Re: Finest partition - exercise in Suppes's book


MoeBlee wrote:
> On page 84 of Suppes's 'Axiomatic Set Theory' (Dover), he mentions an
> "intriguing" exercise (and he says, "the problem is to prove it"). But
> the exercise seems trivial to me while there are exercises in the book
> that are much harder but with no mention that they are difficult. So I
> am wondering whether I've missed something in my proof.
>
> Suppes's definition:
>
> Df. (P is a partition of S & Q is a partition of S) ->
> (P is finer than Q <-> (P not= Q & Ax(xeP -> Eb(beQ & x subsetof b)))
>
> For a conditional definition of a predicate symbol, I like to make the
> antecedent part of the definens, and I think that 'finer than' is a
> 3-place predicate symbol, so for precision, but without any important
> effect on the problem, I modiy to:
>
> Df. P is finer per S than Q <->
> (P is a partition of S & Q is a partition of S & P not= Q & Ax(xeP ->
> Eb(beQ & x subsetof b)))
>
> Df. P is a finest partition of S <->
> (P is a partition of S & AQ((Q not= P & Q is a partition of S) -> P is
> finer per S than Q))
>
> Proposition. AS EP P is a finest partion of S. The proof seems trivial
> to me, so I want to make sure I haven't missed what Suppes claims to be
> "intriguing" about it:
>
> {{y}| yeS} exists since it is a subset of the power set of S.
>
> {{y}| yeS} is a partition of S, as follows:
>
> {{y}| yeS} is pairwise disjoint since if z is an element of {y} and z
> is an element of {y'}, then y = y' (lest z not= z), so {y} = {y'}.
>
> U{{y}| yeS} = S since if zeU{{y}| yeS}, then z = y for some yeS; and if
> zeS, then {z}e{{y}| yeS}.
>
> Ax(xe{{y}| yeS} -> Ey yex) since for some yeS, x = {y}.
>
> Moving on, (Q not= {{y}| yeS} & Q is a partition of S) -> Ax(xe{{y}|
> yeS} -> Eb(beQ & x subsetof b)), as follows:
>
> Let xe{{y}| yeS}. So for some yeS, x = {y}. Since Q is a partition of
> S, there exists a beQ such that yeb. So x subsetof b.
>
> Further, ({y}| yeS} is unique as a finest partition of S, as follows:
>
> Suppose P not= ({y}| yeS} & P is a partition of S & AQ((Q not= P & Q is
> a partition of S) -> P is finer per S than Q). Then Ax(xeP -> Eb(be{y}|
> yeS} & x subsetof b). If P not= {{y}| yeS}, then there is some
> non-empty x that's a member of Q but x is not a singleton. So x cannot
> be a subset of a singleton.
Your next to final sentence starting "Then" shows that every member
of P is a singleton.
Your last sentence is a little unclear -I do not know what Q is
(Probably the partion of all the singletons )It doesn't matter ,it
still follows that for any yeS y is an element of a member of P which
you know is a singleton so must be {y} .Thus P is the partition into
singletons.I just mention this in passing.
> So, have I made a mistake or missed proving something that is
> intriguing and a problem to prove? If not, I don't know why Suppes says
> this exercise is "intriguing" or why he chooses to mention it, in
> particular, as a "problem" to prove.

No,you haven't missed a thing ,and Suppes does say(also on page 84 of
my 1960 Van Nostrand edition) that finest means finer then any other
partition .He forgot to mention a point (which you saw -your
uniqueness part) that the relatiion "finer" is an irreflexive partial
ordering .This means with "finer"just a general relation- 1) if P is
finer than Q then Q is not finer then P and 2) if P is finer then Q and
Q finer then R then P is finer then R. ( 3) P is not finer then P(
follows from 2) ) .Without this a finest partition would not be
necessarily unique (for a different notion of f "finer") .Somewhat
intriguing ,maybe.Regards,smn

.

Relevant Pages

  • Re: Finest partition - exercise in Suppess book
    ... > Your next to final sentence starting "Then" shows that every member ... Then Ax(xeP -> Eb(be| yeS} & x subsetof b). ... P but x is not a singleton. ... > finer than Q then Q is not finer then P and 2) if P is finer then Q and ...
    (sci.logic)
  • Re: Finest partition - exercise in Suppess book
    ... > the exercise seems trivial to me while there are exercises in the book ... > For a conditional definition of a predicate symbol, ... > Eb(beQ & x subsetof b))) ... > finer per S than Q)) ...
    (sci.logic)
  • Re: Finest partition - exercise in Suppess book
    ... MoeBlee wrote: ... >> Your last sentence is a little unclear -I do not know what Q is ... > P but x is not a singleton. ... >> finer than Q then Q is not finer then P and 2) if P is finer then Q and ...
    (sci.logic)
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