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?

  • $\begingroup$ Okay, give your definition of isomorphism on sets. Since usually this notion defined on sets with operation. $\endgroup$ – Jihad Jan 11 '15 at 15:55
  • $\begingroup$ @Jihad Ok, maybe a better question would be why does set equality automatically imply the preservation of the structure, whatever that structure may be? How do we know set equality implies elements behave in the same way? $\endgroup$ – layman Jan 11 '15 at 15:57
  • $\begingroup$ @Jihad I mean my question in the most abstract sense possible. I don't want to say the set is endowed with $+$ and $\cdot$, for example. We could have different operations, or maybe just one operation, or who knows what else. But set equality is simply defined as $A = B$ if $A \subseteq B$ and $B \subseteq A$, and this somehow implies that $A$ and $B$ have the same structure. $\endgroup$ – layman Jan 11 '15 at 16:00

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.

  • $\begingroup$ I agree with you completely. I don't want to impose a structure on the two sets $A$ and $B$ because, I think, in all of these different structures, when $A \subseteq B$ and $B \subseteq A$ (a purely set theoretic statement), it follows that $A$ has the same structure as $B$. How can a purely set theoretic statement imply anything about additional structure on the sets? $\endgroup$ – layman Jan 11 '15 at 16:04
  • 2
    $\begingroup$ @user46944: Your assumption is not true. Consider the example of $\langle\mathbb R,{+}\rangle$ versus $\langle\mathbb R,{\times}\rangle$. It is certainly the case that $\mathbb R\subseteq \mathbb R$ and $\mathbb R\supseteq \mathbb R$, but the two structures are different. Purely set-theoretic containment cannot imply anything about the additional structure -- for the trivial reason that "purely set-theoretic" means that we're ignoring all additional structure. $\endgroup$ – Henning Makholm Jan 11 '15 at 16:12
  • 3
    $\begingroup$ @user46944: When we use the word "set", this implies that there is no additional structure to check. If you want to speak about a situation where you do check some additional structure, you don't say "set", but instead "group" or "ring" or "poset", or some other word that will tell the listener which structure it is that you're considering. $\endgroup$ – Henning Makholm Jan 11 '15 at 16:18
  • 2
    $\begingroup$ @user46944: In that case the text you've been reading is sloppy. Typically one would consider, for example, equality among subgroups of some group $G=\langle G,{*}\rangle$. Then the group $\langle H,{\circledast}\rangle$ is a subgroup of $G$ if and only if $H\subseteq G$ and $\circledast$ is the restriction of $*$ to $H\times H$. So whenever $H$ is given there is only one possible $\circledast$ that will make it a subgroup. ... $\endgroup$ – Henning Makholm Jan 11 '15 at 16:24
  • 2
    $\begingroup$ ... Thus if we already know that $\langle H_1,{\circledast}\rangle$ and $\langle H_2,{\otimes}\rangle$ are both subgroups of $G$, then just knowing that $H_1=H_2$ as sets also tells us that $\langle H_1,{\circledast}\rangle$ and $\langle H_2,{\otimes}\rangle$ are the same group, because then ${\circledast}$ and $\otimes$ are the same operation because each of them must be the restriction of $*$ to $H_1\times H_1$ (which is the same set as $H_2\times H_2$). $\endgroup$ – Henning Makholm Jan 11 '15 at 16:25

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.

  • 1
    $\begingroup$ No. Saying that two sets are equal means that they are equal as sets. Saying that two structures are equal means that their underlying sets, and their interpretations of the symbols relevant to the structure are also equal. Sets are sets. $\endgroup$ – Asaf Karagila Jan 11 '15 at 16:19
  • 1
    $\begingroup$ I don't know about "usually". If you want to argue that the structures are the same one, literally, then you show that the identity function is an isomorphism. Or you show that one is a substructure of the other, and vice versa, in which case the operations and whatnot must agree with each other. $\endgroup$ – Asaf Karagila Jan 11 '15 at 16:23
  • 1
    $\begingroup$ I don't know how you define $\Bbb Z_p(u)$, I'd think it's the transcendental extension of $\Bbb Z_p$ by $u$. But this surely not the case. Either you define it in such way that $\Bbb Z_{p^n}$ is literally a substructure of $\Bbb Z_p(u)$; or that this is an abuse of notation and language. And this abuse is completely normal in mathematics, despite the fact that it can be misleading at first. $\endgroup$ – Asaf Karagila Jan 11 '15 at 16:31
  • 2
    $\begingroup$ @user46944: In that case it is implicit that the structure on $\mathbb Z_p(u)$ is the one it inherits from $\mathbb Z_{p^n}$, because $\mathbb Z_p(u)$ means the smallest subfield of the field we got $u$ from that contains $\mathbb Z_p$ and $u$. So if we can show that $\mathbb Z_p(u)$ has all the elements of $\mathbb Z_{p^n}$, then by definition the operations on those elements behaves the same as the operations of $\mathbb Z_{p^n}$ -- because that's where the operations came from in the first place! $\endgroup$ – Henning Makholm Jan 11 '15 at 16:31
  • 1
    $\begingroup$ Okay, so you define it as a substructure of $\Bbb Z_{p^n}$. So indeed you only have to show equality of sets here, since if $M$ is a substructure of $N$, they agree on all the relations and operations (whenever they can agree, of course), so to show equality of $M$ and $N$ as structures you just need to show that they have the same underlying sets. But here you already have the additional information needed for this inference. $\endgroup$ – Asaf Karagila Jan 11 '15 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.