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- Prove A $\times$ A equinumerous with A, for well ordered A without AC? [duplicate] June 2, 2023For everything below we assume A is infinite For sets A and B, let $A \equiv B$ denote there is a bijection between A and B. I know $\forall A, A \times A \equiv A \iff \text{Axiom of Choice is true}$ However, for particular A, such as the natural numbers, there is a way to […]wsz_fantasy
- Formalising the sentence "someone is plotting against me" June 2, 2023Someone is plotting against me. Can the above sentence be translated into predicate logic without using the existential quantifier? If not, is it because the sentence is self-referential?Richard Gasquet
- Cardinal number of the iterated set $A^{*}$, where $A=\{ a,b,c\}$. Why can't I use Cantor's diagonal argument? [duplicate] June 2, 2023I have a question which may sound silly to you, but I'm confident that I don't understand Cantor's diagonal argument very well to use it. Any provided insight would be appreciated. I was tasked with finding the cardinal number of the set $$A^{*} = \{\epsilon, a, b, c, aa, ab, ac ... \}$$ Which is […]THE_CRANIUM
- A question on existence of ultrafilters and cardinality. June 2, 2023Let $S$ be an infinite set with cardinality $k$. Let $U$ be an ultrafilter containing the filter of cofinite sets. Then, for any set $P \subset U$ such that the cardinal of $P$ is $Nulhomologous
- Benacerraf's identification problem and PA categoricity June 2, 2023Lately I have been interested in mathematical philosophy, and especially structuralism. In this setting, Benacerraf's famous paper is a classic and works as follows: Take the Zermelo ordinals ($x \to \{x\}$) and the Von Neumann ordinals ($x \to x \cup \{x\}$), those two are models of First order Peano Arithmetic $PA$, but they do not […]vigoux
- If you can prove $\bot$, why are you allowed to conclude anything? June 2, 2023I understand that $\bot \rightarrow A$ is a tautology, but I dont fully understand what it means that you can conclude anything after proving/assuming $\bot$. For example: Premise: $(A\rightarrow B)\rightarrow\bot$ Goal: $A\land(B\rightarrow\bot)$ You can prove this by considering the cases $B$ and $B\rightarrow\bot$. So consider the case where $B$. Now, all conclusions that we make […]lightyourassonfire
- full second order arithmetic: the truth and axiomatizations June 1, 2023What of the natural subsystems of the full second order arithemtic $\mathsf{Z}_2$ is not recursive ? What does it mean for a sentence of the full SOA to be true. I'm interested here in the notion of $\beta_k$ models where we do speak about truth. I would say, it depends, sometimes the set of natural […]user122424
- A syntactic deduction on Monoids' language June 1, 2023Let $\mathcal{L}=\{e,\cdot\}$ be the monoids' language. Give a syntactic deduction about: $$\forall x (x \cdot e = x \land e \cdot x = x) \vdash \forall z (\forall x \hspace{2mm} z \cdot x = x) \rightarrow z=e$$ I am stuck with this, I tried the following: \begin{matrix} \forall x (x \cdot e = x \land […]Superdivinidad
- How to prove that { ¬, →,$\forall$ } is a functionally complete set of connectors? June 1, 2023I already know how to prove that a set is functionally complete in propositional language without having the quantifier in the set. Now how can I prove tha this set { ¬, →,$\forall$ } is functionally complete in laguage of predicates ?Jazmine
- What are the order types of computable pseudo-ordinals with no c.e. descending chains? May 31, 2023The notion of a “computable pseudo-ordinal”, i.e. a computable linearly ordered set with no hyperarithmetical descending chains, is an old one going back to Stephen Kleene. Joe Harrison wrote the definitive paper on them in 1968, showing that any such linear order is either well-ordered or has order type $\omega_1^{CK}(1+\eta)+\alpha$ for some computable ordinal $\alpha$, […]Keshav Srinivasan
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