Second countable means the topology on the space has a countable basis (you don't say "second countable basis"). In the discrete topology, the set of singleton sets is the smallest basis.
I have this problem:
S is a set. P is the discrete topology on S. Prove that (S,P) is second countable iff S is a countable set.
Can somebody expand on this? (I assume it's basically what I'm trying to prove):
"Every discrete space is first-countable; it is moreover second-countable if and only if it is countable."
Is this way of thinking correct (I'm an undergrad):
A topology is second countable if it has a second countable base.
A base is second countable if it contains a countable number of open sets.
The discrete topology on S is the power set of S, where, by definition, every subset of S is open.
Am I on the right track? Do I just say something along the lines of S being a set of open balls and show that it's countable?
Suppose that is discrete topological space. i.e. .
Then for each the set is an open set.
Therefore, at each point there is a local basis at the point, first countable.
Now if the discrete space were second countable then there could only be countably many singleton sets. So the space itself is countable.
Is this excessive?
is a discrete topological space if every subset of is open. In particular, for each we have is an open set.
is first-countable, if for any , there is a countable sequence of open sets containing , such that for any neighborhood of there exists a member of this sequence, such that is contained in .
Given a neighborhood of , we have that . But is an open set. So, we can choose the sequence to consist of just one set . For every neighborhood of we then have an open set contained in this neighborhood.
is second-countable, if its topology has a countable base. More precisely, there's a countable collection such that any open set can be written as a union of some (finite of infinite) number of elements of .
Suppose is discrete and countable. Then, since every one-point set is open, any can be written as a union of , where .
Suppose, is second-countable and discrete. Let be a countable collection of open sets with the property that every open subset on can be written as a union of some members of .
Assume that is uncountable and get a contradiction:
Let . Since is an open set, can be written as a union of some elements of . But the only way to write a one-point set as a union of two sets is to say Thus, the set belongs to . Since this should be true for all , and is uncountable by our assumption, we conclude that contains uncountably many open sets . Contradiction. We conclude that must be countable.