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let $(X,\leq)$ and $(Y,\preceq)$ be well-order sets and they are isomorphic.

Prove that there is one isomorphism between $X$ and $Y$.

Notice that $X$ and $Y$ are sets, so by isomorphism I mean isomorphism between sets.

kickstart
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    As far as I can see, this looks like Prove that $A\Rightarrow A$ – Daniel Jan 24 '18 at 00:29
  • I don't understand the question. An isomorphism between well ordered sets is an isomorphism on the set level which preserves the ordering. If you just forget about the ordering, you have an isomorphism on the underlying sets. Or are you asking for something else? – lulu Jan 24 '18 at 00:32
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    I would think it should be isomorphism of ordered sets. And the question would be asking to show that if there is at least one isomorphism, then that isomorphism is unique. – Daniel Schepler Jan 24 '18 at 00:33
  • Voting to close the question as it is not clear what you are asking. If you can, please edit your post for clarity. – lulu Jan 24 '18 at 01:04

3 Answers3

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I would be surprised if you meant "isomorphism between sets" and not "isomorphism between ordered sets" (cause otherwise the claim of uniqueness wouldn't be true... which I assume is what you mean when you say 'prove there is one isomorphism').

If $f$ and $g$ are isomorphisms, are then they must respect least elements. So $f$ and $g$ both map the least element of $X$ to the least element of $Y.$ And so on, by induction. Will leave the details to you.

spaceisdarkgreen
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Every homomorphism between ordered sets $(X, \leq) \to (Y, \preceq)$ has an underlying homomorphism of sets $X \to Y$.

The usual definition of what it means for a homomorphism of ordered sets to be an isomorphism almost literally contains the condition that the underlying homomorphism of sets is an isomorphism.

(that the orders are well-orders is irrelevant)

(also note that the obvious proof proves that one exists; but once you know one exists it's easy to show that there are many more, at least in most cases)

So, I imagine one of the three things are true:

  • You are overthinking it, and making the problem more complicated than it really is.
  • You are using unusual definitions for which this is not immediate — if so you really need to include those definitions so that the rest of us know the actual problem you're facing
  • You haven't written the question you mean to ask.
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We do need an isomorphism between ordered sets here...

One method would be a proof by induction of the following statement: Let $f: X \rightarrow Y$ and $g \rightarrow Y$ be two isomorphisms. Then $\forall x \in X$ $f(x)=g(x)$

It's worth filling in the details yourself as a good exercise, but one key step would be the following:

If $\perp$ denotes the bottom element, then $f(\perp)=g(\perp)=\perp$. This is true because if either of them were a greater element then nothing less, including $\perp$, could be hit by them because the functions preserve order and all remaining inputs are greater than $\perp$.

Fill in the induction step of this proof and you'll have a proof of the entire proposition.

Chessanator
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