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July 3, 2017 / porton

A new easy proposition about funcoids

I have proved (see new version of my book) the following proposition. (It is basically a special case of my erroneous theorem which I proposed earlier.)

Proposition For f \in \mathsf{FCD} (A, B), a finite set X \in \mathscr{P} A and a function t \in \mathscr{F} (B)^X there exists (obviously unique) g \in \mathsf{FCD} (A, B) such that \langle g\rangle p = \langle f \rangle p for p \in \mathrm{atoms}^{\mathscr{F} (A)} \setminus \mathrm{atoms}\, X and \langle g\rangle @\{ x \} = t (x) for x \in X.

This funcoid g is determined by the formula

g = (f \setminus (@X \times^{\mathsf{FCD}} \top)) \sqcup \bigsqcup_{x \in X} (@\{ x \} \times^{\mathsf{FCD}} t (x)) .

and its corollary:

Corollary If f \in \mathsf{FCD} (A, B), x \in A, and \mathcal{Y} \in \mathscr{F} (B), then there exists an (obviously unique) g \in \mathsf{FCD} (A, B) such that \langle g\rangle p = \langle f \rangle p for all ultrafilters p except of p = @\{ x \} and \langle g \rangle @\{ x \} = \mathcal{Y}.

This funcoid g is determined by the formula

g = (f \setminus (@\{ x \} \times^{\mathsf{FCD}} \top)) \sqcup (\{ x \} \times^{\mathsf{FCD}} \mathcal{Y}) .
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