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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 $\beta$ ?

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$\emptyset$ 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.

(Also note that some people tend to neglect the empty set, because we already know it has to be there, e.g. "cofinite topology" is all the cofinite sets and the empty set, which need not be cofinite. People often omit, or forget, the empty set. But it's fine because we know it has to be there anyway.)

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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
When you say contained, do you mean subset or element? – Asaf Karagila Sep 11 '13 at 6:59
I mean $\varnothing$ is an element of $\beta$. – Matt Elson Sep 11 '13 at 7:08
Matt, with very finite sets like here I doubt it would be omitted. You should contact the author and ask, if you have doubts as to what they meant. Return to the definition of basis as given in the text this was taken from, and see if the reasoning I gave in my answer fits that. – Asaf Karagila Sep 11 '13 at 7:33

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