By Veblen O., Whitehead J. H.

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**Example text**

This implies there are E > 0 and N 2 1 such that n,, - W, = A. Thus We now define a new sequence of open symmetric neighborhoods of A by vo = wo, vn = Wl+(*-l)N, Obviously, n 2 1. V, = A. 11 we must prove that V,+l o Vn+l o V,+l for n 2 0. c V, It is clear that V1 o V1 o V1 c VOand by the choice of E and N , V2 o Vz o VZ C V1. Let k 2 2 and take (z,y) E vk+1Ovk+1Ovk+1. Then we prove that (z,y)E v k . Let z , w E X be points such that (z,w), (w,z),(z,y) E Vk+1. If ( p , q ) is any of these three points, then we have from which d(fi(z),fi(y)) < e, lil < 1 + (k - l)N.

Here d(A,B) is defined by d ( A , B )= inf{d(a,b) : a E A,b E B}. To show (4) let 1 > 0. For 1 5 i 5 m there is 0 < ~i < X such that if d(f(z),f(z)) < ~i (f(z), f(z) E f(cl(Ui))) then d(z,z ) < 7. This follows from the fact that f:cl(Ui) --f f(cl(Ui)) is injective. Put E = min{si : 1 5 i 5 m } . If diam(D) < E , then we have diam(Di) < 1 for 1 5 i 5 k. e. f is k-to-one. - . (n;=, u,"==, In the remainder of this section we describe well known theorems that will be used in the sequel. Let X be a topological space.

If this is false, for any n > 0 there exists a subset A, such that B for all B E a. diam(A,) = sup{d(a,b) : a , b E A,} 5 l / n and A, Take x , E A,, for n 2 1. Since X is compact, there is a subsequence {x,,} such that x,, 4 x as i -+ 00. Then x E B for some B E a. Since X \ B is compact, we have a = d ( x , X \ B ) = inf{d(x,b) : b E X \ B} > 0. Take sufficiently large ni satisfying ni > 2 / a and d(x,, ,x ) < a/2. Then, for y E Ani we have d ( y , ~ ) I d ( ~ i x n i ) + d ( z n i , l~/ n ) Ii + a / 2 < a < and thus y E B.