A new proposition proved
I’ve proved the following lemma: Lemma Let for every $latex X, Y \in S$ and $latex Z \in \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y)$ there is a $latex T \in S$ such that $latex T \sqsubseteq Z$. Then for every $latex X_0, \ldots, X_n \in S$ and $latex Z \in \mathrm{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n)$ there […]
I proved a conjecture
After prayer in tongues and going down anointment of Holy Spirit I proved this conjecture about funcoids. The proof is currently located in this PDF file. Well, the proof is for special cases of distributive lattices, but more general case seems not necessary (at least now). It seems easy to generalize it for more general […]
New conjecture about funcoids
New conjecture: Conjecture $latex \mathrm{up} (f \sqcap^{\mathsf{FCD}} g) \subseteq \{ F \sqcap G \mid F \in \mathrm{up}\, f, G \in \mathrm{up}\, g \}$ for all funcoids $latex f$, $latex g$ (with corresponding sources and destinations). Looks trivial? But how to (dis)prove it?
Attempt to generalize filter bases for more general filtrators
In this draft I present some definitions and conjectures on how to generalize filter bases for more general filtrators (such as the filtrator of funcoids). This is a work-in-progress. This seems an interesting research by itself, but I started to develop it as a way to prove this conjecture.
My proof was with an error
I claimed that I have proved the following conjecture: Conjecture $latex \forall H \in \mathrm{up} (g \circ f) \exists F \in \mathrm{up}\, f, G \in \mathrm{up}\, g : H \sqsupseteq G \circ F$ for every composable funcoids $latex f$ and $latex g$. The proof was with an error. So it remains a conjecture.
A conjecture about funcoids proved
WARNING: The proof was with an error! I have proved the following theorem: Theorem $latex \forall H \in \mathrm{up} (g \circ f) \exists F \in \mathrm{up}\, f, G \in \mathrm{up}\, g : H \sqsupseteq G \circ F$ for every composable funcoids $latex f$ and $latex g$. This theorem (being a conjecture at that time) was […]
A new counter-example
Example There is such a non-symmetric reloid $latex f$ that $latex (\mathsf{FCD})f$ is symmetric. Take $latex f=((\mathsf{RLD})_{\mathrm{in}}(\mathord{=})|_{\mathbb{R}})\sqcap (\mathord{\geq})_{\mathbb{R}}$. I have added this to my online book.
I replaced semicolons with commas
I’ve released my math research book and all supplementary materials free with semicolons replaced with commas to denote tuples: $latex (a;b)$ → $latex (a,b)$, in order to comply with usual math notation of other mathematicians.
An informal open problem in mathematics
Characterize the set $latex \{f\in\mathsf{FCD} \mid (\mathsf{RLD})_{\mathrm{in}} f=(\mathsf{RLD})_{\mathrm{out}} f\}$. (This seems a difficult problem.)
A new theorem proved
I have proved $latex (\mathsf{RLD})_{\mathrm{in}} \Omega^{\mathsf{FCD}} = \Omega^{\mathsf{RLD}}$ (where $latex \Omega^{\mathsf{FCD}}$ is a cofinite funcoid and $latex \Omega^{\mathsf{RLD}}$ is a cofinite reloid that is reloid defined by a cofinite filter). The proof is currently available in this draft. Note that in the previous draft there was a wrong formula for $latex (\mathsf{RLD})_{\mathrm{in}} \Omega^{\mathsf{FCD}}$.