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Apr
12
accepted The idea of “generators” for arbitrary categories
Apr
11
comment The idea of “generators” for arbitrary categories
I am not able to understand what you mean with "$\{ x \circ f \mid x \in X \}$"; can you tell me what is $f$?
Apr
11
comment The idea of “generators” for arbitrary categories
@espen180: Be careful because the upset generated is not the same thing than the filter generated. The notion of filter is isually used in the context of lattices (a particular kind of partial orders), and filters are always closed under finite meets, while upsets might not be closed under finite meets.
Apr
11
comment The idea of “generators” for arbitrary categories
@Mariano: I am afraid that this wikipedia notion does not coincide with the one I have written above for partial orders (although the same word "generator" is used), so it is not adequate for what I really want to generalize.
Apr
11
asked The idea of “generators” for arbitrary categories
Mar
31
revised Need help locating a paper
deleted 4 characters in body
Mar
31
answered Need help locating a paper
Mar
28
comment nonisomorphic graph drawings
You can try to use some software for drawing them, this is always helpful (see for instance the "sage" code in the question ask.sagemath.org/question/3473/… )
Mar
28
comment How to explain the power of PA to non-logicians
This essentially means that all $\Sigma_1$-setences that are true in the standard model are provable in PA. This result can be find in most of logic books dealing with PA (and Robinson's Q). In particular, you can take a look at the notes by Peter Smith logicmatters.net/igt/godel-without-tears (in the current version is Theorem 17 in Episode 5, page 40)
Mar
28
comment How to explain the power of PA to non-logicians
Do you consider the fact that "PA (indeed Robinson Q is enough) is $\Sigma_1$-complete" strong enough?
Mar
27
comment Mathematical intro to Turing machines
One example is the following "classic" written by Martin Davis amazon.com/Computability-Unsolvability-Prof-Martin-Davis/dp/…
Mar
19
comment Show that a recursively inseparable pair of recursively enumerable sets exists
Yes, there is a standard proof (for two sets whose definiton only involves Turing machines). One place where you can find this is in Theorem 3.3.5 of the wonderful notes "Syllabus Computability Theory" written by Sebastiaan A. Terwijn math.ru.nl/~terwijn/teaching.html After you know these two sets are inseparable you can easily prove the inseparability of sets with a "logic" flavour (like in Biderman answer)
Mar
18
comment Books (and supporting material) that are useful in deconstructing one's intuition?
@Sabysachi: torus solution?
Mar
15
comment Book on the first-order modal logic
@user132181: The standard translation only works for propositional modal logic (and it translates these formulas into first-order, non-modal, formulas). Thus, right now I am afraid that your question does not make sense: could you clarify your question?
Mar
13
comment When is a Decidable Set Decidable?
I suspect Russell is confusing two different uses of the term "undecidable". We have "undecidability in a formal theory" which applies to a sentence and refers to its unprovability. And then we also have "undecidability in a computability setting" which applies to a family (set) of finite objects and refers to the fact that this set cannot be computed by a Turing machine. For this second use there are some people which have advocated for using the word "uncomputable" instead of "undecidable", in order to avoid these misunderstandings.
Feb
15
comment Complexity of Recursively Inseparable Sets
@Easterly: What do you mean here with the word "Complexity"?
Jan
31
answered Famous Math Texts in Spanish?
Jan
20
answered List naturals in ascending product order
Jan
10
awarded  Yearling
Jan
10
comment Why are mathematical proofs that rely on computers controversial?
@DumpsterDoofus (and the upvoters): What is the statement you talk about?