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Let $\omega_0$ be the first countable ordinal and $\omega_1$ be the first uncountable ordinal. Use $[0, \omega_1)$ to denote the set of all countable ordinals and hence $\omega_1 = [0, \omega_1)$. Could someone construct a set (other than the set of ordinals) that is order isomorphic to the set $[\omega_0, \omega_1)$. In this post, Arthur (in the top post) provided an example, which is the following:

For each $\alpha\in[0, \omega_1)$, let $X_{\alpha}$ be the set of subsets of $\mathbb{Q}$ , each of which is order isomorphic to $\alpha$, and define $X_{\alpha} < X_{\beta}$ iff $\alpha < \beta$. Then it is claimed that the following set $A$, equipped with the order "<"

$$A = \{X_{\alpha}\,\vert\,\alpha\in[0, \omega_1)\}\hspace{1cm}(\subseteq\mathcal{P}[\mathcal{P}(\mathbb{Q})])$$

is order isomorphic to $[0, \omega_1)$. If we can find a uncountable family of sets (ideally not subsets of the set of ordinals) so that the supremum is not in the family and no countable union of elements will be equal to the supremum, will that work for this question?

Sanae
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    Isn't $[\omega_0, \omega_1)$ order isomorphic to $[0, \omega_1)$? – Nate Eldredge Nov 12 '21 at 19:32
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    In fact, $[\alpha,\omega_1)\cong[0,\omega_1)$ for every countable ordinal $\alpha$. – Noah Schweber Nov 12 '21 at 19:54
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    "However, I believe $\omega_1\in A$ in this example because $\mathcal{P}(\mathbb{Q})\in A$ is the max element." It's hard to parse what you mean there, but assuming I'm interpreting it right, no, there are no subsets of $\mathbb Q$ of order type $\omega_1$ (for many reasons, the most notable being that $\mathbb Q$ is countable). $\mathcal{P}(\mathbb{Q})$ has subsets of many different order types, and those that are well-ordered have the order type of some $\alpha < \omega_1.$ So it is very far from meeting the criteria for being in $A.$ – spaceisdarkgreen Nov 15 '21 at 05:02
  • @spaceisdarkgreen That is my mistake and I should instead say "because $\mathcal{P}(\mathbb{N})\in A$, then $\omega_1\in A$". – Sanae Nov 15 '21 at 05:16
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    @SanaeKochiya "$\mathcal{P}(\mathbb{N})\in A$" is also false ... – Noah Schweber Nov 15 '21 at 05:16
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    @SanaeKochiya and $\omega_1\in A$ makes no sense at all taken literally, though I gather what you are trying to say that is that $A$ has something representing that order type, which is false... again, because the rationals don’t have an uncountable subset (let alone an uncountable well ordered subset), – spaceisdarkgreen Nov 15 '21 at 05:26
  • @NoahSchweber Since $\mathcal{P}(\mathbb{Q})$ is uncountable and $\bigcup_{\alpha < \omega_0}X_{\alpha}$ is countable, can I then conclude $X_{\omega_0}$ is uncountable? I should state $\mathcal{P}(\mathbb{N})$ is a subset of $A$ instead of an element. – Sanae Nov 15 '21 at 05:31
  • @spaceisdarkgreen When saying $\omega_1\in A$, I believed $X_{\omega_0}$ is uncountable because $\mathcal{P}(\mathbb{Q}) = X_{\omega_0}\cup\bigcup_{\alpha<\omega_0}X_{\alpha}$ is uncountable, and $\bigcup_{\alpha<\omega_0}X_{\alpha}$ is countable. Hence $X_{\omega_0}$ is uncountable but I am not sure about this. I cannot see different between, for instance, $X_{\omega_0}$ and $X_{\omega_0+1}$ – Sanae Nov 15 '21 at 05:37
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    @SanaeKochiya $X_{\omega_0}$ is the set of all subsets of the rationals that are order isomorphic to $\omega_0$. This is indeed uncountable, but I don't see how that is relevant. $X_{\omega_0+1}$ is the set of all subsets of the rationals that are order isomorphic to $\omega_0+1$ which is a completely different set (in fact disjoint from $X_{\omega_0}$), and is also uncountable. For instance $\mathbb N$ is in the first and ${1-1/(n+1): n\in \mathbb N}\cup{1}$ is in the second. But again, the fact that these sets are uncountable is completely irrelevant. – spaceisdarkgreen Nov 15 '21 at 05:47
  • @spaceisdarkgreen I got what you mean and I shouldn't have said "$\omega_1\in A$". The mapping in the original example should map $X_{\alpha}$ to $\alpha$ but not $\vert,X_{\alpha},\vert$. Then I will need to show the mapping $X_{\alpha}\mapsto\alpha$ maps $A$ onto $[0, \omega_1)$ – Sanae Nov 15 '21 at 06:00
  • @SanaeKochiya Sure, the idea is just that any countable well-ordered set can be order-embedded into $\mathbb Q$. (In fact, any countable ordered set whatsoever can be embedded.) – spaceisdarkgreen Nov 15 '21 at 06:27
  • Let us continue this discussion in chat. – Sanae Nov 15 '21 at 06:32
  • You can find arbitrarily many such sets. $[\omega_0,\omega_1)$ is of cardinality $\omega_1$. So any set $A$ of cardinality $\omega_1$ implies a bijection between $A$ and $[\omega_0,\omega_1)$. This bijection then defines an order on $A$ so that these two sets are order isomorphic. – Lazy Nov 21 '21 at 22:13

1 Answers1

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The map $f(x) = \omega_0 + x$ is an order isomorphism $[0, \omega_1) \to [\omega_0, \omega_1)$.

First, let's prove that it's a bijection. Note that in general, if we have ordinals $a \geq b$, then there exists a unique $c$ such that $a = b + c$. So for all ordinals $a \in [\omega_0, \omega_1)$, there exists a unique $c$ such that $\omega_0 + c = a$. This proves that $f$ is a bijection.

From here, we must show that $f$ is an order isomorphism. Suppose that $x \leq y$. Then write $y = x + z$. Then $f(y) = \omega_0 + y = \omega_0 + x + z = f(x) + z$, so $f(x) \leq f(y)$. The fact that both are total orders proves it's an order isomorphism and not just order-preserving.

Obviously, this result generalises a lot.

Mark Saving
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  • Thank you for your answers. In fact, I am looking for something other than ordinals and that is why I quote Arthur's answer in my question .... I am sorry to tell you I forget to mention this in my question. – Sanae Nov 12 '21 at 21:05