Can we either add the empty set $\varnothing$arbitrarily to a topology, or is $\varnothing$ contained in any collection $B$ of subsets of a set $S$? In my reading material (Appendix A of Introduction to Manifolds, Loring, pp. 322) the “only if” part for the claim

A collection $\mathcal{B}$ of subsets of a set $S$ is a basis for some topology $\mathcal{T}$ on $S$ if and only if i.) $S$ is the union of all the sets in $\mathcal{B}$, and ii.) given any two sets $B_1, B_2 \in \mathcal{B}$ and a point $p \in p \in B_1 \cap B_2$, there is a set $B \in \mathcal{B}$ such that $p \in B \subset B_1 \cap B_2$.

begins by defining $\mathcal{T}$ to be the set of all sets that are unions of the sets in $\mathcal{B}$. Then the author states that:

Then the empty set $\varnothing$ and the set $S$ are in $\mathcal{T}$ and $\mathcal{T}$ is clearly closed under arbitrary union.

By the property i.) $S$ is evidently in $\mathcal{T}$. But what about $\varnothing$? Is it tacitly assumed that $\varnothing$ belongs to all collections of elements, or can we just insert the empty set to $\mathcal{T}$ as we please?
 A: The correct phrasing would be that $\mathcal T$ is the family of unions of subfamilies of $\mathcal B$. You can check that $\varnothing$ is the union of the empty family.
A: Don't confuse the basis $\mathcal{B}$ with the topology $\mathcal{T}$.  Neither $S$ nor $\emptyset$ need to be elements of $\mathcal{B}$ but both need to be elements of $\mathcal{T}$.
As guest points out $\mathcal{T}$ it the set of all unions of elements of $B$.  As $S$ is the union of all elements of $\mathcal B$ we must have $S\in \mathcal T$. But all other partial unions must be included... and that includes the empty union (which is always the empty set).
[This answer isn't meant to supercede guests answer which should be the accepted answer, but to offer more clarity.]
An example of a basis for a topology for the set $S= (0,3)$ and $\mathcal B= \{(0,2),(1,3),(1,2)\}$.  That satisfies the definition of basis.  There are $3$ elements so there are $2^3$ unions.  So the topology $\mathcal T$ will have $8$ (not nesc distinct) elements of all possible unions.  They include the empty union $\emptyset$ and the full union $(0,2)\cup (1,3)\cup(1,2)=S$.  As well they include $(0,2),(1,3),(1,2), (0,2)\cup(1,3)=S; $ not a distinct element $; (0,2)\cup (1,2)=(1,2);$ and $ (1,2)\cup (2,3)$.
