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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!