A 1.71 . 7 \quad B1.12

Logic, Computation and Set Theory | Part II, 2002

(i) State the Knaster-Tarski fixed point theorem. Use it to prove the Cantor-Bernstein Theorem; that is, if there exist injections ABA \rightarrow B and BAB \rightarrow A for two sets AA and BB then there exists a bijection ABA \rightarrow B.

(ii) Let AA be an arbitrary set and suppose given a subset RR of PA×AP A \times A. We define a subset BAB \subseteq A to be RR-closed just if whenever (S,a)R(S, a) \in R and SBS \subseteq B then aBa \in B. Show that the set of all RR-closed subsets of AA is a complete poset in the inclusion ordering.

Now assume that AA is itself equipped with a partial ordering \leqslant.

(a) Suppose RR satisfies the condition that if baAb \geqslant a \in A then ({b},a)R(\{b\}, a) \in R.

Show that if BB is RR-closed then cbBc \leqslant b \in B implies cBc \in B.

(b) Suppose that RR satisfies the following condition. Whenever (S,a)R(S, a) \in R and bab \leqslant a then there exists TAT \subseteq A such that (T,b)R(T, b) \in R, and for every tTt \in T we have (i) ({b},t)R(\{b\}, t) \in R, and (ii) tst \leqslant s for some sSs \in S. Let BB and CC be RR-closed subsets of AA. Show that the set

[BC]={aAba(bBbC)}[B \rightarrow C]=\{a \in A \mid \forall b \leqslant a(b \in B \Rightarrow b \in C)\}

is RR-closed.

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