By L. Badescu, D. Popescu

ISBN-10: 3540129308

ISBN-13: 9783540129301

**Read or Download Algebraic Geometry Bucharest 1982. Proc. conf PDF**

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**Extra info for Algebraic Geometry Bucharest 1982. Proc. conf**

**Example text**

TD is completely regular. Proof. Let x ∈ A ∈ TD . Assume for simplicity that x = 0. We must deﬁne a deep-I-approximately continuous function f : R → R such that f (0) = 1 and Ac ⊂ f −1 (0). 1 we can ﬁnd an open interval set V ⊂ A and an interval set E ⊂ V composed of closed intervals such that 0 is an I-density point of E. Deﬁne 1 x=0 f (x) = dist(x, V c ) x=0 dist(x, V c )+dist(x, E) It is easy to see that f (0) = 1, Ac ⊂ V c ⊂ f −1 (0) and that f is continuous on R\{0}. Moreover, f is deep-I-approximately continuous at 0 as E ⊂ f −1 (1).

This implies (c, d) ∩ nkp A = ∅ for every p ∈ N. 2(viii) is proved. To prove the converse, assume that condition ( ) is false for some s > 0. Then for every n ∈ N and Ds = Rs = n1 there are numbers 0 < D ≤ Ds and 1 tn = D and an interval (an , bn ) ⊂ (−1, 1) \ {0} such that bn − an ≥ s and for every interval (c, d) ⊂ (an , bn ) with the property that (c, d) ∩ tn A = ∅ we have d − c ≤ Rs (b − a) ≤ n1 . 4, we can choose an increasing sequence {nk }k∈N of natural numbers and a nonempty interval (a, b) such that (a, b) ⊂ (ank , bnk ) for every k ∈ N and the sequence {tnk }k∈N is increasing and diverging to inﬁnity.

TI ∩ TN ⊂ TD . Let E = R \ Q. 3(ii), Q is nowhere dense in TI and TN and so E ∈ TI ∩ TN . 8(iii), Q is dense in TD . Hence, E ∈ TD . TN ⊂ TI . Let C be a nowhere dense set of positive Lebesgue measure. By the Lebesgue Density Theorem, D = C ∩ ΦN (C) ∈ TN and D = ∅. 2(ii), D is nowhere dense in TI , so D ∈ TI . TD ⊂ TN . Let C ⊂ [ 12 , 1] be a closed nowhere dense set with positive Lebesgue measure and let {bn }n∈N be a decreasing sequence of positive numbers such that lim bn+1 /bn = 0. 1, 0 is a deep-I-dispersion point of E = E c ∈ TD .

### Algebraic Geometry Bucharest 1982. Proc. conf by L. Badescu, D. Popescu

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