Why axiom 3) of topology is redundant? I'm learning topology and I'm reading Dugundji's General Topology book. He gives  the definition of axioms of a topology $\tau$ on $X$:


*

*1) & 2) (a topology is closed under arbitrary union and finite intersection )

*3) $\emptyset$ and $X$ belong to topology $\tau$.


then he says: "*Observe that since the union (resp. intersection) of an empty family of sets in X is $\emptyset$ (resp. X) axiom 3 is actually redundant. *"
My question are, 
why he consider a empty family of sets of $X$ in the observation? 
whitout axiom 3, how do we know  that in fact the empty set belongs to the topology $\tau$?
I really don't understand his observation. Any help would be appreciated.
 A: Property 1 says an arbitrary union of open sets is open. An arbitrary union of sets $$\bigcup_{A\in \mathcal{A}} A$$ is the union of all the sets in some collection $\mathcal{A}$ of sets. In particular, you could take $\mathcal{A}=\emptyset,$ the empty collection of sets. The union over the empty collection is the empty set (what else?). Thus property 1 implies that the empty set is open.
Likewise the intersection over the empty collection is naturally defined as $X.$ This makes sense since intersecting with more and more sets restrict the elements more and more. It makes sense to 'start out' with all the elements, so the intersection of the empty collection of sets is $X.$ Also, it is a finite intersection (since the empty collection is finite). Thus, property 2 implies that $X$ is open.
EDIT:
As commenters have emphasized, at the end of the day, assigning a value to an intersection of an empty collection of sets is a convention. It only makes sense (to my knowledge) if you have a notion of a total space $X$ that every set under consideration is a subset of, as is the case in topology. In this scenario, the notion is well-defined and I tried to make it intuitive in my answer.
Formally, in the scenario with a total space $X,$ the definition of the union and interection are given by $$\bigcup_{A\in \mathcal{A}} A = \{x\in X\mid \exists A\in\mathcal{A}\mbox{ such that } x\in A \}\\\bigcap_{A\in \mathcal{A}} A = \{x\in X\mid \forall A\in\mathcal{A},\; x\in A \}.$$ Applying the definition for the intersecton in the case $\mathcal{A}=\emptyset$ gives the result $X$ as indicated when you apply the notion of vacuous truth to the quantified statement  $\forall A\in \emptyset,\ldots$, in which case it is always true regardless of what $\ldots$ is.
So while it's true that the 3rd property is redundant, this fact is somewhat formal and not all that interesting. Thus why pretty much everyone retains the 3rd property for clarity.
