Greetings, please explain the theorem and give your PROOF with steps so I may understand and follow your work. I cannot follow the proof given. Thank you.

open Q-neighborhood, which is quasi-coincident with H at some point w. Since B C 4, B(x)

THEOREM 5.2. The derived set of each fuzzy set is closed iff the derived set of each point is closed. Proof. The necessity is obvious. We shall now show its sufficiency. Let H be an arbitrary fuzzy set. In the light of the corollary of Theorem 5.1, in order to show that H' =D is closed, it suffices to show that for an arbitrary accumula- tion point x; of D, x's D. Since x, E D = (HU)C(H) = H, by Theorem 4.1', X, is an adherence point of H. If x, H then x, is an accumulation point of H, i.e., d'y e D. We may assume that x; E H, i.e., ^ 0. From Lemma 5.1, P > A(x) = H(x), and hence *'07 H. But since X., 4CCH, X., is an accumulation point of H, X, D. Moreover, 1 1 - 1, and C is also an open Q-neighborhood of xj. Because x; is an accumulation point of D, C is quasi-coincident with D at some point s, i.e., D() + C(x) > I. Owing to the fact that D is the union of all the accumulation points of H, there is an accumulation point 2, such that he + C(x) > I. Therefore C is also an open Q-neighborhood of z.. The proof will be carried out, according to the three possible cases concerning fu, as follows: (1) When z = x and u

p, then H. From (3) of Proposition 5.1, there is an open set such that B(x) = 1 -p>I-. Therefore G =CU is also an open Q-neighborhood of %, and G and H are quasi-coincident at some point w. Since G(x) C(x) > 1 - u, by Definition 4.1, there exists an open set BC ' such that (1)() > B(x) > I - . Therefore Bn B is also an THEOREM 5.2. The derived set of each fuzzy set is closed iff the derived set of each point is closed. Proof. The necessity is obvious. We shall now show its sufficiency. Let H be an arbitrary fuzzy set. In the light of the corollary of Theorem 5.1, in order to show that H' =D is closed, it suffices to show that for an arbitrary accumula- tion point x; of D, x's D. Since x, E D = (HU)C(H) = H, by Theorem 4.1', X, is an adherence point of H. If x, H then x, is an accumulation point of H, i.e., d'y e D. We may assume that x; E H, i.e., ^ 0. From Lemma 5.1, P > A(x) = H(x), and hence *'07 H. But since X., 4CCH, X., is an accumulation point of H, X, D. Moreover, 1 1 - 1, and C is also an open Q-neighborhood of xj. Because x; is an accumulation point of D, C is quasi-coincident with D at some point s, i.e., D() + C(x) > I. Owing to the fact that D is the union of all the accumulation points of H, there is an accumulation point 2, such that he + C(x) > I. Therefore C is also an open Q-neighborhood of z.. The proof will be carried out, according to the three possible cases concerning fu, as follows: (1) When z = x and u

p, then H. From (3) of Proposition 5.1, there is an open set such that B(x) = 1 -p>I-. Therefore G =CU is also an open Q-neighborhood of %, and G and H are quasi-coincident at some point w. Since G(x) C(x) > 1 - u, by Definition 4.1, there exists an open set BC ' such that (1)() > B(x) > I - . Therefore Bn B is also an