no uncountable set can be a subset of a countable set I was reading Rudin's "Principles of Mathematical Analysis, 3rd Ed."  I don't quite follow the author's remark after stating and proving Theorem 2.8.  Specifically, on page 26, we have  
2.8 Theorem $\quad$ Every infinite subset of a countable set A is countable. 
$\quad$ Proof $\quad$ Suppose $E \subset A$, and $E$ is infinite.  Arrange the elements $x$ of $A$ in a seqeuence $\{x_n\}$ of distinct elements.  Construct a sequence $\{n_k\}$ as follows: Let $n_1$ be the smallest positive integer such that $x_{n_1}\in E$.  Having chosen $n_1, ..., n_{k-1}$ $(k=1,2,3,...)$, let $n_k$ be the smallest integer greater than $n_{k-1}$ such that $x_{n_k}\in E$.  Putting $f(k)= x_{n_k}\:(k=1,2,3,...)$, we obtain a 1-1 correspondence between $E$ ane $J$ (the set of positive integers). 
(And then comes the following remark that I don't see how it is implied by the theorem above.)  
The theorem shows that, roughly speaking, countable sets represent the "smallest" infinity: No uncountable set can be a subset of a countable set.
I can understand the theorem and its proof, but I don't quite see the connection between the theorem and the remark.  I'd appreciate if someone can point it out for me.
 A: There are two kinds of 'infinite': (1) countably infinite, and (2) uncountably infinite.  
These are the only two kinds of infinite sets, since the second is simply "all infinite sets which aren't countable".
We have the implication 
$$\text{$A$ an infinite subset of countable set} \implies \text{$A$ is countable}$$
which is equivalent to
$$\text{$A$ an infinite subset of countable set} \implies \text{$A$ not uncountable}$$
and the contrapositive of this is
$$\text{$A$ uncountable} \implies \text{$A$ not infinite subset of countable set}$$
or simply
$$\text{$A$ uncountable} \implies \text{$A$ not subset of countable set}$$
A: Also roughly speaking, think about $\mathbb{Q}\subset\mathbb{R}$.  You know that $\mathbb{Q}$ contains countably infinitely many elements. All of these elements are contained in $\mathbb{R}$, but so are the irrationals. They both contain infinitely many elements, but it is like $\mathbb{R}$ has "more." This theorem implies that in a more rigorous way.  It is saying, if a set is countably infinite, it can only contain infinite subsets that are countable, not subsets that are uncountably infinite. In a manner of speaking, $\mathbb{R}$ would be too "big" to fit in $\mathbb{Q}$.
