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.

  • 1
    $\begingroup$ Do you know how to show that the product of two countable sets is countable? Then you can use induction on $n$. (Also, your definition of the generating set for the product topology is wrong; the different coordinates don't need to be the same, e.g. you should have $T_1\times T_2\times . . . \times T_n$ instead of $T\times T\times . . . \times T$.) $\endgroup$ – Noah Schweber Sep 12 '16 at 2:07
  • $\begingroup$ I don't. But I could read it up. Where do I take it after this? Edit: I had different indices but decided at the last minute they ought to be the same.. $\endgroup$ – Mathematicing Sep 12 '16 at 2:08
  • $\begingroup$ @NoahSchweber How do I know if $\tau_{i}$ for all i from 1 to n has countably many elements? $\endgroup$ – Mathematicing Sep 12 '16 at 2:23
  • 1
    $\begingroup$ That's not quite the right question. You're looking at the basis, not the topology (the topology $\tau$ has uncountably many elements, of course - there are uncountably many open sets!). So let $B$ be a countable basis for $\tau$. Then (exercise) $B\times B\times . . . \times B$ is a basis for the product topology on $\mathbb{R}^n$. Now, how many elements does $B\times B\times . . . \times B$ have? Well, it's a product of $n$-many countable sets; can you prove (by induction on $n$) that the product of $n$-many countable sets is countable? $\endgroup$ – Noah Schweber Sep 12 '16 at 2:31
  • $\begingroup$ Got it @NoahSchweber. I can take it from here. $\endgroup$ – Mathematicing Sep 12 '16 at 2:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.