After removing an erroneous theorem I posed two new open problems to take its place: Conjecture If $latex f$ is a complete funcoid and $latex R$ is a set of funcoids then $latex f \circ \bigcup {\nobreak}^{\mathsf{FCD}} R = \bigcup {\nobreak}^{\mathsf{FCD}} \langle…

read moreNow Funcoids and Reloids online article contains the section “Connectedness regarding funcoids and reloids” which previously was in a separate article. In this section there are among definitions and theorems a few open problems.

read moreI have said that there were several errors in my draft article “Connectedness of funcoids and reloids” at Algebraic General Topology site. I have corrected the errors, but now some of what were erroneous theorems downgraded to the status of conjecture.

read more[I found that my computations below are erroneous, namely $latex \mathrm{Cor} \langle f^{-1}\rangle \mathcal{F} \neq \langle \mathrm{CoCompl} f^{-1}\rangle \mathcal{F}$ in general (the equality holds when $latex \mathcal{F}$ is a set).]

read moreIn my Algebraic General Topology series was a flaw in the proof of the following theorem. So I re-labeled it as a conjecture. Conjecture A filter $latex \mathcal{A}$ is connected regarding a reloid $latex f$ iff it is connected regarding the funcoid…

read moreI updated online article “Funcoids and Reloids”. The main feature of this update is new section about complete reloids and completion of reloids (with a bunch of new open problems). Also added some new theorems in the section “Completion of funcoids”.

read moreConjecture $latex \mathrm{Cor}\bigcup^{\mathfrak{F}} S=\bigcup\langle \mathrm{Cor}\rangle S$ for any set $latex S$ of filter objects on a set. See this wiki site for definitions of used terms.

read moreLet $latex \mathfrak{A}$ is a complete lattice. I will call a filter base a nonempty subset $latex T$ of $latex \mathfrak{A}$ such that $latex \forall a,b\in T\exists c\in T: (c\le a\wedge c\le b)$. I will call a chain (on $latex \mathfrak{A}$) a…

read moreConjecture Let $latex a$ and $latex b$ are filters on a set $latex U$. Then $latex a\cap b = \{U\} \Rightarrow \\ \exists A,B\in\mathcal{P}U: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U).$ [corrected]…

read moreLet $latex {U}&fg=000000$ is a set. A filter $latex {\mathcal{F}}&fg=000000$ (on $latex {U}&fg=000000$) is a non-empty set of subsets of $latex {U}&fg=000000$ such that $latex {A, B \in \mathcal{F} \Leftrightarrow A \cap B \in \mathcal{F}}&fg=000000$. Note that unlike some other authors I…

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