Sets have the same cardinality if and only if there exists a bijective map between them. The lowest infinite cardinality is \(\displaystyle \aleph_0\), or being "countably infinite". An easy way to think about a countably infinite set is that it's a set whose members can be listed in some meaningful way without missing any. That is to say, for the natural numbers, we could write (0, 1, 2, ...). Clearly, I can't write out all the natural numbers, but if you imagine continuing my list infinitely, it shouldn't be hard to see that I would capture every natural number.

Nonetheless, we can give a more formal definition to countably infinite - a set \(\displaystyle A \) is countably infinite if and only if there exists a bijection \(\displaystyle f:A \rightarrow \mathbb{N}\). Note that this definition is just a formal way of saying what I just described (do you see why?). Many familiar sets are also countably infinite - \(\displaystyle \mathbb{Q} \), the set of all rational numbers; \(\displaystyle \mathbb{Z} \), the set of all integers; and the set of all algebraic numbers, to name a few. I should note here that it's not uncommon for the word "countable" to sometimes mean finite sets and countably infinite sets, and others will use countable to strictly mean the countably infinite case. When in doubt, you should always clarify.

We can make an interesting remark here about a situation that will come up again when we discuss the uncountable case. Note that \(\displaystyle \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \), but \(\displaystyle \mathbb{N} \ne \mathbb{Z} \) and \(\displaystyle \mathbb{Z} \ne \mathbb{Q} \). That is, every natural number is an integer, but not every integer is a natural number. This slightly counter-intuitive notion is one that comes up quite often in discussions of cardinality. Finite notions of size do not necessarily translate into the infinite case.

As you are clearly aware from your previous posts, there are also infinite sets for which no such bijection \(\displaystyle f:A \rightarrow \mathbb{N} \) exists. We call these sets "uncountable". One very familiar set of this kind is \(\displaystyle \mathbb{R}\), or the set of real numbers. As previously demonstrated, \(\displaystyle \mathbb{R} \) has the same cardinality as any interval \(\displaystyle (a,b) \) for \(\displaystyle a,b \in \mathbb{R} \) and \(\displaystyle a \ne b \). Also of interest, \(\displaystyle \mathbb{C} \) and \(\displaystyle \mathbb{R}\) x \(\displaystyle \mathbb{R} \) both share the so called cardinality of the continuum, or the cardinality of the reals.

I'll take a quick moment to clarify some notions on functions and their relation to cardinality. Let \(\displaystyle S\) and \(\displaystyle T\) be two sets. If there exists an injective map \(\displaystyle g:S \rightarrow T \), then we say that \(\displaystyle |S| \le |T| \) where the absolute value-looking bars denote cardinality. If there exists a surjection \(\displaystyle h:S \rightarrow T\), then we can conclude that \(\displaystyle |S| \ge |T| \). Since a bijection takes care of both of these cases, it is a necessary and sufficient condition for two sets to have equal cardinality for there to be a bijection between them. So, we've answered one of your questions. No countable set can be surjectively mapped on to an uncountable set. For, if it could, then its cardinality would be at least that of the uncountable set, which we're supposing is false.

The Cantor-Bernstein-Schroeder theorem gives us an interesting tool for comparing cardinality. If there is an injective map \(\displaystyle f:S \rightarrow T \) and an injective map \(\displaystyle g:T \rightarrow S\), then there is a bijection \(\displaystyle h:S \rightarrow T\), or \(\displaystyle |S| = |T| \).

By Cantor's Theorem, we know that a bijection between a set and a set's powerset cannot exist. Thus, \(\displaystyle | \mathbb{R}| < |2^{\mathbb{R}}| \). This shows us that there is no "largest" cardinality and that uncountable sets, unlike countably infinite sets, do not all have the same cardinality.

I think I've answered (though not always explicitly) most of your questions here. If I didn't explain something well enough, or if it's not clear, feel free to ask!