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bio website jdh.hamkins.org
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I am a professor at the City University of New York, at the College of Staten Island and the CUNY Graduate Center, living in midtown Manhattan. My main research interest lies in mathematical logic, particularly set theory, focusing on the mathematics and philosophy of the infinite. A principal concern has been the interaction of forcing and large cardinals, two central concepts in set theory. I have worked in group theory and its interaction with set theory in the automorphism tower problem, and in computability theory, particularly the infinitary theory of infinite time Turing machines. Recently, I am preoccupied with the set-theoretic multiverse, engaging with the emerging field known as the philosophy of set theory.


2d
awarded  Enlightened
Dec
20
comment How can i be distinguished from -i?
Meanwhile, I should say that I don't actually agree with Shapiro's account here, since my view is that we habitually operate, perhaps without even thinking about it much, in a mathematical realm having additional structure in which the objects become discernible.
Dec
20
comment How can i be distinguished from -i?
But on to your larger point, yes, the case of $i$ and $-i$ is just one of many common cases of indiscernibility, and Shapiro is trying to grapple with the general problem, not just the case of $i$. But the case of $i$ is a little special because it seems that we mathematicians do definitely treat $i$ as a definite description, in a way more so than in other cases of indiscernibility.
Dec
20
comment How can i be distinguished from -i?
The theory of the complex numbers (in the language of fields, say) does not distinguish $i$ from $-i$, unless you add a constant symbol for $i$, which of course is the point under discussion. The philosophical issue here is that we all go around talking about $i$ and $2i+3$ and so on, but there is no mechanism at all to ensure that what I call $i$ is the same as what you call $i$, or indeed the same as what I called $i$ yesterday, or two lines earlier in my argument. But meanwhile, we write equations that rely on the identity of those referents, not just on their automorphic nature.
Dec
20
comment How can i be distinguished from -i?
See Stewart Shapiro's article, An "i" for an i (journals.cambridge.org/action/…), in which he analyzes the interesting and perplexing philosophical issue arising here, namely, how it is that mathematicians seem able to treat $i$ as a definite description, in light of the indiscernibility arising by conjugation.
Dec
20
comment In set theories where Continuum Hypothesis is false, what are the new sets?
A sequence of statements $\phi_n$ forms a "ratchet", if each is a button, each necessarily implies the previous, and in any model of set theory in which $\phi_n$ is not true, you can force to make $\phi_n$ true without making $\phi_{n+1}$ true. Thus, you can turn the sound louder and louder, but never quieter.
Dec
20
comment Complete resolutions of GCH
Oh, your answers and questions are outstanding!
Dec
19
comment Complete resolutions of GCH
No problem! It is good to have several answers, even if they overlap.
Dec
19
answered Complete resolutions of GCH
Dec
19
awarded  Enlightened
Dec
19
awarded  Nice Answer
Dec
18
comment Classify set theories whose transitive models sharing the same sets of ordinals are equal
In general, models of ZF with the same sets of ordinals need not agree on sets of sets of ordinals. But if they satisfy V=L(P(Ord)), then they do. If two models of V=L(R) have the same sets of ordinals, then in particular, they have the same real numbers and the same ordinals. And one can prove by induction on $\alpha$ that $L_\alpha(\mathbb{R})$ is the same in each of them, and so they are equal. The same argument works with P(Ord) in place of $\mathbb{R}$.
Dec
18
comment When does Skolemization require the axiom of choice?
@EmilJeřábek, why not post an answer explaining the situation? To Skolemize the theory and have a conservative extension does not require any AC; to get some model of that Skolemized theory generally requires some choice, even when you start with a satisfiable theory; to expand a given model to satisfy the Skolem theory is fully equivalent to AC.
Dec
18
comment Classify set theories whose transitive models sharing the same sets of ordinals are equal
For example, the theory $ZF+V=L(P(\text{Ord}))$ is $P(\text{Ord})$-categorical, for the same reasons that $V=L(\mathbb{R})$ is.
Dec
18
comment Classify set theories whose transitive models sharing the same sets of ordinals are equal
I think you might mean that $X$ is a set of sets of ordinals? I don't think this will be necessary, however, since it could be that $X$ is a proper class of sets of ordinals, and we have $V=L(X)$. But if we make your assertion where $X$ is a class, it seems no longer to be first-order expressible.
Dec
18
comment When does Skolemization require the axiom of choice?
Yes, as I indicated I was referring only to the model-expansion version of conservativity (and this also is often called conservativity--is there another term?), and this is equivalent to AC. But even if you are just looking at the theory version, if we have a collection of finite sets $A_i$, and we write down the corresponding theory $T$ asserting that $f(i)\in A_i$, using constant symbols for the elements of each $A_i$, then it will require choice for finite sets to get a model of this theory. The proofs-are-finite observation shows that the theory is conservative, but it has no models!
Dec
18
comment When does Skolemization require the axiom of choice?
But is the version of the Skolem theorem that one would formalize in a weak fragment of arithmetic really the same theorem as the one we might want to consider in a set theoretic context, with uncountable languages and whatnot? The conservativity result on Skolem functions (asserting that one can expand any model to a model with a Skolem function) is obviously equivalent to AC, for the reasons that have been mentioned.
Dec
18
answered Classify set theories whose transitive models sharing the same sets of ordinals are equal
Dec
18
comment When does Skolemization require the axiom of choice?
Isn't this obviously a choice principle? The function $f$ is making choices for you, choosing for each $x$ a particular witness $y$, namely $y=f(x)$, among all the possible witnesses $y$ that could work. It seems like the very essence of choice, and any instance of choice could be converted into such a case.
Dec
17
comment A question on models of set theory and Lebesgue measure
I edited to give a little more explanation of the complexity. Basically, it is $\Sigma_2$ because you can verify it in any sufficiently large $V_\theta$, and indeed, you don't have to go very high. Statements that are not $\Sigma_2$ must involve set-theoretic properties that stretch up arbitrarily high in the set-theoretic universe. Lebesgue measurability is not like that, since once you have the reals and all the sets of reals, then all the issues about measurability will be determined.