Timeline for An order type $\tau$ equal to its power $\tau^n, n>2$
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 27, 2013 at 11:07 | comment | added | bof | Really? Too complicated for me, I only see it for size $\omega$. | |
| Nov 27, 2013 at 10:35 | comment | added | Garrett Ervin | Ah, I see. I understood you to be enumerating only intervals, without any reference to the underlying points. That's a nice argument. It seems to go through similarly for any order $\alpha$ without endpoints of size $\kappa$, where $\kappa$ is regular, assuming that $\alpha$ has no cofinal sequences in either direction of length less than $\kappa$. | |
| Nov 27, 2013 at 3:15 | history | edited | bof | CC BY-SA 3.0 |
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| Nov 27, 2013 at 2:48 | history | edited | bof | CC BY-SA 3.0 |
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| Nov 27, 2013 at 2:33 | comment | added | bof | @GarrettErvin Yes, we start by choosing enumerations of $A$ and $B$. I didn't think to mention then, I was only trying to describe how I modified the usual back-and-forth, namely, that the domain and range at each stage is a finite union of intervals instead of a finite set of points. Is your question about how to extend the partial isomorphism $f_k$ so as to cover a given point of $A$ or $B$? | |
| Nov 26, 2013 at 22:30 | comment | added | Garrett Ervin | bof, I'm still a bit confused. In the usual back-and-forth argument one fixes an enumeration of the intended domain and range beforehand, to ensure that the $f$ you construct ends up being total. I agree that if you enumerate the $I_j$ inductively as you suggest, by splitting up multiples of $\alpha$ into multiples of $\alpha \xi$ when necessary, you can ensure at every finite stage there will be space between consecutive intervals. But how then do you guarantee that the intervals $\{I_j\}_{j \in \omega}$ you end up with cover $\alpha$ (or, equivalently, that the $f$ you construct is total)? | |
| Nov 26, 2013 at 0:20 | comment | added | bof | @GarrettErvin My argument was terse, perhaps to the point of being cryptic. The intervals are defined inductively, and at each step we make sure that the newly chosen interval has room on both sides, the o.t. of the gap being a "nonzero right multiple of $\alpha$". Note that any nonzero right multiple of $\alpha$ can be expressed in the form $\alpha\theta$ where $\theta$ has no first or last element; this is because $\alpha\theta=\alpha\xi\theta$ and $\xi\theta$ has no first or last element. This is why the new interval can be chosen without crowding the old intervals. | |
| Nov 25, 2013 at 23:40 | comment | added | Garrett Ervin | bof, I may misunderstand your argument, but why for example can you guarantee that the interval between $I_j$ and $I_{j+1}$ will always contain a copy of $\alpha$? If, say, $\xi = \mathbb{Z}$, then in your proof we would be constructing an isomorphism between $\alpha = \alpha \times \mathbb{Z}$ and $\alpha \times \beta$. Might we have chosen our intervals so that for some $n \in \mathbb{Z}$ we have $I_1 =$ the $n$th copy of $\alpha$ in $\alpha \times \mathbb{Z}$, and $I_2 =$ the $(n+1)$st copy of $\alpha$ in $\alpha \times \mathbb{Z}$? Then interval between $I_1$ and $I_2$ would be empty. | |
| Nov 25, 2013 at 21:40 | review | First posts | |||
| Nov 25, 2013 at 21:53 | |||||
| Nov 25, 2013 at 21:21 | history | answered | bof | CC BY-SA 3.0 |