bio | website | comp.nus.edu.sg/~a0078129 |
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location | Singapore | |
age | ||
visits | member for | 2 years, 8 months |
seen | 3 hours ago | |
stats | profile views | 331 |
Problem mining.
Jan 22 |
comment |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
@AndresCaicedo: I did have in mind that the length of an interval equals its measure. |
Jan 22 |
comment |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
@Burak: I mean ~CH is a consequence of ZFC + there is a measure on \mathbb{R}, and this measure is necessarily not Lebesgue. You are right over ZF they are not really orthogonal. |
Jan 22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
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Jan 22 |
comment |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
I meant Lebesgue measure is total otherwise choice is not necessarily ruled out. Sorry for the ambiguity. |
Jan 22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
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Jan 22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
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Jan 22 |
comment |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
Sorry Cantor space and R confusion. I remedied this. Isn't that such measure simply does not exist if R is a countable union of countable sets, by $\sigma$-additivity? |
Jan 22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
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Jan 22 |
comment |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
@AndresCaicedo: thanks! Which survey by Fremlin do you refer to? |
Jan 22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
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Jan 22 |
comment |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
@bof,@Jesse: the negation of CH is indeed a consequence of ZFC+exists a total measure. What I'm really interested is the extension of Lebesgue measure. I'll edit the question. |
Jan 22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
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Jan 22 |
asked | Consequences of ZF+“all subsets of reals are Lebesgue measurable” |
Jan 13 |
comment |
Does PA prove a sentence asserting that all of I-sigma(n) theories are consistent?
Suppose you could formalize $\forall n I\Sigma_n^0$ in PA in the first place, then for any model of PA, $\{n: I\Sigma_n^0\vdash \varphi\}$ is an arithmetic set in the model. Take the least element and it would be a real natural number. |
Jan 11 |
answered | Does PA prove a sentence asserting that all of I-sigma(n) theories are consistent? |
Nov 10 |
comment |
Cohesive set with degree below non-high Martin-Löf random reals
So the question is really for MLR which neither low or high. |
Sep 24 |
awarded | Autobiographer |
Jul 2 |
awarded | Curious |
May 20 |
awarded | Yearling |
Feb 15 |
answered | Can a Decidable Theory Have Non-recursive Models? |