The following is one of a few (possibly non-equivalent) definitions of products of funcoids: Definition Let $latex f$ be an indexed family of funcoids. Let $latex \mathcal{F}$ be a filter on $latex \mathrm{dom}\, f$. $latex a \mathrel{\left[ \prod^{[\mathcal{F}]} f \right]} b \Leftrightarrow…
read moreDefinition $latex a \mathrel{\left[ \prod^{(A 2)} f \right]} b \Leftrightarrow \exists M \in \mathrm{fin} \forall i \in (\mathrm{dom}\, f) \setminus M : \Pr^{\mathsf{RLD}}_i a \mathrel{[f_i]} \Pr^{\mathsf{RLD}}_i b$ for an indexed family $latex f$ of funcoids and atomic reloids $latex a$ and $latex…
read moreI’ve found an easy positive proof of this my conjecture.
read moreConjecture $latex S^{\ast}(\mu(E)) = E$ for every partial order $latex E$. The function $latex S^\ast$ and micronization $latex \mu$ are defined in my research monograph.
read moreI’ve added the following to my research book: Definition Galois surjection is the special case of Galois connection such that $latex f^{\ast} \circ f_{\ast} $ is identity. Proposition For Galois surjection $latex \mathfrak{A} \rightarrow \mathfrak{B}$ such that $latex \mathfrak{A}$ is a join-semilattice…
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read moreI have defined two new kinds of products of funcoids: $latex \prod^{\mathrm{in}}_{i \in \mathrm{dom}\, f} f = \prod^{(C)}_{i \in \mathrm{dom}\, f} (\mathsf{RLD})_{\mathrm{in}} f_i$ (cross-inner product). $latex \prod^{\mathrm{out}}_{i \in \mathrm{dom}\, f} f = \prod^{(C)}_{i \in \mathrm{dom}\, f} (\mathsf{RLD})_{\mathrm{out}} f_i$ (cross-outer product). These products…
read moreI’ve added to my book the following conjecture: Conjecture For every composable funcoids $latex f$ and $latex g$ $latex (\mathsf{RLD})_{\mathrm{out}}(g\circ f)\sqsupseteq (\mathsf{RLD})_{\mathrm{out}}g\circ(\mathsf{RLD})_{\mathrm{out}} f.$
read moreAfter noticing an error in my math book, I rewritten its section “Funcoids and filters” to reflect that $latex (\mathsf{RLD})_\Gamma = (\mathsf{RLD})_{\mathrm{in}}$. Previously I proved an example demonstrating that $latex (\mathsf{RLD})_\Gamma \ne (\mathsf{RLD})_{\mathrm{in}}$, but this example is believed by me to be…
read moreI proved both $latex (\mathsf{RLD})_\Gamma \ne (\mathsf{RLD})_{\mathrm{in}}$ and $latex (\mathsf{RLD})_\Gamma = (\mathsf{RLD})_{\mathrm{in}}$. So there is an error in my math research book. I will post the details of the resolution as soon as I will locate and correct the error. While the…
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