Keinstein
• Member for 10 years, 4 months
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• Dresden

Most of the answer has already been given with the reference to the auditory system. It dissolves the sound into a certain base functions. I don't know of a good description of their behaviour, but ...

In general I would see the two main classes, that can are distinguished in German language. The term Verband refers to an algebra (in the sense of universal algbera) with a join and a meet operation. ...

Warning: This anwer should be treated with caution. For further information see the comments. As a partial anwer consider the set $\mathcal L$ of order types of linear orders. Then $\bar{\mathcal L}$ ...

If you don't want to use $\parallel$ you might try $\not\lessgtr$ (\not\lessgtr) or something that looks better, maybe $\,|\!\!\!\!\!\lessgtr$. This would complement the usage of $\lessgtr$ for ...

Actually you should differentiate between a dense order relation and a dense subset of an ordered set. The order relation is dense when (3) holds. A $Y$ is dense in an ordered set $(X,≾)$ iff a ...

The first thing you should keep in mind is that there is a difference between the Hasse diagram or the corresponding neighbourhood relation and the order itself. The next problematic term is the word “...

For the sake of contradiction, take an antichain $C \subseteq A$ in the well-ordered set $(A, \leq)$, that is a set of pairwise non-comparable elements ($\forall x,y \in C, x \not \leq y \text{ and } ... View answer 1 votes As Gabriel Romon commented: Lagrange interpolation does the trick. The polynomials form a vector space (which works very similar to the 3D coordinate space) where the coefficients work as ... View answer 1 votes Just as you didn't mention groups: There has been done much work on po-groups as well. As far as I know, some people are working on quasiorders (preorders) on algeraic structures. Clone theory tells ... View answer 1 votes In case of an infinite dimensional vector space there exists a basis that represents the order. Nevertheless the separating “Hyperplane” may be the whole vector space itself. For example take$\...

There are several methods to do that. The main question is how the lattices are given and which properties they have. In case the lattice is doubly founded, it is sufficient to consider all bijective ...

Additionally to he two answers given so far: I think the main link between maths and music is hidden behind the scenes: Rules: Composition as well as Improvisation heavily relies on a reduced ...

Let $\equiv = \leq \cap \geq$ the equivalence relation of the preorder. Then, ${(P,\leq)_/}_\equiv$ is an ordered set. Let $C$ be a maximal chain in ${(P,\leq)_/}_\equiv$. And $c:C\to P$ a choice ...

The main idea is: Distributivity can be (theoretically) destroyed by a single counter example. Embedding means injectivity: if $x\neq y$ then $e(x)\neq e(y)$. So if an equation has a counterexample in ...

I interpret „being shaky“ as: The differences between nearby samples are constantly swapping their sign. This leads me to the suggestion: Model it using ARMA (autoregressive moving averages) models. ...

As you already stated there are orbit partitions into 2 2-element sets and into a singleton and a 3-element set. Furthermore a group action can be considered a group homomorphism of on group into ...

As Matt Samuel stated, $ℤ\cup\{-∞,∞\}$ is a lattice that consists only of irreducible elements. It is even a complete lattice with this property. On the other hand in the set of extended reals ...

As already said in the comments: every pair means every unordered pair, that can be formed of elements of the given base set, in our case the lattice or ordered set. An upper bound of a set $S$ is an ...

Well, I assume, you mean $(X,Y) ∈ R$ iff $X⊂Y$ or $X=Y$. Let us further assume that $⊆$ is a partial order relation, while $⊂$ is irreflexive ($∀X:X\not⊂X$). Then $⊂$ is called “strict (partial) order“...

To avoid confusion, you should provide a definition of a filter. I'm using the more general definition of an order filter that is sometimes called an upset. But your suggestion looks not very ...

Another proof is based on the fact that a join irreducible element has a unique lower neighbour. Let $A⊆X$ and $B⊆X$ with $A≠B$ be two irreducibles in $\mathcal L$. Let further for any element \$A∈\...

If it is a partially ordered group then it is also a lattice ordered group. The usual way is to prove $$a≤b⇒xay≤xby.$$ Edit: Your task is to prove that it is a partially ordered group. Your ...