📚 Recommended Mathematics Books
Topology (Munkres) | General Topology (Engelking) | Counterexamples in Topology | Rudin's AnalysisAs an Amazon associate, I earn from qualifying purchases.
This conjecture appeared to be false.
Now I propose an alternative conjecture:
Let $latex A$, $latex B$ be sets.
Conjecture Funcoids $latex f$ from $latex A$ to $latex B$ bijectively corresponds to the sets $latex R$ of pairs
$latex (\mathcal{X}; \mathcal{Y})$ of filters (on $latex A$ and $latex B$ correspondingly) that
- $latex R$ is nonempty.
- $latex R$ is a lower set.
- every $latex \left\{ \mathcal{X} \mid (\mathcal{X} ; \mathcal{Y}) \in R \right\}$ is a dcpo for every $latex \mathcal{Y} \in \mathfrak{F}B$ and every $latex \left\{ \mathcal{Y} \mid (\mathcal{X} ; \mathcal{Y}) \in R \right\}$ is a dcpo for every $latex \mathcal{X} \in \mathfrak{F}A$
by the mutually inverse formulas:
$latex (\mathcal{X} ; \mathcal{Y}) \in R \Leftrightarrow \mathcal{X} \times^{\mathsf{FCD}} \mathcal{Y} \sqsubseteq f \quad \mathrm{and} \quad f = \bigsqcup^{\mathsf{FCD}} \left\{ \mathcal{X} \times^{\mathsf{FCD}} \mathcal{Y} \mid (\mathcal{X} ; \mathcal{Y}) \in R \right\}$.
🔬 Advanced Mathematics References
- Sheaves in Geometry and Logic
- Categories for the Working Mathematician
- Stone Spaces
- Algebraic Topology (Hatcher)
- Concrete Mathematics
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