# Any collection of subsets of $X$ can serve as a sub-base for a topology

In my lecture notes there is this observation:

"Any collection $S$ whatsoever of subsets of the non-empty set $X$ can serve as sub-basis for a topology on $X$."

So, taking the set $X:=\{1,2,3,4,5\}$ and an arbitrary collection of subsets: $S:=\{ \{1,2\}, \{1,2,3\}, \{3,4\} \}$

Any topology on $X$ will include the element $X$ (by the definition of topology) and by the definition of sub-base, every element of the topology on $X$ is a union of a set of finite intersections of elements of $S$. However, no such union of finite intersections of elements of $S$ will give me $X$. So how can such a topology be constructed?

Thanks Tal

• en.wikipedia.org/wiki/Subbase – joriki Apr 11 '13 at 17:03
• Take the empty intersection $\bigcap_{A\in\emptyset}A$. This is defined as the underlying set. Intuitively, the more sets intersect, the smaller the resulting set becomes. So if you intersect not a single set, then you get just everything. So this definition makes sense. – Stefan Hamcke Apr 11 '13 at 17:34
• See the answers to this question. – Brian M. Scott Apr 11 '13 at 19:20

The intersection of a family of subsets $\{E_j: j\in J\}$ of some set $X$ is defined as $$\bigcap_{j\in J}E_j= \{x\in X: \forall j\in J\ \ x\in E_j \} \tag1$$ If the index set $J$ is empty, the condition $\forall j\in J \dots$ is vacuously true, and (1) simplifies to $$\bigcap_{j\in \varnothing}E_j= \{x\in X\} =X$$