Noah Schweber
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 Feb 4 comment Size of a point Just to be clear, this is a good question, just not a good question for this site. Math.stackexchange would be a somewhat better fit, but I think that philosophy.stackexchange might be the best. Feb 1 comment Axiom of countable choice need for the cantor-bernstein theorem For the OP: The reason this question was downvoted and put on hold was that it is not appropriate for this site, which is for research mathematics. Math.stackexchange is a better site for such questions (I would not have answered, except that I forgot which site I was on. :P). As a side note, though, you should explain what you've tried and where you got stuck - in particular, for a question like this (where the standard proof goes through in ZF), explain what step(s) seem like they invoke choice. Feb 1 comment Help with math Real Analyis This question is not appropriate for this site, which is for research mathematics. Math.stackexchange is a forum for math questions at all levels, but if you ask this there you should provide motivation: what have you tried? Where did you get stuck? (You shouldn't just ask people to do your homework for you.) Feb 1 answered Axiom of countable choice need for the cantor-bernstein theorem Jan 25 revised a variant of the Kleene tree added 45 characters in body Jan 25 comment a variant of the Kleene tree @RobertLubarsky Easily arranged! Given a tree $T$, let $\hat{T}=\{\sigma: (\sigma(0), \sigma(2), \sigma(4), . . . , \sigma(2k), . . )\in T\}$; then every path through $\hat{T}$ computes a path through $T$, and $\hat{T}$ is of the same complexity as $T$, but $\hat{T}$ has continuum-many paths. Jan 25 comment a variant of the Kleene tree @RobertLubarsky Sorry, I should have said "$i$th c.e. tree." The point is, for a fixed Turing machine $\Phi_i$, I keep $\sigma_i$ on the tree I'm building - until $\Phi_i(\sigma_i)\downarrow=1$ (as well as for every predecessor of $\sigma_i$). then I kill $\sigma_i$, and only bring it back if for some $n$, I see $\Phi_i(\tau)\downarrow=0$ for every extension $\tau$ of $\sigma_i$ of length $n$. Jan 21 answered a variant of the Kleene tree Jan 21 comment Statements that Could be Forced by Ultrapowers I disagree with the votes to close - I don't think it's too broad for a useful answer. Jan 18 accepted Reverse-engineer forcing: am I reinventing the wheel? Jan 18 comment Constructive compactness for countable models? I believe both the proof of compactness for countable models from WKL and the reversal are constructive, according to most definitions of the term; so my understanding is that the status of the two are identical. (But I could be wrong.) Jan 17 comment “Partial-computably isomorphic” sets @bof No - every pair of sets satisfies that property (take $\hat{A}=\hat{B}=\mathbb{N}$). You need the computable partial maps to send $A$ to $B$ and $B$ to $A$, and invert each other when restricted to those sets. Jan 17 comment Existence of Spanning Tree implies Well Ordering Principle @ToddTrimble Ah, yes, I should have seen that. Jan 17 awarded lo.logic Jan 16 revised Compactness for countable models? deleted 35 characters in body Jan 16 comment Compactness for countable models? @CarlMummert Quite right, I was being silly: I took the last sentence of en.wikipedia.org/wiki/PA_degree#Properties as attributing the result to Simpson's survey paper (derp :P) "Degrees of unsolvability: a survey of results". Fixed! Jan 16 comment Existence of Spanning Tree implies Well Ordering Principle Note also that a well-order is very different from a total order. I do not know whether "every set can be totally ordered implies the axiom of choice. Jan 16 comment Existence of Spanning Tree implies Well Ordering Principle This is a good question, just not for this site - you should ask it at math.stackexchange. Jan 15 comment Is there an uncountable Borel almost disjoint family? @AndrésCaicedo I presume the construction is the Zorn's Lemma one: first show that there is a maximal almost disjoint family via Zorn, and then show that no countable almost disjoint family is maximal almost disjoint. (And in fact there is no Borel, or even analytic, maximal almost disjoint set - this was proved by Mathias in "Happy Families," see the first page of math.uni-hamburg.de/home/khomskii/papers/…) Jan 14 comment Is there an uncountable Borel almost disjoint family? @GeraldEdgar That's easy, though. Being a complete theory (coded appropriately as a subset of $\omega$) is a $\Pi^0_1$ property - in fact, the set of codes for complete theories is closed!