I have just had my first week of topology, and I have a question that is rather basic.
Why must the empty set be an element of any given topology?
(For reference, the definition of a topology T I am working with, for a set X:
1. X and the empty set must be in T.
2. the union of elements of any subset in T, is also in T.
3. the intersection of elements of any finite subcollection in T, is also in T.)
Best Answer
It's more convenient this way. $\def\less{\smallsetminus}$
Usually in math when you have definitions that involve the empty set, the reason they do is that it's more elegant, convenient, economical to phrase things this way.
Suppose you didn't want to deal with the empty set. You could define topology to be a collection of nonempty subsets of X that satisfy a few axioms. But then your axioms would have to be more complicated and include special cases.
You'd have to say that only a nonempty finite intersection of open sets is open. You could modify the 3rd axiom to say that. But the reason you need that axiom in the first place is that very frequently in proofs you take a few open sets, and rely on the fact that their intersection is still open (and therefore has some qualities useful to you). In the new version, every time you need to intersect two open sets in a proof, you'd have to list two cases: a) if they're disjoint, then their intersection is empty and therefore... b) if they're not disjoint, then their intersection is open and therefore... Your proofs would be longer and more tedious.
Similarly, normally you define a subset $Y$ of $X$ to be closed if $X \less Y$ is open. Then under the usual definition the whole space $X$ is a closed set. There're many reasons why you want that to be the case and why that corresponds to the intuition of "closed set". But if you don't consider the empty set to be open by definition, you don't get $X$ to be closed. Again, you can patch it up by modifying the definition of closed set: "A subset $Y$ of $X$ is closed if $Y=X$ or $Y$ is the complement of an open set". This will work. But, again, your proofs will become more complicated: when you have a closed set $Y$ about which you know little, you can't just assume anymore that $X \less Y$ is open and work with that, now you have to consider two cases.
These are just two examples, but there're more. Basically, it turns out that while making the empty set open by definition may not look a priori very natural to you, it makes all kinds of objects and properties "click together" more naturally and economically than otherwise.