Hello, quick question: Can a model interpret a formula both true and false or not interpret it in first order logic? Looking at the definition, it looks like the model is inheriting the internal logic of the meta-object/model. Just one or two sentence of guidance would be nice.
Let $X$ be a finite, commutative, boolean monoid with a $0$ element.
Let $A = \{ x \subset X : 0 \in x$ and $x^2 \subset x\}$ where $x^2 = \{ab: a,b \in x\}$. Define for $x, y \in A$, $x + y = x \Delta y \cup \{0\}$. Then since $x^2 \subset x$ for $x \in A$, we actually have $x \subset x^2$ as...
@LeakyNun Where I am confused: Don't we define an interpretation of a formula recursively, so doesn't your statement needs a proof, it's not a direct property through definition?
@LeakyNun So can't we get a case where a formula is both true and false? I am trying to imagine that a contradiction in the model can lead to a case where the interpretation is both true and false.
so by exactly this part of the definition, we argument that an interpretation of a formula under a model cannot be both true and false, and because this interpretation is for every formula defined, we know that it must be true or false, am I right?
Thank you! I want to clarify one more thing, when we say that a formula is independent of an axiom system, we mean that it has models which model this system and the formula and also this system and the negation of the formula. Am I right?
@LeakyNun Aren't they the same thing? A system can prove a formula if this formula is satisfied in every model of this system, right? So, we also cannot prove the negation of this formula and this directly implies that there must be models for each case. Am I right?
So we put law of excluded middle to this kind of first order logic through using it in our metatheory if necessary, one example being completeness theorem?
And I read a phrase "we believe that first-order logic is consistent". What does it mean?
Truth is something specific to a particular structure, or a particular model of a theory. Sometimes, when a theory has a canonical model, we abuse the terminology and when we say "true" what we mean is "true in the canonical model". This is the case with the natural numbers and with the real numb...
If x is in a baire space and there's an injection function mapping one set of natural numbers omega to another set of natural numbers omega...is it open? I think it is because there is an unique mapping so we have interior points and open balls and neighborhood
Let $f: [a,b] \rightarrow \mathbb{R}$ be continuous. Suppose $f(x)\geq 0$ for each $x\in [a,b]$ and $f(c)>0$ for some $c\in (a,b)$. Then $\int^{b}_{a}f>0$
My Attempt: Suppose $f$ is continuous for for $\epsilon=f(c)>0$ $\exists$ $delta_1>0$ such that whenever $x \in (c-\delta_1,c+\delta_1)$, we have $f(x)>f(c)$.
Now, how do I show that $\int^{b}_{a} f(x)$ $\geq \delta_1 f(c)
@TedShifrin that's the correct statement: Let $f: [a,b] \rightarrow \mathbb{R}$ be continuous. Suppose $f(x)\geq 0$ for each $x\in [a,b]$ and $f(c)>0$ for some $c\in (a,b)$. Then $\int^{b}_{a}f>0$
okay, so $f(x)>\frac{f(c)}{2}$ for all x in a delta neighbourhood of c, so would the right way of approaching it would be by constructing a partition such $a_0,a_1,a_2....,b$ such that $a_i-a_{i-1}$ < $\delta$ for each i?
if G is a finite group, a group ring k[G] for k a field is artinian, so has a decomposition into a direct sum of indecomposable projective modules. how do you actually find these modules? in practice i mean
i would have said the free module generated by $[0]+[1]$, but explicitly $$(a[0]+b[1])([0]+[1])=a[0] + b[1] + a[1]+b[0]=(a+b)[0]+(a+b)[1]$$ which i think is still a free module generated by $[0]+[1]$ and i could even treat it as a R-module of rank 1 if i wanted
and when a module has only finitely generated submodules we call it noetherian. is there some nice conditions of R that make R-modules always finitely generated?
@usukidoll so that's a good place to start with this question. Figure out what the open sets in $^\omega\omega$ look like, and then you can better understand whether or not $I$ is open or not.
@LeakyNun because when you treat k[G] as a module over itself, this corresponds to the standard k-linear representation of G, and this decomposition is somehow related to the decomposition of that representation into its irreducible subrepresentations?
> For the group algebra of a finite group, the (isomorphism types of) projective indecomposable modules are in a one-to-one correspondence with the (isomorphism types of) simple modules: the socle of each projective indecomposable is simple (and isomorphic to the top), and this affords the bijection, as non-isomorphic projective indecomposables have non-isomorphic socles.
Let $X$ be a finite, commutative, boolean monoid with a $0$ element.
Let $A = \{ x \subset X : 0 \in x$ and $x^2 \subset x\}$ where $x^2 = \{ab: a,b \in x\}$. Define for $x, y \in A$, $x + y = x \Delta y \cup \{0\}$. Then since $x^2 \subset x$ for $x \in A$, we actually have $x \subset x^2$ as...
