Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the set $\mathbb{A}$ of algebraic numbers in $[0, 1]$ of first category?

Actually, I want to prove that the set $T$ of transcendental numbers in $[0, 1]$ is of second category using Baire Category Theorem,

I can prove that $\mathbb{A}$ is countable, so I think $\mathbb{A}$ is of first category, but I cannot prove it.

share|cite|improve this question

Yes, a countable set is of first category. This is immediate from the fact that every singleton set is nowhere dense in $\mathbb{R}$, and a countable set is a countable union of singletons.

share|cite|improve this answer
but closure of rational is real line, and its interior is non empty, does it satisfy the definition of nowhere dense? – pras Jul 11 '14 at 23:26
The rationals are not nowhere dense, but a first category set need not be nowhere dense. It merely needs to be a countable union of such sets. – Ian Jul 11 '14 at 23:29
@pras No: A set is nowhere dense if its closure contains no open intervals, and $\mathbb{R}$ certainly contains an interval. But sets of first category aren't always nowhere dense. – user61527 Jul 11 '14 at 23:31
thanks, now I can prove it – pras Jul 11 '14 at 23:33
You're very welcome. – user61527 Jul 11 '14 at 23:34

Every countable set is of the first category. This follows immediately from the usual definition ("countable union of nowhere dense sets").

share|cite|improve this answer
rational numbers also is of first category? – pras Jul 11 '14 at 23:20
@pras Are the rationals countable? – Shawn O'Hare Jul 11 '14 at 23:21
Yes, certainly, countable union of singletons. – André Nicolas Jul 11 '14 at 23:22
The algebraics are dense in the interval, which may be the reason you doubt. But the definition says countable union of nowhere dense sets, and $1$-element sets are certainly nowhere dense. – André Nicolas Jul 11 '14 at 23:30

Notice that $T\cup A=[0,1]$, $A=\cup_{n=1}^\infty\{x_n\}$, beacuse $A$ is countable so is union of singleton sets, that is no where nowhere dense(first category).

If $T$ were of first category, $[0,1]$ would be countable union of nowhere dense set, which is not true because $[0,1]$ is complete metric space. So $T$ must be of second category.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.