11 '15 at 11:30
• So that's the thing, that it's not closed under arbitrary unions, since it's not closed under the empty union; as Mark's and my answers explain. – Asaf Karagila Aug 11 '15 at 11:35

Note that included in the union of every subfamily of $\tau$ is the union of the empty (void) subfamily. The union of no sets contains no elements and is therefore the empty set.
This is included in the definition by the expression "the union of every subfamily". And because every family has an empty subfamily you can always take the union of the empty subfamily - and the definition required this (the empty set) to be included in $\tau$.
If $\tau$ is a family of sets such that $\varnothing\notin\tau$, then $\tau$ is not closed under arbitrary unions.
Recall that being closed under arbitrary unions means that if $\cal U\subseteq\tau$, then $\bigcup\cal U\in\tau$.
Certainly it is true that $\varnothing\subseteq\tau$, and $\bigcup\varnothing=\varnothing$. So if $\varnothing\notin\tau$, it is impossible that $\tau$ is closed under unions.