Timeline for Explicit ordering on set with larger cardinality than R
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2012 at 17:30 | comment | added | Joel David Hamkins | Andreas, yes, I agree. | |
| Jul 29, 2012 at 3:01 | comment | added | Andreas Blass | @Joel: The original question has a slightly cheaper answer than the product of $\mathbb R$ and its Hartogs ordinal. You could just use the disjoint union of $\mathbb R$ and its Hartogs ordinal, ordered by putting all of $\mathbb R$ before all the ordinals. | |
| Jun 2, 2010 at 21:12 | comment | added | Joel David Hamkins | Sune, in general it is a weak choice principle that every set admits a linear order. I'm not sure if this extends down to $R^R$, but I think it might. Perhaps $2^R$ is also a natural case: Can you linearly order $2^R$ without AC? I think that in the usual model with $\neg AC$, the set $2^R$ has no linear order... | |
| Jun 2, 2010 at 21:06 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Jun 2, 2010 at 21:01 | comment | added | Sune Jakobsen | "there are linear orders that do not map order-preservingly into R that are not larger than R in cardinality". Yes, the example I thought of was R^2 with lexicographical ordering. | |
| Jun 2, 2010 at 20:54 | comment | added | Sune Jakobsen | Doh! I was only trying to define a total ordering on the powers set of R or on R^R. Is this possible? | |
| Jun 2, 2010 at 20:52 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Jun 2, 2010 at 20:51 | vote | accept | Sune Jakobsen | ||
| Jun 2, 2010 at 20:45 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |