For proving that A ⊆ B, if A is uncountable, then B is also uncountable, I simply would like to start from the definitions of bijection and countable sets and then proving by contradiction that the requirements mentioned have consequences for the uncountable sets as well.
I wonder if I can start the proof from the assumptions stated in this way:
From the definition of bijection, we know that if A bijects in B, B bijects in A (I don't know how to do tilde on keyboard). From the definition of countable sets, we know that if A is countable and B is a substet of A, then B is also countable (I wonder if I should give more evidence about this last statement in bold characters or if I may write it more clearly, showing that it is a consequence of the definitions of countable and bijection).
After stated these postulates, I might prove "Let A ⊆ B. Prove that if A is uncountable, then B is also uncountable" by contradiction in simple way:
Assume that subset A is uncountable and superset B is countable. For B to be countable means that bijects in N and that its subsets are also countable. But A is uncountable. This disprove our assumption about B, therefore B is also uncountable.
PS. However, am I wrong if I say that the whole theory tells us only that an uncountable set cannot be a subset of a countable set but, on the other hand, an uncountable set can be the superset of countable sets. i.e. isn't it the case of R with respect to Q, Z and N?
Best Answer
Take the contrapositive statement (which is equivalent): if $A \subseteq B$ and $B$ is countable, then $A$ is countable. This is much easier, don't you think?
If $f: B \to \mathbb{N}$ is a bijection, then you construct a bijection $f: A \to f(A)$ simply restricting $f$. So you have to show that $f(A) \subseteq \mathbb{N}$ is countable.
Define recursively $$x_0 = \min f(A)$$ and $$x_{n+1} = \min f(A) \setminus \{ x_0, \dots , x_n\}$$
The map $\mathbb{N} \to f(A)$ defined by $n \mapsto x_n$ is a bijection (check).