bio | website | |
---|---|---|
location | ||
age | ||
visits | member for | 4 years, 8 months |
seen | 3 hours ago | |
stats | profile views | 3,980 |
I'm a graduate student at UC Berkeley, studying mathematical logic. I'm especially interested in reverse mathematics and abstract model theory.
Mar 23 |
awarded | Enlightened |
Mar 23 |
awarded | Nice Answer |
Mar 23 |
revised |
Does PA+Con(PA) entail the existence of non-standard models of PA?
added 188 characters in body |
Mar 23 |
comment |
Does PA+Con(PA) entail the existence of non-standard models of PA?
For what it's worth, I disagree with the vote to close. |
Mar 23 |
answered | Does PA+Con(PA) entail the existence of non-standard models of PA? |
Mar 18 |
comment |
Surjectivity from union of a set system to the set system
I don't think this works - as I wrote in my comment, I don't think there's an injection $f$ from $\bigcup\mathcal{A}$ to $\mathcal{A}$ satisfying $a\in f(a)$, since 0 and 1 must both be sent to the same place. |
Mar 18 |
comment |
Surjectivity from union of a set system to the set system
If you mean $\{\{0, 1\}\}\cup . . .$, I don't think this works - what is the $f$? Otherwise - if we treat numbers as ordinals - 0 can't be in $\mathcal{A}$ since $\mathcal{A}$ should consist of nonempty sets only. |
Mar 18 |
answered | Surjectivity from union of a set system to the set system |
Mar 17 |
awarded | Taxonomist |
Mar 15 |
comment |
Surjective (strong) reducibility of Borel equivalence relations
If a Borel equivalence relation $E$ has only countably many equivalence classes, isn't each class Borel? |
Mar 15 |
awarded | Nice Answer |
Mar 15 |
revised |
Independence of the countable axiom of choice
added 58 characters in body |
Mar 15 |
comment |
Independence of the countable axiom of choice
Yes, exactly: without the axiom of choice, it is possible that a countable union of countable sets - in fact, a countable union of finite sets - might not be countable! |
Mar 14 |
revised |
Independence of the countable axiom of choice
deleted 2 characters in body |
Mar 14 |
comment |
Independence of the countable axiom of choice
A fun exercise - in the argument I gave above, it would not obviously work if we replaced $A_n$ by the set of subsets of $A$ of size $n$ - it is important that we work with sequences, or we need a little bit extra justification. Why? |
Mar 14 |
comment |
Independence of the countable axiom of choice
BTW, for basic exercises in proofs involving choices, mathstackexchange is maybe a better fit. |
Mar 14 |
comment |
Independence of the countable axiom of choice
Yes, that hint is basically the whole problem - given an infinite set $A$, for each $n$ the set $A_n=\{X: X$ is an injection from $n$ into $A\}$ is nonempty, so - by countable choice - the family $\{A_n: n\in\omega\}$ has a choice function. What is this choice function? Well, it's essentially a sequence of sequences of distinct elements of $A$, of increasing length. Concatenating these sequences gives an infinite sequence of elements of $A$ - a single element might occur multiple times in this sequence, but infinitely many elements will occur. Delete repetition, and we're done. |
Mar 14 |
comment |
Independence of the countable axiom of choice
That's a very good point - I like the embedding theorems partly because they lead to a nice combinatorial question, "What's the lowest rank where we can shove this construction in?" |
Mar 14 |
comment |
Independence of the countable axiom of choice
To elaborate on (2), the issue of how to transfer those core ideas from the context of urelements to genuine ZF results was seriously investigated by Jech and Sochor, and later by Pincus, who developed powerful machinery to say: "Oh hey, we can do this thing with urelements, so we must be able to do it without them." |
Mar 14 |
comment |
Independence of the countable axiom of choice
You may also be interested in the following: another question mathoverflow.net/questions/104450/road-to-solovays-land; Timothy Chow's article arxiv.org/abs/0712.1320; Cohen's account projecteuclid.org/download/pdf_1/euclid.rmjm/1181070010; and many others. |