Cardinality of open sets of $\mathbb{R}^d$

Question

Let $d \in \mathbb{N}$, then count the number of open sets in $\mathbb{R}^d$.

I just don't know how to start. I think I understand how I should count sets, but for this one, I really don't see it... How should I do this?

Answer 1

If $B$ is a countably infinite base (basis) for a topology $T$ then the cardinal $|T|$ (the cardinal of the set of all open sets) is at most $|\Bbb R|=2^{\aleph_0}=2^{|B|}$ because $T$ is the functional image of P$(B)$ (the Power-set of $B$, the set of all subsets of $B$) via the function $f(A)=\cup A$ for all $A\subset B$.

For $n\in \Bbb N$ the set $B=\{ \prod_{i=1}^n(a_i,b_i): a_1,...,a_n,b_1,...,b_n\in \Bbb Q\}$ is a countable base for $\Bbb R^n.$

So the cardinal $|T_n|$ of the topology $T_n$ on $\Bbb R^n$ is at most $|\Bbb R|.$

And $|T_n|\geq |\Bbb R|$ because $T_n$ has a subset $\{(r,r+1)^n:r\in \Bbb R\}$ that has cardinal $|\Bbb R|.$

By the Cantor-Schroeder-Bernstein Theorem (a.k.a. Cantor-Bernstein & Schroeder-Bernstein) and the Axiom of Choice, therefore $|T_n|=|\Bbb R|=2^{\aleph_0}.$