Where $\mathbb T $ is the set of transcendental numbers, and $\mathbb I $ is the set of irrational numbers and $\mathbb A $ is the set of algebraic numbers. The sets $\mathbb Q$ and $\mathbb R$ have their usual meanings.
I'm having trouble understanding how the union of the transcendental $\mathbb T $ and algebraic numbers $\mathbb A $ is the same as the real numbers $\mathbb R$ since $\mathbb I ,\mathbb Q\subset\Bbb R$ as well.
Put in another way, I think it should be $( \mathbb T \cup \mathbb A) \subset \mathbb R $ and $(\mathbb Q \cup \mathbb I) \subset \mathbb R$
So I'm basically questioning the equality in $\mathbb T \cup \mathbb A = \mathbb Q \cup \mathbb I =\mathbb R$.
The reason I'm asking this question is because those unions I mentioned above don't seem to 'cover' all the Reals $\mathbb R$ .
Thanks in advance.
Kindest Regards.
Given a real number, if it is rational then it is in $\Bbb Q$, otherwise it is irrational and in $\Bbb I$. So $\Bbb{Q\cup I=R}$.
Similarly, if a real number is not algebraic then it is transcendental.
The set $\Bbb I$ is essentially defined as $\Bbb {R\setminus Q}$ and $\Bbb T$ is defined as $\Bbb{R\setminus A}$.
This is analogous to the following situation: Let $P$ be the set of prime numbers, and $Q$ the set of composite natural numbers. Let $E$ and $O$ denote the sets of even and odd name natural numbers respectively.
Then $\Bbb N=P\cup Q=E\cup O$. Just two partitions of the same sets.