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I'm a graduate student at UC Berkeley, studying mathematical logic. I'm especially interested in reverse mathematics and abstract model theory.

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9h
comment Understanding Mathematics
You might also add some more details, and then ask the question at math.stackexchange (which is a different site).
1d
comment Induction and nonstandard halting times of standard machines
No, they should be the same - that's a much clearer way to phrase it.
1d
comment Fun math puzzle
One such place is math.stackexchange - this site is specifically for research-level questions (see the FAQs).
1d
revised Induction and nonstandard halting times of standard machines
edited title
1d
asked Induction and nonstandard halting times of standard machines
2d
comment Will this be a case of self plagiarism or will it annoy the referee?
This might be a better fit in academia.stackexchange?
May
18
comment ${\frak b}$ and ${\frak d}$ defined with $\leq$ instead of $\leq^*$
You don't need the "$+\aleph_0$," since $\mathfrak{d}$ is infinite.
May
18
answered ${\frak b}$ and ${\frak d}$ defined with $\leq$ instead of $\leq^*$
May
15
accepted Minimal degrees of structures
May
15
revised Minimal degrees of structures
added 28 characters in body
May
15
comment Minimal degrees of structures
For those, like me, who have not heard of pb-genericity before: ac.els-cdn.com/S0168007298000141/…
May
15
asked Minimal degrees of structures
May
15
comment standard deviation of an equation
This question is not appropriate for this forum, which is for research mathematics; math.stackexchange would be a better fit.
May
15
revised standard deviation of an equation
edited tags
May
8
comment Knuth's intuition that Goldbach might be unprovable
Well, Andres Caicedo and Terry Tao both give answers which explain why a $\Pi^0_1$ sentence $P$, if undecidable (we need $P$ to be non-disprovable, not unprovable) in ZFC, is true. Moreover, contrary to your answer, no consistency or soundness assumption is needed here - if ZFC is inconsistent then the hypothesis "$P$ is unprovable in ZFC" is never satisfied, and regardless of whether ZFC is sound it cannot fail to prove a true $\Sigma^0_1$ sentence. So the answers I mentioned are completely accurate, and there's no need for further elaboration.
May
8
comment Knuth's intuition that Goldbach might be unprovable
But this itself is non-absolute: a still larger model $O\supseteq N$ might contain a bijection between $X$ and the set $N$ thinks is $\mathbb{R}$, and moreover might contain no new reals - so $O$ would think $X$ is not a counterexample to CH. And this can keep going forever.
May
8
comment Knuth's intuition that Goldbach might be unprovable
Absoluteness is a kind of non-verifiability: let's say I have a model $M$ (nice and well-founded) of ZFC and a set $X\in M$. Now, $M$ might think $X$ has size continuum, in the sense that $M$ contains a bijection between $X$ and a set $A\in M$ which $M$ thinks is the set of all reals. But, there might be real numbers not in $M$ - for instance, if $M$ is countable. So there might be a "larger" model of ZFC, $N\supseteq M$, such that there is no bijection - in $N$ - between $X$ and the set $N$ thinks is $\mathbb{R}$. In $N$, $X$ might be a counterexample to CH.
May
8
comment Knuth's intuition that Goldbach might be unprovable
The point is that a counterexample to Goldbach, if it exists, will be both easy to find and easy to prove a counterexample. By contrast, for something like the Continuum Hypothesis, a counterexample might be (1) impossible to effectively describe, so ZFC wouldn't be able to even refer to it, and (2) impossible to prove a counterexample even if it were definable. Here, we learn nothing from the statement "ZFC doesn't prove CH," since ZFC just doesn't have the ability to build and verify a counterexample if one exists. (This is vague, but hopefully helpful?)
May
8
comment Knuth's intuition that Goldbach might be unprovable
Well, the point is that (1) a counterexample to Goldbach can be verified: if $a\in\mathbb{N}$ is a counterexample to Goldbach, then $PA$ (in fact, just the ordered semiring axioms) can prove that (the numeral corresponding to) $a$ is a counterexample to Goldbach; and (2) Goldbach quantifies over a set which is appropriately enumerable (the natural numbers), as opposed to a more wild set (such as the set of sets of reals, as the Continuum Hypothesis does).
May
8
comment Knuth's intuition that Goldbach might be unprovable
That's not quite true - you need the property $P$ to be appropriately simple (barring additional assumptions, this means $\Sigma^0_1$). That is, you need "being a counterexample" to be absolute.