Finite subsets of countable and uncountable sets If I take all the subsets of exactly size $n$ (where $n$ is some fixed natural number) from the set of natural numbers, and put them into a set, then the result is a countable set (a set containing all subsets of size $n$).
If I do the same, but this time taking the subsets (of size $n$ - I'll call them $n$-subsets) from the reals, then the result is an uncountable set.
Can someone please explain to me why this is the case? And perhaps how to show it?
I guess it's because the natural numbers contain a countable number of $n$-subsets, and the reals contain an uncountable number of $n$-subsets. But I don't know how to formalize this.
Thanks in advance
 A: We can (injectively) map the set of $n$-subsets of a set $X$ into $X^n$, hence there at most as many $n$-sets as therre are $n$-tuples - and in case of infinite $X$ we have $|X|=|X^n|$. This shows that there are at most as many $n$-sets as elements if $X$ is infinite.
If $X=\mathbb N$, we can also injectively map $X$ to the set of $n$-subsets of $X$ by mapping $a\mapsto \{a,a+1,\ldots,a+n-1\}$. More generally, if $X$ is infinite, we can write it as disjoint union of countable subsets and repeat this construction piecewise. This shows that there are at least as many $n$-sets as elements if $X$ is infinite.
In summary, the cardinality of the set of $n$-subsets of $X$ is the same as the cardinality of $X$ (if $X$ is infinite).
A: Let $n$ be only $1$. The set of all subsets of one element of the reals is already uncountable since it is equivalent to the set of reals.
So the harder question is why the set of $n$-touples of natural numbers is countable. It is easy! Take $n$ different prime numbers: $p_1,p_2,...,p_n$ and observe that an $n$-tuple of naturals $\{n_1,n_2,...n_n\}$ uniquely determine one natural number
$$N_{n_1,n_2,...,n_n}=p_1^{n_1}\times p_2^{n_2}\times \cdots \times p_n^{n_n}.$$
This why you can construct only a subset of the naturals... 
A: If the set $A_i$ are such that $|A_i|=n$ and $$\mathbb R=\bigcup_{i\in\mathcal I}A_i$$
then the union is not countable since $\mathbb R$ is not countable. 
