Cardinality of the Euclidean topology and the axiom of choice

Question

It is relatively straightforward to prove that the Euclidean topology has the same cardinality as the space itself. I have sketched a proof below. The proof seems to rely quite heavily on the axiom of choice. I'm perfectly fine with (countable) dependent choice — I want to be able to define sequences recursively, for example — but one doesn't usually need more choice for topology in Euclidean spaces.

Is dependent choice enough to show that the topology of $\mathbb R^n$, $n\geq1$, has the same cardinality as $\mathbb R^n$? If yes, how to prove it? If not, how much do we know about the cardinality of the topology? It seems that the cardinality of the topology lies between $\mathfrak{c}$ and $2^{\mathfrak{c}}$, but are there tighter bounds?

Proof of equal cardinalities with choice: Let $T$ be the topology of $\mathbb R^n$ and $B$ a countable basis for it (rational balls, for example). Since $|2^B|=|\mathbb R^n|$, it suffices to find an injection $f:\mathbb R^n\to T$ and a surjection $g:2^B\to T$. These together will establish $|\mathbb R^n|\leq|T|\leq|2^B|=|\mathbb R^n|$ and the conclusion follows. (Alternatively, the surjection $g$ has an injective right inverse, and we thus have injections both ways between $\mathbb R^n$ and $T$. The existence of a bijection follows from the Schröder–Bernstein theorem.) We can simply take $f(x)=B(x,1)$ and $g(A)=\bigcup_{U\in A}U$.

Injecting $\mathbb R^n$ to $T$ requires no choice and always $T\subset 2^{\mathbb R^n}$, so at least $\mathfrak{c}\leq|T|\leq2^{\mathfrak{c}}$ without AC. Without choice, I can't compare cardinals so easily or produce injective right inverses for surjections, so my proof falls apart.

Answer 1

The argument does not really use the axiom of choice.

Fix a countable basis $\{B_n\mid n\in\Bbb N\}$, now note the following is true.

If $U,V$ are open subsets of $\Bbb R^n$, then $U=V$ if and only if $\{n\mid B_n\subseteq U\}=\{m\mid B_m\subseteq V\}$.

This gives us an injection from the open sets to subsets of $\Bbb N$. Now using Cantor-Bernstein, which holds without choice, we're done.

Answer 2

You don't need choice here at all - every open set $U$ has a canonical description in terms of basic open sets, $code(U)=\{B_\eta: B_\eta\subseteq U\}$ (where $\{B_\eta: \eta<\omega\}$ is the set of all balls with rational radius and center). We clearly have $code(U)=code(V)\iff U=V$, and the set of codes has size continuum.