Is it true that the cardinality of the topology generated by a countable basis has at most cardinality $|P(\mathbb{N})|$?

Question

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?

Answer 1

The answer is yes, almost trivially.

Prove the following lemma first:

Suppose that $B$ is a basis for a topology $\tau$. Define for $U\in\tau$ the set $B_U=\{V\in B\mid V\subseteq U\}$, then $U=U'$ if and only if $B_U=B_{U'}$.

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.

Answer 2

The surjection $f: \mathcal{P}(\mathbb{N}) \to \mathcal{T}$ as you described ($I \to \bigcup_{n \in I} B_n$) indeed shows (assuming choice) that $|\mathcal{T}| \le 2^{\aleph_0}$.

You will never prove exact equality, there are spaces with a countable topology, or spaces with a finite topology, even. The inequality is the best you can get.