$\Bbb Z_6$ would be a counterexample (nilradical is $\{0\}$ which is not prime because it doesn't contain $2$ or $3$ despite the fact that $2\times3=0$)
So, clearly nilpotents are contained in prime ideals 'cause $a^n=0\in P\implies a\in P$.
Hello. Suppose $R\to S$ is a ring homomorphism. When is there a factorization $R\to \mathrm{End}(A)\to S$ through the endomorphism ring of some abelian group?
If $M$ is an (oriented) Riemannian manifold and $U$ a non-empty subset of $M$, does $U$ have positive measure with respect to the volume measure on $M$?
I knew the very basics, including how they are recusively constructed and how to notate them, but other than that, I have not worked much with them yet
Heredity finite sets are those sets whoose subsets are finite sets, whoose subsets are finite sets and so on
The above musing can be more precisely stated as follows:
$S$ is a set that has finite cardinality but contains subsets of infinite cardinality
I came up with $S$ when trying to comprehend this line in this MSE:
> In particular, P(ω) (and therefore ω) injects into P(P(X)) for any infinite set X. That is to say that P(P(X)) is always either finite, or D-infinite.
Is the powerset of every Dedekind-finite set Dedekind-finite?
I think this statement can be written in $\textbf{Set}$: If every mono (=injection) $f: A \to A$ is iso (=bijection), then every mono $g: 2^A \to 2^A$ is iso.
(Please edit if necessary)
Does the answer depend on some Choice princip...
@Secret In the second sentence, they mean P(P(X)) is always either finite (which happens when X is finite) or D-infinite (which happens when X is infinite).
hmm... I thought D-infinite and infinite are equivalent even in the absence of choice, as I recall the weird sets that you get when discarding choice are the infinite dedekind finite sets?
Another thing I am pondering about infinite D-finite sets $D$ are their intersection with countable sets, in that it is known that it can only be a finite set. But then if $D$ have nonempty intersections with countably many countable sets and all these intersections are disjoint, then it will mean $D$ has a subset consists of countably many elements and thus cannot be D-finite
Hm. I didn't realize that Dedekind infinite was equivalent to the existence of an injection from $\omega$ into it. But I think I see how to prove it
(Not a reply to what you just wrote)
If $A$ is Dedekind-infinite there's a bijection from $A$ to a proper subset $B$. Let $x$ be in $A\setminus B$; then the injection from $\omega$ into it is $n\mapsto f^n(x)$
o wait nvm. If my infinite dedekind finite subset is not amophous to begin with, then it is trivial to partition this set union with any finite number of finite sets into two infinite sets
I'm trying to find any reference that defines what a "conjugate direction field" is. I think this is a topic of differential geometry, but I can't find any reference for this, I have many papers that mention this concept but none of them provide a good definition or reference I can look up.
Is t...
@Bubaya The tensor product functor is adjoint to the Hom functor. But now Hom(-, M) $\cong$ Hom(-, N) implies M $\cong$ N as R-modules by the Yoneda lemma.
Uncountable is some uncountable set. I am not sure if I need to make it dedekind finite in order to avoid the possibility of picking up a countable subset from it, but I think without the axiom of countable choice, it should be impossible for me to pick countably many i from an uncountable set
(also the uncountable set is not linearly ordered, as otherwise I would have introduced a linear ordering on $A$ and it will no longer be amorphous, I think...)
Assuming the Axiom of Choice, every cardinal is either finite (i.e., an element of $\omega$) or Dedekind-infinite (i.e., in bijection with a proper subset of itself). This dichotomy is not true in ZF, however; in particular, it is consistent with ZF that there exist sets which are non-finite but ...
I am suspecting an amorphous set can also be produced by a countable union of finite sets, as long the index set is only a poset to prevent obtaining a countable sequence
And it also seemed that partitions are an important tool in studying the structures and properties of amorphous sets
But I think the point is now to give a heads up to all the hate messages and harsh criticisms youtube is getting for their shitty algorithms prior to this
Also since this is machine learning, I suspect there's going to be a lot of appeals that has to be manually dealt with at the start for the algorithm to learn monetizability
Yea you can tell Patreon is starting to fancy themselves a 'big deal' because theyre starting to kick all the 'adult content' creators on the platform to the curb
of course i first assume it is , but this assumption means there is some combination of the above operations that gives $\neg$ , i dont know nothing about the combinaiton.
I'm asking because there's no minimal functionally complete set of connectives with more than $3$ elements so one of those can be thrown away and that seems weird
connectives means $n$-ary connectives with $n\le 2$ here
@Liad Try thnking about it this way. You need a function $f$ where $f(1) = 0$ and $f$ can be written as a composition of $\vee, \wedge, \rightarrow, \leftrightarrow$. What happens where you make $p$ true in $p \wedge p, p \vee p, p \rightarrow p, p\leftrightarrow p$?
$\land,\lor,\rightarrow,\leftrightarrow$ are truth-preserving, in the sense that they return true under a valuation that assigns true to all variables, while $\neg$ isn't
Wait, why not? if $\neg$ is a combination of the above, then by the "truth-preserving" of those operation we will get that putting true to "p" will give true, and it supposed to be false, or have i did not understand what you were saying?
Think about what we've said so far as a base case. What happens then if we do another connection? So let $q$ be any of $ \{p \wedge p, p \vee p, p \rightarrow p, p \leftrightarrow p \}$. If $p$ is true, what is the truth value of $q \wedge p, q \vee p, q \rightarrow p, q \leftrightarrow p$?
I've been dabbling with a medical dataset i've managed to get my hands on and i've built a simple feed-forward network using baseline characteristics.. I've been measuring the models performance using the area under the curve. I'm calculating the auc on both the training set & test set but for some reason i'm getting a much higher auc on the test set. train-.89 test=..94
I'm a little concerned as to why theres a higher auc on the test set? If anything i'd expect it to be a little lower than the train because of overfitting
@liad So we can do a proof by induction then. We know the base case, start with $p$ and connect it t any of the connectives we're using here. If p is true it remains true. And for the inductive step we need something like if $q$ and $r$ are well-formed formulas of $p$ and the connectives that are true when $p$ is true, then using any of our connectives to connect $q$ and $r$ still returns a true result.
Is the powerset of every Dedekind-finite set Dedekind-finite?
I think this statement can be written in $\textbf{Set}$: If every mono (=injection) $f: A \to A$ is iso (=bijection), then every mono $g: 2^A \to 2^A$ is iso.
(Please edit if necessary)
Does the answer depend on some Choice princip...
