Are there any infinite sets that have a lower cardinality than the natural numbers? Is there a proof of this?
[Math] Infinite sets with cardinality less than the natural numbers
cardinalselementary-set-theory
Related Solutions
First of all, I suggest you find a good article/book on Cantor's work on multiple infinities. Reading that will enlighten you and answer probably most of the questions you have concerning this subject. Try these lecture notes.
But to give a brief answer to your current questions:
Per definition, two sets have the same cardinality if there exists a bijection between them. This definition was introduced by Cantor because for infinite sets, you could not just count the elements. Not all infinite sets have the same cardinality. For example, the natural numbers $\mathbb{N}$ and the real numbers $\mathbb{R}$ do not have the same cardinality. This was proven with the diagonal argument.
$\mathbb{N}$ and $\mathbb{Z}$ are both infinite sets. I suggest you check out their definitions on wikipedia (the natural numbers, the integers). They also, like Ali said in his post, have the same cardinality. The bijection that was given by him is probably the easiest one to grasp.
As for $\mathbb{Q}$, the rational numbers, this is also an infinite set that has the same cardinality as $\mathbb{N}$ (and thus also $\mathbb{Z}$). I suggest you check out the bijection that is given in the lectures notes that I linked above, that one is probably the clearest one.
You can find encodings in the natural numbers of certain collections of subsets of $\mathbb{N}$. For example, as was pointed out by Henning Makholm, there is a very nice encoding of the collection of finite subsets of $\mathbb{N}$.
More generally, let $A$ be a recursively enumerable subset of $\mathbb{N}$. Let $T_A$ be a Turing machine that, on input $n$, halts if $n\in A$, and does not halt otherwise. Then we can encode $A$ by using the usual index of the machine $T_A$.
This encoding is not fully satisfactory. There are infinitely many Turing machines that halt on input $n$ iff $n\in A$. Thus every recursively enumerable set has infinitely many encodings. Moreover, there is no algorithm for telling, given two natural numbers, whether they encode the same set.
For more modest collections than the collection of recursively enumerable sets, there are far more satisfactory encodings. As a small and not very interesting example, consider the collection of subsets of $\mathbb{N}$ that are either finite or cofinite (their complement is finite). A small modification of the encoding of finite subsets will take care of the finites and cofinites. Basically, just use the even natural numbers for the finites. If you have used $2k$ to encode a finite, use $2k+1$ to encode its complement.
In addition, the elements of any countably infinite subset of the power set of $\mathbb{N}$ can, by the definition of countably infinite, be encoded using the natural numbers. However, by cardinality considerations, we cannot encode all the subsets of $\mathbb{N}$ using elements of $\mathbb{N}$.
Best Answer
No there are none. If $A$ has cardinality of at most the natural numbers, we may assume that it is a subset of the natural numbers.
One can show that a subset of the natural numbers is either bounded and finite, or unbounded and equipotent to the natural numbers themselves.