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Apr 9, 2011 at 19:59 vote accept porton
Apr 8, 2011 at 15:19 comment added Emil Jeřábek It does not, products (restricted or otherwise) of filters satisfy the condition automatically. In fact, the second condition fixed an oversight in the first part of my argument: I didn’t notice that I needed the condition (in particular, $S_i\ne\varnothing$ for every $i$) in order to make the restricted product included in $f$ nonempty (otherwise it would be pointless).
Apr 8, 2011 at 14:44 comment added porton I did an error, I forgot the second condition in the definition of multifuncoids (now added). Happily this does not break the counter-example by Emil Jeřábek.
Apr 8, 2011 at 14:43 history edited porton CC BY-SA 3.0
Added forgotten second condition in the definition of multifuncoids
Apr 7, 2011 at 17:26 history edited porton CC BY-SA 2.5
Cleared the definition of multifuncoid
Apr 7, 2011 at 16:53 comment added Todd Trimble But I still think Andreas and Emil are kind! :-)
Apr 7, 2011 at 16:52 comment added Todd Trimble @Daniel, @Emil: perhaps you are right and I was being unfair to porton. I had some difficulty understanding what a multifuncoid was supposed to be at first reading, which is when I clicked on the link. On another more careful reading, I think it is indeed decipherable -- my apologies.
Apr 7, 2011 at 16:41 comment added porton Oh, sorry, my "additional condition" was too strong and I commented it out. We may make up some other condition if it will be necessary.
Apr 7, 2011 at 16:39 history edited porton CC BY-SA 2.5
Deleted "the additional condition"
Apr 7, 2011 at 16:39 comment added Daniel Litt (To be fair, I still am in awe of the kindness of both responders, and have upvoted their answers.)
Apr 7, 2011 at 16:37 comment added Emil Jeřábek I found the question quite self-contained, I certainly didn’t read any porton’s paper.
Apr 7, 2011 at 16:34 comment added Daniel Litt @Todd Trimble: While I am of course not thrilled by this question, it seems to me that porton does say what a multifuncoid is, in his fourth paragraph. Or are you referring to another unfamiliar term?
Apr 7, 2011 at 16:33 comment added porton @Andreas Blass: I a little reordered the text of the question. For "the additional condition" it does not matter what are ultrafilters $a_0$ and $a_1$. $A_i$ and $B_i$ in "the additional condition" are completely arbitrary. For your values of $A$ and $B$ "the additional condition" takes the form $f(U;U)\Leftrightarrow f(U;\emptyset) \vee f(\emptyset;U)$. Hopefully this answers your question. If not please ask more specifically.
Apr 7, 2011 at 16:26 history edited porton CC BY-SA 2.5
Little reorder
Apr 7, 2011 at 16:15 comment added Todd Trimble I am in awe of the kindness of the mathematicians who responded. The question makes no attempt to define the totally unfamiliar terms, save for a link to porton's web page, where the reader has to pick a paper, take a deep breath, and wade into porton's stuff. I don't think the way the question is worded really deserves such kindness.
Apr 7, 2011 at 15:22 answer added Andreas Blass timeline score: 10
Apr 7, 2011 at 15:21 answer added Emil Jeřábek timeline score: 22
Apr 7, 2011 at 14:31 comment added Andreas Blass Is the additional condition formulated correctly? It seems to me that it fails even for the two-element index set $I=\{0,1\}$ when $a_0$ and $a_1$ are ultrafilters. Take $A_0=B_1=U$ and $A_1=B_0=\emptyset$.
Apr 7, 2011 at 13:40 history edited porton CC BY-SA 2.5
Little clarification
Apr 7, 2011 at 11:51 comment added porton I added a possible (guessed) necessary condition.
Apr 7, 2011 at 11:50 history edited porton CC BY-SA 2.5
Again grammar
Apr 7, 2011 at 11:45 history asked porton CC BY-SA 2.5