Timeline for Is the measurable space $(\omega_1,\mathcal{P}(\omega_1))$ separable?
Current License: CC BY-SA 4.0
6 events
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| Jun 15, 2018 at 20:23 | comment | added | Ruy | Thanks, @Will. As you say, (CH) implies (LOG), if you allow me to refer to the property "$2^X=2^Y \Rightarrow X=Y$" by that acronym. Thus (LOG) may be seen as a weaker form of (CH) and it would be interesting to decide whether it is equivalent to any well known axiom in Set Theory. | |
| Jun 15, 2018 at 17:09 | comment | added | Will Brian | @Ruy: The Continuum Hypothesis (CH) suffices to prove that if $|2^X| = 2^{\aleph_0}$ then $|X| = \aleph_0$. But these two things are not equivalent: there are models where CH fails, and yet it is still true that if $|2^X| = 2^{\aleph_0}$ then $|X| = \aleph_0$. Similarly, the Generalized Continuum Hypothesis (GCH) implies that if $|2^X| = |2^Y|$ then $|X| = |Y|$. But again, these two things are not equivalent: it is consistent to have the GCH fail, and yet to have that if $|2^X| = |2^Y|$ then $|X| = |Y|$. So the (G)CH is enough for what you're asking about, but it's not exactly the same. | |
| Jun 15, 2018 at 16:56 | comment | added | Ruy | Will, thanks for clarifying that $2^{\aleph_0}=2^{\aleph_1}$ is also independent from ZFC. This made me curious as to what exactly one needs to prove that $2^X=2^Y \Rightarrow X=Y$. | |
| Jun 14, 2018 at 20:49 | vote | accept | Héctor | ||
| Jun 14, 2018 at 18:35 | history | edited | Will Brian | CC BY-SA 4.0 |
added 161 characters in body
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| Jun 14, 2018 at 18:26 | history | answered | Will Brian | CC BY-SA 4.0 |