In a general topology exercise I am asked to prove the following:
A topological space $(X,\tau)$ is said to satisfy the second axiom of countability if there exists a basis $B$ for $\tau$, where $B$ consists of only a countable number of sets.
- Prove that $\mathbb R^n$ satisfies the second axiom of countability for every positive integer $n$.
But instead I came up with a proof that proves the opposite, that $\mathbb R^2$ does not satisfies the axiom:
My proof:
The basis for the euclidean topology is $B=\{\alpha_i<x_i<\beta_i,i \in \{1,…,n\}\}$.
If let $A\in B$. Then we can define a function $f:B \to \mathbb R^n \times\mathbb R^n$, such that $f(A)=\left((\alpha_1,…;\alpha_n),(\beta_1,…,\beta_n)\right)$. This function is a bijection, so we have that $B \sim \mathbb R^n \times\mathbb R^n$. $\mathbb R$ is uncountable, so $R^n$ is also uncountable. Because of that $\mathbb R^n \times\mathbb R^n$ is also uncountable. This $B$ must also be uncountable since $B \sim \mathbb R^n \times\mathbb R^n$. So $\mathbb R^n$ does not satisfies the axiom of countability.
What did I do wrong in this proof? Where is the mistake?
Best Answer
You are right that that particular base $B$ is uncountable; your mistake is thinking that because $B$ is uncountable, every base for the usual topology on $\Bbb R^n$ must be uncountable. If you keep only those members of $B$ for which the endpoints $\alpha_i$ and $\beta_i$ are all rational, you’ll have a countable family that is still a base for the topology.