Difference between equality and isomorphism I don't really understand the difference between sets being equal and isomorphic.  I know that two sets $A$ and $B$ are equal of $A \subseteq B$ and $B \subseteq A$.  I know two sets are isomorphic is there is a homomorphic bijection between the two sets, i.e., they have the same cardinality and the elements of each can be identified in such a way that they behave in exactly the same way.  (In other words, the two sets have the same structure.)
Is equality stronger?  I guess my main question is:
If two sets $A$ and $B$ are equal, does this imply they are isomorphic?  How do $A \subseteq B$ and $B \subseteq A$, which are purely set theoretic statements, imply that the structure of the two sets is the same, i.e., their elements can be identified in such a way that they act the same way?
 A: Isomorphism means "similar structure", so you have to be more accurate and say what type of structure. If you want to talk about every possible structure, then only way two sets are "always isomorphic" is if they are both the empty set (which carries no structure other than empty sets).
Equality as sets, as Henning writes in his excellent answer, means that the sets are in fact equal. But note that when we say that the group $G$ is equal to the group $H$ we mean not only that the sets are equal, but also that the operation is equal (as a set of ordered triplets, or so). We can easily define two non-isomorphic group structures on an infinite set.
If you only consider them as unstructured sets, then isomorphism just says that there is a bijection between the two sets; but then you can't say that the elements "act the same way" since there are no actions. Equality means that the two sets are the same set. Note that $\{\varnothing\}$ and $\{\{\varnothing\}\}$ are both singletons, and are therefore isomorphic as sets (i.e. they have the same cardinality), but they are certainly not equal.
A: There seem to be two problems here.
First, if $A$ and $B$ have the same elements (every element of $A$ is an element of $B$ and vice versa), then they are the same set, the strongest possible form of equivalence.
And the identity function (which sends every $x\in A$ to itself) is certainly a bijection $A\to A$ and preserves whichever structure we care to equip $A$ with. So $A$ is always isomorphic to itself.
Second, the word "isomorphic" denotes many different concepts, depending on which kind of structure we require the isomorphism to preserve. If we say that $A$ and $B$ are isomorphic as sets, we only require that it's a bijection, and "isomorphic" is then the same as "has the same cardinality".
But we can also speak about being isomorphic as groups, or as rings, or as partially ordered sets, or as graphs, or as a lot of other things. In each of those cases we're strictly speaking using sloppy language to speak not only of $A$ and $B$ in themselves as sets (that is, which elements they have), but also additionally (and implicitly) about some structure we have chosen to consider for each of $A$ and $B$. The structure might be a binary operation on the set (when we're speaking of isomorphic-as-groups), or two binary operation (for isomorphic-as-rings), or an order relation (for isomorphic-as-posets), and so forth.
For example, if we say that $A$ and $B$ are isomorphic as groups, then what we really mean is that we have chosen operations $*:A\times A\to A$ and $\circledast:B\times B\to B$ such that $\langle A,{*}\rangle$ and $\langle B,{\circledast}\rangle$ are groups (and having made those choices is neccessary before we can even ask whether $A$ and $B$ are isomorphic groups) and that there is an $f:A\to B$ that is a bijection and satisfies $f(a_1*a_2)=f(a_1)\circledast f(a_2)$.
What we really should be saying is "the groups $\langle A,{*}\rangle$ and $\langle B,{\circledast}\rangle$ are isomorphic". But we're employing an informal shorthand where we can say $A$ when we mean $\langle A,{*}\rangle$, provided that it is clear which ${*}$ we mean.
Note that it is possible for $A$ and $B$ to be the same set, yet because we have chosen different $*$ and $\circledast$ they are not isomorphic by structure. For example, $\langle \mathbb R,{+}\rangle$ and $\langle \mathbb R,{\times}\rangle$ are not isomorphic as monoids, even though the underlying set $\mathbb R$ is the same.
