# Base and empty set of a topology

Given space $X = \{a, b, c\}$, $\beta$ is a basis for a topology $\tau$ on X.
$\tau = \{ \varnothing, X, \{a\}, \{b\}, \{a,b\}\}$, $\beta = \{\{a\}, \{b\}, X\}$.

$\beta$ can't union its elements to get empty set $\varnothing$ contained in $\tau$ , but the definition of basis require that every open set can be expressed as a union of basis elements.

So why $\varnothing$ is not an element of $\beta$ ?

-
$\emptyset$ is the empty union. – William Sep 11 '13 at 6:09
as you said, topology generated by a basis is (equivalently) collection of "all" unions of elements of basis $\beta$... in particular empty union gives me $\emptyset$.. Is that alright?? – Praphulla Koushik Sep 11 '13 at 6:17

The definitions may vary a bit from place to place, but note that $\varnothing$ is the union of no element from the basis. That is to say, there is some $A\subseteq\beta$ such that $\bigcup A=\varnothing$.
If you require that "for all $U\in\tau$ there is $A\subseteq\beta$ such that $U=\bigcup A$", then this fine.
if $\varnothing$ is contained in $\beta$, is it also okay? because we already know it has to be there and just express it in an implicit way. – Matt Elson Sep 11 '13 at 6:53
I mean $\varnothing$ is an element of $\beta$. – Matt Elson Sep 11 '13 at 7:08