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Just a few minutes ago I’ve formulated a new important conjecture about funcoids:

Let $A$, $B$ be sets.

Conjecture Funcoids $f$ from $A$ to $B$ bijectively corresponds to the sets $R$ of pairs $(\mathcal{X}; \mathcal{Y})$ of filters (on $A$ and $B$ correspondingly) that

1. $R$ is nonempty.
2. $R$ is a lower set.
3. $R$ (ordered pointwise) is a dcpo

by the mutually inverse formulas: $(\mathcal{X} ; \mathcal{Y}) \in R \Leftrightarrow \mathcal{X} \times^{\mathsf{FCD}} \mathcal{Y} \sqsubseteq f \quad \mathrm{and} \quad f = \bigsqcup^{\mathsf{FCD}} \left\{ \mathcal{X} \times^{\mathsf{FCD}} \mathcal{Y} \mid (\mathcal{X} ; \mathcal{Y}) \in R \right\}$.

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#### 4 Comments

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1. porton / Apr 23 2016 01:16 Oh, a trivial counter-example: $f = ([0 ; 2] \times [0 ; 1]) \cup ([0 ; 1] \times [0 ; 2])$. I will try to figure another axioms.

2. porton / Apr 23 2016 02:04 Oh, sorry, the counter-example is wrong. However it inspired me with another similar conjecture.

3. porton / Apr 23 2016 20:30 No, it is wrong. For a counter-example take $R=\{\uparrow\emptyset,\uparrow\{0\},\uparrow\{1\}\}$.

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