Are $1+ω$ and $ω+1$ isomorphic?

Question

In set theory, $1+ω$ is defined as the ordinal number of ordinal sum of sets $\{a\}$ and ℕ. Also $ω+1$ is defined as the ordinal number of ordinal sum of sets $\Bbb N$ and $\{a\}$. You know what the ordinal sum means: the ordinal sum of well-ordered sets $A$ and $B$ is the set $A \cup B$ ordered as you know. We know that the sets $\{a\} \cup ℕ$ and $ℕ \cup \{a\}$ are equipotent and well-ordered. And also we know by a theorem in set theory, any two equipotent well-ordered set are isomorphic (similar).

Doesn't this argument contradict with the fact that in set theory, the ordinal numbers $1+ω$ and $ω+1$ are not equal to each other?

Pinter mentioned in his book following theorem and he proved it: "Theorem Let A and B be well-ordered classes; exactly one of the following three cases must be hold:

i) A is isomorphic with B.

ii) A is isomorphic with an initial segment of B.

iii) B is isomorphic with an initial segment of A.".

And he concluded " well-ordered classes do not differ from one another except in their size". I remind you the "isomorphism" concept is equivalent to "similar" concept; and it is a one to one correspondence between two sets and order-preserving.

Answer 1

The term "isomorphism" means similar structure. But it doesn't specify the structure. The structure needs to be inferred from context, or stated explicitly in the text.

If the context is just sets, then isomorphisms are not required to preserved any structure, in which case $\Bbb N$ and $\Bbb Q$ are isomorphic, since both are countably infinite.

If the context is ordered sets, then the isomorphism must also preserve the order. Not just be a bijection. In this context $\Bbb N$ and $\Bbb Q$ are of course not isomorphic. And also $\omega+1$ and $\omega$ (or $1+\omega$, which is isomorphic to $\omega$ in this context).

To see why two structures are not isomorphic, we can sometimes look at their "inherent properties", things like being a dense set or having a minimum or maximum, are things which must be preserved by isomorphism of ordered sets. So if one ordered set has a maximal element, and another does not then there is no way they are isomorphic.

Now you can easily see that while $1+\omega$ and $\omega+1$ is isomorphic as sets; they are not isomorphic as ordered sets, since $1+\omega$ does not have a maximal element, whereas $\omega+1$ does.

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EDIT:

Note that the theorem Pinter quotes does not require that $1+\omega$ and $\omega+1$ are isomorphic. It is still possible, and that much is in fact true, that $1+\omega$ is isomorphic to a proper initial segment of $\omega+1$.

Moreover, a Cantor-Bernstein "like" theorem for well-ordered sets is in fact true: If $A$ and $B$ are two well-ordered sets such that there are order-embeddings from $A$ into $B$ and from $B$ into $A$, then $A$ and $B$ are isomorphic.

But this doesn't contradict my above answer either. Simply because there is no order-preserving injection from $\omega+1$ into $\omega$.

Answer 2

A well-ordering exists on both, but the ordinal orderings on $1+\omega$ and $\omega+1$ are different. The former has no maximal element, the latter has one.

Answer 3

$\omega+1$ and $1+\omega=\omega$ have the same cardinality and thus there's a bijection $f$ from $\omega+1$ onto $\omega$. $\in$ is the canonical well-ordering on $\omega+1$ and so via $f$ you'll get a well-ordering $<_f$ on $\omega$, namely $\alpha <_f \beta$ iff $f^{-1}(\alpha) \in f^{-1}(\beta)$ for $\alpha,\beta\in\omega$. However, this well-ordering is not the same as the well-ordering $\in$ on $\omega$. Likewise, you can "project" the well-ordering $\in$ from $\omega$ to $\omega+1$ via $f^{-1}$ but what you'll get will not be $\in$ on $\omega+1$.

Answer 4

There is no contradiction here. While it's certainly true that $\omega+1$ and $1+\omega$ have the same number of elements, and there's therefore a bijection between their elements, isomorphism of wellorders is more demanding.

Let $(A,\leq_A)$ and $(B,\leq_B)$ be wellorders. They are isomorphic iff there's a bijection $f:A\to B$ between them such that $f(x)\leq_B f(y) \iff x\leq_A y$. Being in bijection and also being a bijection on sets that have wellorders is a necessary condition, but far from a sufficient one, for isomorphism of ordinals; in particular, there is no bijection $f$ between the two ordinal sums you mention that satisfies the condition above.