algebraic number and transcendental number

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.

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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.

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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").

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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.

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