1 answer

Axiomatic Systems Problem solving, touch question!! will give lifesaver review!

Question:

The town of Greendale has an interesting club system. Each club is just a list of townspeople and no two clubs have exactly the same membership list. Here are theimportant rules regarding the clubs that are followed by every person in town and every club: a) Every two people have a club that they both belong to, but it isunique, meaning that for each pair of people, there is only one such club. b)every club has at least two members. c)No club contains all the townspeople. d)If you namea club and then a person who is not a member of that club, there will be one and only one club that person belongs to that has no members in common with the firstclub.
Now, suppose that we have a valid club system in Greendale(where there are 4 or more people). Let X be a club with memebers {x1,...xn} and let Y1 be any other clubthat has x1 as a member. We know from Axiom (a) that none of x2,x3....,xn can be in Y1. So using axiom (d), we know that for each xi(i=2,....,n), there is a club Yisuch that xi is in Yi and Yi and Y1 have no members in common.
a)Draw a picture that depicts this situation.
b)Carefully prove that every person in Greendale belongs to exactly one of the clubs Y1, Y2,...,Yn.

Answers

In the space S (=Smithville), there are points (=people) and lines (=clubs).
We use “p∈L”, meaning that a person p belongs to club L, and “X∩Y”,
meaning the members being in common both clubs X and Y.
We are to follow the next axioms:
0)Clubs L and L’ are identical if and only if L=L’ as a set;
a)∀p1,p2(distinct)∈S, ∃!(unique)L⊂S;
b)∀L⊂S, |L|≥2;
c)L≠S for any club L;
d)∀L⊂S, ∀p∉L, ∃!(unique)L’⊂S such that p∈L’ and L∩L’=φ.

Solution:
(a)
x1 x2 x3 xn
*___*____*_______* :X
| | |
y1 y2 : L
*___*___

(b)
(*)By axiom (c), there exists y1∉X.
By axiom (a), there exists unique club Y1 (≠X) such that x1 and y1 in Y1.
By axiom (d), there exists unique club L such that y1∈L and X∩L=φ.
By axiom (b), there exists y2 in L with y2≠y1.
By axiom (a), there exists unique club Yi (2≤i≤n) such that xi, y2 in Yi,
say i=2,∃!Y2. Since x1∈Y2 ⇒x1,x2∈Y2 ⇒Y2=X. But y2∉X.
Let R be the set of remaining persons.
If R≠φ, for p in R we have two possibilities:
(1)p∈Y2;
(2)p∉Y2.
In case (1), we continue the procedure by taking y3 (y3≠y2,y1) in Y2 and obtain Y3.
In case (2), by axiom (d), we obtain new unique club L’ such that p∈L’ and X∩L’=φ,
L∩L’=φ. And we restart from the part(*).
We continue these steps until becoming R=φ.
And finally we obtain required Yi(i=1,2,..).

.

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