I'm curious about what the cardinalities of second countable spaces can be at most. I have an idea as to how to show that the topology generated by a countable basis (which, for basis $B = \{B_i\}_{i=1}^\infty$, we will denote $\tau_B$) is at most cardinality $|P(\mathbb{N})|$, but I can't seem to find the right steps to do so.
Note that I am assuming the axiom of choice here.
The "intuitive idea" in my head is that you could attempt to create a correspondence between the elements of $\tau_B$, which are of the form $U = \bigcup_{i\in I\subseteq\mathbb{N}} B_i$, to the set $I\in\mathcal{P}(\mathbb{N})$ indexing the sets in $B$ used to construct $U$. The problem is that this map, in general, isn't well defined (since, for example, $B(0, 1) = B(0, 1)\cup B(0, 1/2)$ in the Euclidean topology). I suppose if we could define it properly it would be an injection from $\tau_B$ to $\mathcal{P}(\mathbb{N})$, but then we have the further problem of constructing an injection from $\mathcal{P}(\mathbb{N})$ to $\tau_B$. The simplest idea, taking $I\in\mathcal{P}(\mathbb{N})$ to $\bigcup_{i\in I}B_i$, is clearly not an injection (though it is a surjection).
If I could find an injection from $\mathcal{P}(\mathbb{N})$ to $\tau_B$, and properly formulate the injection from $\tau_B$ to $\mathcal{P}(\mathbb{N})$, then Cantor-Schroeder-Bernstein would give us the result. Alternatively, if I could find a surjection from $\tau_B$ to $\mathcal{P}(\mathbb{N})$,then we could use the dual Cantor-Schroeder-Bernstein theorem to get the same result. However, my attempts at either have so far been fruitless.
Can anyone share any insights into this?
Best Answer
The answer is yes, almost trivially.
Prove the following lemma first:
Therefore you immediately get an injection from $\tau$ into $\mathcal P(B)$. And therefore if $|B|=|\Bbb N|$, you get that $|\tau|\leq|\mathcal P(\Bbb N)|$, as you wanted. Because you wanted at most, rather than "exactly".
The key point is to remember that there is often a lot of redundancy when you take a basis for a topology, and this is a feature rather than a bug. You want basis elements to become smaller and smaller, rather than having $U\mapsto B_U$ a bijection.
Here is a nice example, by the way, for "less". Take $\Bbb N$ with the topology that an open set is an initial segment of the order $\leq$. There are exactly $\aleph_0$ of those, and you need all of them (except $\varnothing$ and $\Bbb N$ themselves) to create a basis, too.
Or, you know, any topological space with finitely many open sets.