The set of funcoids is a co-frame (without axiom of choice)
A mathematician named Todd Trimble has helped me to prove that the set of funcoids between two given sets (and more generally certain pointfree funcoids) is always a co-frame. (I knew this for funcoids but my proof required axiom of choice, while Todd’s does not require axiom of choice.) He initially published his proof here […]
Lattices ComplFCD and ComplRLD are co-brouwerian
I’ve found today earlier stated conjecture that lattices $latex \mathrm{Compl}\mathsf{FCD}(A;B)$ and $latex \mathrm{Compl}\mathsf{RLD}(A;B)$ are co-brouwerian. Exercise: Prove this fact.
A conjecture about multifuncoids and ultrafilters is proved
I’ve proved the following conjecture: Theorem Let $latex f$ be a staroid such that $latex (\mathrm{form}\, f)_i$ is an atomic lattice for each $latex i \in \mathrm{arity}\, f$. We have $latex \displaystyle L \in \mathrm{GR}\, f \Leftrightarrow \mathrm{GR}\, f \cap \prod_{i \in \mathrm{dom}\, \mathfrak{A}} \mathrm{atoms}\, L_i \neq \emptyset $ for every $latex L \in \prod_{i […]
What are hyperfuncoids isomorphic to?
Let $latex \mathfrak{A}$ be an indexed family of sets. Products are $latex \prod A$ for $latex A \in \prod \mathfrak{A}$. Hyperfuncoids are filters $latex \mathfrak{F} \Gamma$ on the lattice $latex \Gamma$ of all finite unions of products. Problem Is $latex \bigsqcap^{\mathsf{FCD}}$ a bijection from hyperfuncoids $latex \mathfrak{F} \Gamma$ to: prestaroids on $latex \mathfrak{A}$; staroids on […]
Funcoids are Filters article
I have completed preliminary error checking for my online article Funcoids are Filters. This article is a major step forward in the theory of funcoids.
Category theoretical generalization of reloids and funcoids
While walking home from McDonalds I conceived the following idea how we can generalize reloids and funcoids. Let $latex C$ be a category with finite products, the set of objects of which is a complete lattice (for the case of funcoids as described below it is enough to be just join-semilattice). One can argue which […]
Two new theorems
I’ve proved the theorem: Theorem $latex f \mapsto \bigsqcap^{\mathsf{RLD}} f$ and $latex \mathcal{A} \mapsto \Gamma (A ; B) \cap \mathcal{A}$ are mutually inverse bijections between $latex \mathfrak{F} (\Gamma (A ; B))$ and funcoidal reloids. These bijections preserve composition. (The second items is the previously unknown fact.) and its consequence: Theorem $latex (\mathsf{RLD})_{\Gamma} g \circ (\mathsf{RLD})_{\Gamma} […]
Yahoo! I’ve proved this conjecture
Theorem $latex (\mathsf{RLD})_{\mathrm{in}} (g \circ f) = (\mathsf{RLD})_{\mathrm{in}} g \circ (\mathsf{RLD})_{\mathrm{in}} f$ for every composable funcoids $latex f$ and $latex g$. See proof in this online article.
Two conjectures proved
In this online article I’ve proved: Theorem $latex \mathrm{dom}\, (\mathsf{RLD})_{\mathrm{in}} f = \mathrm{dom}\, f$ and $latex \mathrm{im}\, (\mathsf{RLD})_{\mathrm{in}} f = \mathrm{im}\, f$ for every funcoid $latex f$. and its easy consequence: Proposition $latex \mathrm{dom}\, (\mathsf{RLD})_{\Gamma} f = \mathrm{dom}\, f$ and $latex \mathrm{im}\, (\mathsf{RLD})_{\Gamma} f = \mathrm{im}\, f$ for every funcoid $latex f$.
Yet three conjectures
Conjecture $latex \mathrm{dom}\, (\mathsf{RLD})_{\Gamma} f = \mathrm{dom}\, f$ and $latex \mathrm{im}\, (\mathsf{RLD})_{\Gamma} f = \mathrm{im}\, f$ for every funcoid $latex f$. Conjecture $latex (\mathsf{RLD})_{\Gamma} g \circ (\mathsf{RLD})_{\Gamma} f = (\mathsf{RLD})_{\Gamma} (g \circ f)$ for every composable funcoids $latex f$ and $latex g$. Conjecture For every funcoid $latex g$ we have $latex \mathrm{Cor}\, (\mathsf{RLD})_{\Gamma} g = […]