Timeline for An order type $\tau$ equal to its power $\tau^n, n>2$
Current License: CC BY-SA 3.0
17 events
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| Oct 4, 2016 at 23:00 | history | edited | Garrett Ervin | CC BY-SA 3.0 |
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| Nov 27, 2013 at 10:48 | comment | added | Garrett Ervin | Credit where it's due. I've edited. | |
| Nov 27, 2013 at 10:47 | history | edited | Garrett Ervin | CC BY-SA 3.0 |
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| Nov 27, 2013 at 6:31 | comment | added | bof | That's a nice application of the "Schroeder-Bernstein theorem for linear orders". As long as you're mentioning Schroeder and Bernstein, you might as well mention poor Adolf Lindenbaum (1904-1941) who actually proved it. | |
| Nov 27, 2013 at 1:44 | history | edited | Garrett Ervin | CC BY-SA 3.0 |
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| Nov 24, 2013 at 10:17 | comment | added | Garrett Ervin | Mohammad: I actually considered that idea myself, but it seems difficult to adapt the argument. For larger $\kappa$ it seems we would want to be considering the space of sequences $\beta^{\kappa}$ instead of $\beta^{\omega}$. A sequence $s \in \beta^{\kappa}$ would correspond to a $\kappa$-length nested intersection of intervals. But it is not at all clear in that case that even after taking an $\omega$-length intersection of nested $\tau$'s, we would be left with an interval isomorphic to (or even containing a copy of) $\tau$. | |
| Nov 24, 2013 at 4:33 | comment | added | Mohammad Golshani | Does the argument work for all uncountable cardinals if we assume GCH, as then for any uncountable cardinal $\kappa,$ we can consider a saturated dense linear order of size $\kappa$ | |
| Nov 24, 2013 at 2:38 | comment | added | Harry Altman | Oh, right -- equivalence is defined by tails, not by isomorphism. Silly me! OK yes that makes sense now. So does the rest. Yes it was often unclear whether "dense" meant "dense" in the order theory sense, or dense inside some larger ambient ordering which I think contributed to the confusion there. | |
| Nov 24, 2013 at 1:42 | comment | added | Garrett Ervin | Harry: every $\bar{s}$ is countable: for $s \in \beta^{\omega}$, $\bar{s}$ is the set of sequences of the form $r^{\frown}s'$, where $r \in \beta^{<\omega}$ and $s'$ is some tail sequence of $s$. There are only countably many $r$ (since $\beta$ is countable), and countably many $s'$. On your first comment: you're right, that sentence is unclear. The point is that each $\bar{s}$ is dense in $\beta^{\omega}$. In particular, if $a, b \in S = \cup_i \overline{s_i}$ and $a < b$ then there are $c, d, e \in S$ (we may even take $c, d, e$ all from $\bar{s_0}$) s.t. $c<a<d<b<e$. So $S \cong \eta$. | |
| Nov 24, 2013 at 0:28 | comment | added | Harry Altman | One thing I'm confused about -- why does the $\overline{s_i}$ all having order type $\eta$ imply that their union does? It doesn't seem that in general a countable union of things of order type $\eta$ again does; for instance, suppose you had $S=A+B+C+D$, where $A\cong D\cong \eta$, $B\cong \eta+1$, and $C\cong 1+\eta$. Then $A+C\cong B+D\cong \eta$, but $S$ itself fails to be dense. (Also, a nitpick: I don't see why $\overline{s}$ will be countable -- unless you meant to restrict to when $I_s\ne\emptyset$. But adding that restriction fixes that, I think.) | |
| Nov 23, 2013 at 20:32 | history | edited | Garrett Ervin | CC BY-SA 3.0 |
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| Nov 23, 2013 at 19:57 | comment | added | Garrett Ervin | Joel, you're quite right. I was thinking of $\times$ as being lexicographic from the left while writing my answer. Hopefully now it reads correctly | |
| Nov 23, 2013 at 19:47 | history | edited | Garrett Ervin | CC BY-SA 3.0 |
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| Nov 23, 2013 at 14:36 | comment | added | Joel David Hamkins | I'm reading your answer, but I am a bit confused in some places. Have you reversed the usual meaning of $\times$ on linear orders? That is, usually $\alpha\times\beta$ means $\beta$ copies of $\alpha$, but in your remarks on $\eta_i\times L_i$, it seems that you intend the other order. | |
| Nov 23, 2013 at 14:12 | comment | added | Joel David Hamkins | Garrett, I think I've lost confidence in the comments I had made at math.SE, since I no longer see why the forcing to make $\tau$ countable should not also create an isomorphism from $\tau$ to $\tau^2$. So I'm no longer certain whether the question is equivalent in the countable case or not. | |
| Nov 23, 2013 at 13:26 | history | edited | Garrett Ervin | CC BY-SA 3.0 |
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| Nov 23, 2013 at 13:21 | history | answered | Garrett Ervin | CC BY-SA 3.0 |