Question:
Show that the standard topology on $\mathbb{R}^{n}$ has a countable basis.
Let $\tau$ be a standard topology on $\mathbb{R}$. By definition, the standard topology on $\mathbb{R}\times\cdot \cdot \cdot \times\mathbb{R}$ is the product topology $\tau \times \cdot \cdot \cdot \times \tau$. Thus, we have the product space.
Recall: A topology $\tau$ has a countable basis IFF there is at least one basis that generates $\tau$ and has countably many elements. It does not matter whether $\tau$ has uncountable basis.
Indeed, by definition of product space, the product topology $\tau \times \cdot \cdot \cdot \times \tau$ is generated by $B=\left \{ T \times\cdot \cdot \cdot \times T \mid T \in \tau \right \}$
Recall: a set B is countable IFF there exists a bijection
$f:\mathbb{Z}^{+}\rightarrow B=\left \{ T \times\cdot \cdot \cdot \times T \right \}$
How should I determine if B has countably many elements?
Any help s appreciated.
Thanks in advance.