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Formally: [Il V:c E 11 :Jy E c :Jz El1::c -< y 1\ Y co z. Clearly, :c =I z and :c -< z. Let:co El l ' By [Il, :Jyo E c :J:Cl E 11 such that :Co -< Yo co Xl' Again, Xl E 11 implies :JYl E c :JX2 E 11 such that :Cl -< Yl co X2. Note that, clearly, Yo =I Yl' Continuing with this construction we get two subset of X, Xl = {:CO,:Cll' .. 17). 10(i) can be embedded into the poset (XI, -< IXlxxl) where XI = Xl U Yll a contradiction with the hypothesis. , assume there exists X E 11 for which it is not possible to construct a triangle structure.

Let y E c(A) n 1. Clearly x ~ y, but y E c(A) which means that the proof is done. This proves Part (b). Z Elc(A), and 40 2. Partially Ordered Sets ({=) We have to prove that VA E D V[ E L:c(A) n [ # 0. Let A E D and 1 E L be arbitrary. Clearly c(A) # 0 since (X, -<) is strongly cut-bounded and A # 0. If 3c E C such that c ~ c(A) then the result follows immediately since (X, -<) is K-dense. We shall prove that this is always the case. , Vc E C: c C£. c(A); we will arrive at a contradiction. Let c' ~ c(A) be a maximal co-set in c(A); c' exists and is non-empty because c(A) # 0.

Since y E Min(A) then Zn-1 EA. Again (Zn-1,y) E-<' ~ (B x E) U (E x B). For the same reasons as before, it follows that either y E Obmin(A) ~ c(A) or Zn-1 E Obmaz(A) ~ c(A). In both cases the proof is again done. JE. Then (X, -<) is D-continuous <¢=:::? (X, -<) is K-dense, cui-bounded and Ve E E: I"el i=- 1 i=- le"l. 7. 8 Exercises. 1. 6(ii). 2. We can define two order relations on the set of all cuts of a poset: Definition Let (X, -<) be a poset and C1 [;:;1 C2 C1 [;:;2 C2 (i) iff iff Vx E Vy E C1 C2 C1, C2 3y E 3x E C2: E C (X, -<).

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