# Prove the set of transcendental number is dense in $\mathbb{R}$ using Baire category theorem

I want to show the set of transcendental number $T$ is dense in $\mathbb{R}$ using Baire category theorem.

The fact is easier to show directly using definition and Archimedean property. But I was asked to use Baire category theorem specifically. Since $\mathbb{R}$ is complete metric space, I think I should express $T$ as a countable intersection of open dense set, which is also dense. But I don't know how to construct it.

I have found a proof using Liouville number, that is show the set of Liouville number is dense using Baire category theorem and also show Liouville number is transcendental. But since it's an optional exercise from 2nd year undergraduate course, I don't expect the proof will be so complicate. Is there any easier way?

HINT: The set of algebraic numbers is countable.

• So it's of first category. Hence transcendental number is of 2nd category, and then? – John Nov 16 '14 at 18:10
• @John: Consider the open sets of the form $\Bbb R\setminus\{r\}$, where $r$ is algebraic. – Brian M. Scott Nov 16 '14 at 18:11
• Ah, I see! Thanks, you are right. – John Nov 16 '14 at 18:13
• @John: Youâ€™re welcome. – Brian M. Scott Nov 16 '14 at 18:14