In my general topology exercise I have 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.
- Show that $\mathbb R$ satisfies the second axiom of countability
And then they give the tip "use exercise 3", which is:
Let $B$ be the collection of all open intervals $(a,b)$ in $\mathbb R$, with $a < b$ and $a,b \in \mathbb Q$. Prove that $B$ is a basis for the euclidean topology.
I already proved exercise 3, and I think that the tip basically says to use the basis from exercise 3. So now I have to prove that $B$ has a countable number of sets.
My approach:
We know that the set $\mathbb Q$ is countable so, let $U_\alpha=\{(\alpha,n), \forall n \in \mathbb{Q}\}$. Then $U_\alpha \sim \mathbb Q$. So every $U_\alpha$ is countable. Now, the family of sets $\mathcal F= \{U_\alpha, \forall \alpha \in \mathbb Q \}$ is also countable because $\mathcal F \sim \mathbb Q$, So $B = \bigcup \mathcal F$ is the union of a countable number of sets.
My question is, is this proof right and is this the most correct way to prove it?