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July 4, 2017 / porton

A new unexpected result (ERROR!)  

The below is wrong! The proof requires \langle g^{-1}\rangle J to be a principal filter what does not necessarily hold.

I knew that composition of two complete funcoids is complete. But now I’ve found that for g\circ f to be complete it’s enough f to be complete.

The proof which I missed for years is rather trivial:

\bigsqcup S \mathrel{[g \circ f]} J \Leftrightarrow J \mathrel{[f^{- 1} \circ g^{- 1}]} \bigsqcup S \Leftrightarrow \langle g^{- 1} \rangle J \mathrel{[f^{- 1}]} \bigsqcup S \Leftrightarrow \bigsqcup S \mathrel{[f]} \langle g^{- 1} \rangle J \Leftrightarrow \exists \mathcal{I} \in S : \mathcal{I} \mathrel{[f]} \langle g^{- 1} \rangle J \Leftrightarrow \exists \mathcal{I} \in S : \mathcal{I} \mathrel{[g \circ f]} J

Thus g\circ f is complete.

I will amend my book when (sic!) it will be complete.

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