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August 13, 2017 / porton

New simple theorem in my book

I added to my online research book the following theorem:

Theorem Let \mathfrak{A} be a distributive lattice with least element. Let a,b\in\mathfrak{A}. If a\setminus b exists, then a\setminus^* b also exists and a\setminus^* b=a\setminus b.

The user quasi of Math.SE has helped me with the proof.

July 4, 2017 / porton

A wrong result

I’ve published in my blog a new theorem.

The proof was with an error (see the previous edited post)!

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.

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}) .
June 30, 2017 / porton

Error in my theorem – found

I found the exact error noticed in Error in my theorem post.

The error was that I claimed that infimum of a greater set is greater (while in reality it’s lesser).

I will delete the erroneous theorem from my book soon.

June 30, 2017 / porton

Error in my theorem

It seems that there is an error in proof of this theorem.

Alleged counter-example:

f=\bot and z(p)=\top for infinite sets A and B.

I am now attempting to locate the error in the proof.

June 29, 2017 / porton

New theorem about funcoids (ERROR!)

I have proved (and added to my online book) the following theorem:

Theorem Let f \in \mathsf{FCD} (A ; B) and z \in \mathscr{F} (B)^A. Then there is an (obviously unique) funcoid g \in \mathsf{FCD} (A ; B) such that \langle g\rangle x = \langle f\rangle x for nontrivial ultrafilters x and \langle g\rangle @\{ p \} = z (p) for p \in A

After I started to prove it, it took about a hour or like this to finish the proof.