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By Karakhanyan M.I., Khor'kova T.A.

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Example text

The following result can be proved as a matter of routine, hence left as an exercise for the reader to prove. 39: Let {Aα | α ∈ Ω} ∪ {A} be a collection of fuzzy subrings of R. Suppose that L has the finite intersection property. If A = ⊕ A α then for all x ∈ A∗, α∈Ω x= ∑ xα α∈Ω for unique   xα ∈ A *α and A(x) = inf  Aα ( xα ) x = ∑ xα , xα ∈ Aα*  . 40: Let A and B be fuzzy subrings of R. Suppose that L has the finite intersection property. Then (AB)∗ = A∗B∗. Proof: Straightforward, hence left for the reader to prove.

The following theorem which can be proved by a routine computation is left as an exercise for the reader to prove. THEOREM [36]: Let µ be a fuzzy subset of a ring R with card Im µ < ∞. Define subrings Ri of R by R0 = 〈 {x∈ R  µ (x) = sup µ (z)}〉 and z∈R Ri = 〈{Ri–1 ∪ {x ∈ R  µ (x) = sup µ( z ) 〉, 1 ≤ i ≤ k z∈R − Ri −1 where k is such that Rk = R. Then k < card Im µ. Also the fuzzy subset µ∗ of R defined by sup µ ( z )  z ∈ R µ*( x ) =   sup µ ( z )  z∈R\ Ri−1 if x ∈ R0 if x ∈ Ri \ Ri −1 , 1 ≤ i ≤ k is a fuzzy subring generated by µ in R.

For all s, t ∈ Im(A), s ≥ t, At ∩ Bs = As, then A has an extension to a fuzzy subset Ae of S such that (Ae)t ⊇ Bt for all t ∈ Im (A). 57: Let R be a non-empty subset of a set S and let A be a fuzzy subset of R such that A has the sup property. If B = {Bt t ∈ Im (A)} is a collection of subsets of S which satisfies (i) to (iii) of conditions given in the above theorem then A has a unique extension to a fuzzy subset Ae of S such that (Ae)t = Bt for all t ∈ Im (A) and Im(Ae) = Im(A). Now we proceed on to define extension of fuzzy subrings and fuzzy ideals.

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A Characteristic Property of the Algebra C(Q)B by Karakhanyan M.I., Khor'kova T.A.


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