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I like math and a few other things...


2d
comment Is it consistent that $\frak{d} < 2^{\aleph_0}$?
See the reference to Hechler in this answer: mathoverflow.net/a/29626
Dec
22
comment A question regarding $ZFC^{-}$
In that case, the powerset axiom is really hard to beat!
Dec
17
comment Classifying set theories whose standard models sharing the same ordinals are equal
This is my way of giving another plus one: thanks Ali!
Dec
15
comment Existence of internal toposes/inner models in a topos
An (elementary) boolean topos with nno is certainly able to construct the initial model of an essentially algebraic theory. Note that this model may be degenerate (equal to the terminal model) even if this is not true for the same theory in the real world. Booleanness shouldn't be that essential. Some parts of the classical theory rely on symbols of languages being distinguishable from each other but I don't think that's essential in this case.
Dec
13
revised totally disconnected and zero-dimensional spaces
fixed latex
Dec
13
comment A question regarding $ZFC^{-}$
@ThomasBenjamin: Ah, you don't require $\sigma$ to be a theorem of $\mathrm{ZFC}$? Yes, compactness still applies: if $\mathrm{ZFC}^- \vdash \sigma\to\mathrm{Powerset}$ then this is provable in $\Sigma_n\mbox{-}\mathrm{KP}$ for some $n$.
Dec
13
comment A question regarding $ZFC^{-}$
Replacement is basically a special case of collection. Depending on how it is formulated, collection might give you a set that is larger than the range of your function, but you can always trim it down using comprehension.
Dec
13
answered A question regarding $ZFC^{-}$
Dec
13
comment A question regarding $ZFC^{-}$
There are a great deal of such sets, so you probably need to be more specific.
Dec
13
comment A question on recursion and transfinite recursion in extensions of KP
And by "$\Sigma_n$ transfinite recursion" you mean the restriction to ordinals? Since both are consequences of $\Sigma_nKP$, what do you mean by "equivalent"?
Dec
13
comment A question regarding $ZFC^{-}$
Since powerset is a reasonably simple sentence, I can't really see what you're asking for.
Dec
13
comment A question regarding $ZFC^{-}$
I was using your definition.
Dec
13
comment A question regarding $ZFC^{-}$
Replacement is a consequence of ZFC-.
Dec
12
revised Higher order arithmetic and fragments of ZFC
added 1065 characters in body
Dec
12
comment Higher order arithmetic and fragments of ZFC
@ColinMcLarty This kind of technology, which I presume is similar to Zbierski's, won't do that. To get replacement in $\mathcal{V}$, you need something really close to choice. Basically, you need that for every definable relation $R \subseteq S_{n-1} \times S_n$ contains a definable subrelation $R'$ with size at most $S_{n-1}$. When $n = 1$, this is equivalent to countable choice. More generally, if $S_{n-1}$ is wellordered with size $\kappa$ this is equivalent to choice for families of size at most $\kappa$.
Dec
11
comment Higher order arithmetic and fragments of ZFC
The argument gives a little more than just $\mathrm{Z}_{n-1}^-$. Namely, the Mostowski Beta axiom (every extensional wellfounded relation on a set has a Mostowski collapse) is valid in $\mathcal{V}$. This gives a little bit of replacement but not much.
Dec
11
comment Higher order arithmetic and fragments of ZFC
And, of course, the set sorts are required to satisfy extensionality. Technically, they don't have to be actual sets, we just have membership relations $\in_i$ from $S_{i-1}$ to $S_i$, but with extensionality we may as well assume that they consist of actual sets.
Dec
11
comment Higher order arithmetic and fragments of ZFC
It's debatable but what I wrote in the second paragraph is pretty standard. The sort $N$ satisfies basic arithmetic ($0,1,+,\times$) with induction for all formulas (including formulas that involve quantifiers or parameters of higher sorts). Full comprehension for each set sort $S_1,\ldots,S_n$ (again, for all formulas). I think some would include some choice, in particular choice over $N$, but Colin specifically asked about omitting choice.
Dec
11
answered Higher order arithmetic and fragments of ZFC
Dec
11
comment Higher order arithmetic and fragments of ZFC
@JoelDavidHamkins: You're right about the strong AC, which implies collection. But I'm not sure how your arguments from the ZFC-P paper apply. Specifically, I don't see how to get $\omega_1$ singular and having $\mathcal{P}(\omega)$ exist.