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2d
awarded  Nice Answer
2d
comment How to explain the concentration-of-measure phenomenon intuitively?
It's debatable whether my answer here is really different from the one posted by @foliations while I was writing mine, or whether it's more intuitive than a standard proof. But it uses fewer mathematical symbols and so has a chance of being verbally explainable.
2d
answered How to explain the concentration-of-measure phenomenon intuitively?
Jun
22
comment What lets the Square of Opposition fail in Intuitionistic Logic?
@j4nbur53 My post does answer your question, in the negative. Specifically, it shows that, with an appropriate intuitionistic background logic (a type theory, or just one allowing certain relativizations), one cannot have (Q) without $\phi\lor\neg\phi$. Emil had already answered the question in the case of a very weak background logic (containing only propositional logic).
Jun
19
comment What lets the Square of Opposition fail in Intuitionistic Logic?
I don't think it's wise to delete everything here. Even if it's not what you wanted, your question, Emil's comment, and my answer may be of interest or even of use to others. So I would prefer that these things remain on the site, perhaps with an addition to the question saying that it's not really what you wanted. If you start over, then that should be posted as a new question, perhaps with links to and from this one.
Jun
18
comment What lets the Square of Opposition fail in Intuitionistic Logic?
That I assumed the availability of relativization was explicit in my answer ("working in higher-order intuitinonistic logic") and in my first comment to Emil Jerabek. Nothing was smuggled in, I relativized both quantifiers correctly, and my deduction of $(\neg\neg\alpha)\to\alpha$ is correct. I don't intend to answer any more of your comments, but I suggest you read what I wrote (and what Emil wrote) rather than imagining what you think we ought to have meant to write.
Jun
18
comment What lets the Square of Opposition fail in Intuitionistic Logic?
Your claim that "(Q) holds trivially for domains that are subsets of a singleton" is true in classical logic, where the only subsets of a singleton are the singleton itself and the empty set, but it is false in intuitionistic logic. Your further claim that "so you won't derive anything from it" is wrong in the intuitionistic context because I derived $(\neg\neg\alpha)\to\alpha$ from it. I don't see the relevance of the rest of your comment, since you apparently insist on applying (Q) to an implication $(A\to\phi)$, whereas my proof applied (Q) to $\bot$.
Jun
18
comment What lets the Square of Opposition fail in Intuitionistic Logic?
To do this without types, you'd need something else to serve as the domain over which the quantified variables range. If you allow a domain that is a subset of a singleton $\{a\}$ and contains $a$ iff $\alpha$, then my argument still works. (I don't understand the part of your comment about "if $\phi$ is $\alpha\to\bot$" since that's quite different from the $\phi$ that I suggested.)
Jun
18
comment What lets the Square of Opposition fail in Intuitionistic Logic?
@EmilJeřábek You're quite right that I'm using only a tiny piece of higher-order logic, namely relativizing quantifiers to a subset of a singleton. (Since quantifying over a singleton is vacuous, I'm tempted to call this "subvacuous" quantification.) Also, I wasn't interpreting your comment on the question as saying (or hinting) that (Q) is strong. I mentioned that comment as a reason for my needing to assume something about the background logic.
Jun
18
answered What lets the Square of Opposition fail in Intuitionistic Logic?
Jun
16
answered Non-standard naturals and goodstein sequences
Jun
10
comment Distribution of dot product of two unit random vectors
First, consider the distribution of $|u\cdot v|^2$ when $v$ has one fixed value (and $u$ is uniformly distributed). That distribution is the same, regardless of the fixed value that you chose for $v$. So you also get the same distribution when you first pick $v$ any way you like, for example randomly with respect to any probability distribution you choose.
Jun
5
comment What is $\infty^6$?
I agree with @FranzLemmermeyer that it refers to the number of (independent) real variables needed to specify a motion. But I think the "motions" mentioned here are the members of the Euclidean group (rotations, translations, and maybe reflections). I also think that, in the good old days, people were not so careful about whether those 6 parameters are available globally or only locally. So in the end, it comes down to your guess about a 6-dimensional manifold.
Jun
4
comment Is $\clubsuit_{\omega_1}$ enough to get Suslin tree?
Terminological disaster: The guessing principle named "club" is not the same thing as club-guessing.
Jun
2
comment Minimum cardinality of lower-bounding subset of $[\omega]^\omega$
The bug in Step 1 is merely that the argument you give there proves (correctly) that $\lambda\geq\mathfrak d$ (not the reverse inequality that you claimed). If the argument in Step 2 were correct (i.e., if $I$ were in $\mathcal B$), then it would prove $\lambda\leq\mathfrak d$, which is not in general true.
Jun
2
comment Minimum cardinality of lower-bounding subset of $[\omega]^\omega$
Please un-accept the incorrect answer and accept Ashutosh's correct answer instead.
Jun
2
comment Minimum cardinality of lower-bounding subset of $[\omega]^\omega$
I don't think that the $I$ in the last paragraph is necessarily in $\mathcal B$.
Jun
2
comment Minimum cardinality of lower-bounding subset of $[\omega]^\omega$
I agree with this answer. In more detail: If $\mathcal A$ is an almost disjoint family and if $S$ is lower-bounding, then each element of $\mathcal A$ must have a different subset in $S$ because of the almost-disjointness. So $|S|\geq|\mathcal A|$.
May
26
comment Polish by compact is Polish?
I don't understand why people are upvoting my answer more then yours, since they're essentially the same and yours arrived a little earlier. Maybe they like my long-windedness. (Anyway, I upvoted your answer.)
May
26
answered Polish by compact is Polish?