Nature of D-finite sets. A is called D-finite if A is not containing countable subset. 
With the above strange definition I need to show the following two properties:


*

*For a D-finite set A, and finite B, the union of A and B is D-finite.

*The union of two D-finite sets is D-finite.
By the way, can we construct such D-finite set? 
Only hints...
Thank you.
 A: We can't really construct an infinite D-finite set by defining it, simply because it is consistent that there are no such sets (i.e. all the D-finite sets are finite).
We can construct models in which the axiom of choice fails and there are infinite D-finite sets, but those constructions are difficult and involve forcing in many cases.

First note that a finite set is D-finite. So to solve the second question means to solve the first one as well.
To show that if $A$ and $B$ are D-finite then $A\cup B$ is also D-finite, assume by contradiction that $A\cup B$ is not D-finite then there is some $X\subseteq A\cup B$ such that $|X|=\aleph_0$. Consider $X\cap A$ and $X\cap B$, and show that at least one of those has size $\aleph_0$, in contradiction to the assumption that both sets are D-finite.
A: HINT: Note that a set $S$ is D-finite if and only if there is no injection from $\omega$ into $S$. If there is an injective $f:\omega\to A\cup B$, then ...
A: Hints only:
The first property may be shown directly.
The second however... Try showing what happens when the union of two sets is not D-finite.
Hope it helps.
