Can we write the empty set as a union of open intervals? The set $\mathcal{B}=\{(a,b) : a < b, a,b\in\mathbb{R}\}$ is a base for the usual topology. Hence every open set in $\mathbb{R}$ can be written as a union of elements of $\mathcal{B}$. 
What I don't know is how does that definitions work here with the empty set. I don't see the way you can write the empty set as a union of elements of $\mathcal{B}$
 A: Take the union of zero elements of $B$.
EDIT, responding to comments...
So one way to think of an "arbitrary union of elements of $B$" is to make a subset, $A \subseteq B$, then consider
$$  \bigcup_{a \in A} a  \text{.}  $$
That is, $A$ represents an indicator for which elements of $B$ to include, or even $A$ is an index set for the elements of $B$ to union.  If we do this, then we can take $A = \varnothing \subset \mathbb{R}$, producing a union of zero sets.  By definition, an element of the universe is an element of a union if it is an element of one of the sets in the union.  Since there are no sets in the union, there are no elements in the union.
$$  \bigcup_{a \in A} a = \varnothing  \text{.}  $$
(That is, for any element, $x$, of the universe, there is no set in the union that is a witness to the presence of $x$ in the union.)  This is analogous to the convention that the summation of an empty set of terms is zero.  That is, if $A = \varnothing \subset \mathcal{P}(\mathbb{R})$, is an empty collection of subsets of $\mathbb{R}$, 
$$  \sum_{a \in A} a = 0  \text{.}  $$
Referring to Henning Malcolm's "in-joke", there is another convention: the product of an empty set of terms is one.  That is, if $A = \varnothing \subset \mathbb{R}$, 
$$  \prod_{a \in A} a = 1  \text{.}  $$  By definition, an element of the universe is an element of an intersection if it is an element of every member of the intersection.  For every $x$ in the universe, $x$ is an element of every set in an empty collection of sets.  (That is, there is no witness to the exclusion of $x$ from the intersection.)  Thus, the intersection of an empty collection of sets is the universe.  So if $A = \varnothing\subset \mathcal{P}(\mathbb{R})$ is an empty collection of subsets of $\mathbb{R}$,
$$  \bigcap_{a \in A} a = \mathbb{R}  \text{.}  $$
(In case you're wondering about the "$\subset \mathbb{R}$" in the "$A = \varnothing \subset \mathbb{R}$"s, it's to specify what the universe is.)
A: Given any set $\mathcal F$ of intervals we can take their union:
$$ \bigcup_{I\in\mathcal F} I = \{ x \mid \exists I\in\mathcal F : x\in I \}$$
When $\mathcal F=\varnothing$ this produces the empty set -- so the empty set is the union of zero of the open intervals.
A: Definition: $\cup V=\{x: \exists u\in V\;(x\in u)\}.$
A base $B$ for a topology $T$ on a set $X$ is a subset of $T$ such that every $t\in T$ is equal to $\cup V$ for some $V\subset B.$ In  the formaal language of set theory,  $(B\subset T)\land (\;\{\cup V:V\subset B\}=T\;).$
We have $\phi \subset B$ and $\cup \phi =\phi.$
The problem is a matter of accurate terminology. Rather than saying any $t\in T$ is a union of members of $B$, we should say $t$ is the union of a subset of $B.$ That subset could be empty. 
BTW some authors call it a basis instead of a base. I prefer base. "Basis"  has other uses.
