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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?

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  • $\begingroup$ Note: $\aleph_1$ is not (at least not by definition) the same as cardinality of $\mathcal P(\mathbb N)$. The latter is known as continuum and is usually denoted by $2^{\aleph_0}$ or $\mathfrak c$ (less frequently, $\beth_1$) $\endgroup$ – Wojowu Sep 19 '18 at 12:36
  • $\begingroup$ (1) You're switching from $\aleph_1$ to $2^{\aleph_0}$ (or rather, $\cal P(\Bbb N)$), and those are not the same thing; (2) if you're trying to do topology without choice, where talking about the dual Cantor–Bernstein theorem is somehow important, it is better to start with choice present, especially since it is consistent without choice that $2^{\aleph_0}$ and $\aleph_1$ are not comparable without choice. $\endgroup$ – Asaf Karagila Sep 19 '18 at 12:36
  • $\begingroup$ My bad, I'm awful at cardinals and axiomatic set theory and all that, I'll update the question. $\endgroup$ – user3002473 Sep 19 '18 at 12:41
  • $\begingroup$ Note that the first sentence "I'm curious about what the cardinalities of second countable spaces can be at most." asks a different question, the cardinality of the space can be unbounded, just put the trivial topology on it! $\endgroup$ – Alessandro Codenotti Sep 22 '18 at 22:47
  • $\begingroup$ @AlessandroCodenotti I meant the cardinalities of the topologies themselves. $\endgroup$ – user3002473 Sep 23 '18 at 1:52
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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.

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  • $\begingroup$ Honestly, I fiddled around with that idea in your lemma but I guess I didn't see that it was true immediately. Also, I can't believe I forgot that I wanted "at most" rather than "exactly". Thanks for the help! $\endgroup$ – user3002473 Sep 19 '18 at 12:56
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    $\begingroup$ No problem at all. Anything to procrastinate revising a paper. :-) $\endgroup$ – Asaf Karagila Sep 19 '18 at 13:24
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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.

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  • $\begingroup$ How much does this add to my answer, where I explicitly give example of a countably infinite topology and mention finite topologies? Or to your comment from three minutes before posting this answer? $\endgroup$ – Asaf Karagila Sep 19 '18 at 22:02
  • $\begingroup$ @AsafKaragila You had the nice injection, I just wanted to point out that his own surjection in itself was also good enough (assuming AC as he does). The comment was in draft, then I decided to post it as a short alternative answer, but the comment went out anyway. Now deleted. $\endgroup$ – Henno Brandsma Sep 20 '18 at 3:54

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