By Paul A. Smith, Samuel Eilenberg
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Sample text
Let f be a real-valued function defined for d 0 and such that (i) f ( 0 ) = 0 ; (ii)f(d,) < f ( d 2 ) f o r erlery dl < 4 ;(iii)f(dl+d2) After realizing that the solution of the problem requires more than trying to pick disks one by one until one hits on M , the first thing to do is to collect all subfamilies A of D satisfying requirement (1) into one set d . T h e second step consists of checking that d is not void. Here this is trivial. Third, it is necessary to introduce a reflexive partial in d which i s suitable for the purpose. I n our example the ordering A , if A, is a right partial ordering is given by inclusion, so A, subfamily of A,. T o show that y ( A ) c A’ suppose that x g A’. Then there is an 0,, say 0, = c(B u y ( B ) ) ,such that 0, n A c {x}. Then A - {x} 5 B u y ( B ) and y ( A - {x}) G y ( B u y ( B ) ) = y ( B ) . Since x E 0, = c(B u y ( B ) ) we have x $ y ( B ) and so x $ y ( A - {x}). ] 3. Consider an operator y ( A ) for the set of all subsets of a set X subject to the axioms: (1) y(X) = X . (2) y ( A ) G A for every A. (3) r(r(A))= r(A) for every A. r(A f-7 B ) = A 4 n y ( 4 (4) Find a topology F on X such that y ( A ) = A i where the interior is taken with respect to F , [The family of open sets is 0 = { y ( A ): A E X } .