Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1997. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 24)
CERETKOVÁ, S., FULIER, J. and TÓTH J. T., On the certain subsets of the space of metrics
112 S. Ceretková, J. Fulier and J. T. Tóth Obviously /^(í 1)) = A and g~ l{{0}) = M\H. It is purpose of this paper to establish how large sets /1({í})? ő'1 ({*})» axe given. In what follows if U C A4 , then U is considered as a metrics space with the metric <i* \uxU ( a metric subspace of A4). Main results Let tp(t) — for t £ [0, -f oo). Then (p is increasing and continuous function on [0, +oo). Therefore f(d) = sup <p(d(x,y)) for d E A4. The x,y£X natural question arises wether / is continuous on A4, too. The answer of this question is positive. We have Lemma. The function /, g are uniformly continuous on AA and the function h is uniformly continuous on B. Proof. Let 0 < e < 1 and d\,d 2 6 M such that d*(d x,d 2) < £. We show |/(di)-/(d 2)| <d*(d ud 2), \g(d 1) - g(d 2)\ <d k(d ud 2). We can simply count <p(di(x,y)) < <p{d 2(x,y)) + \<p(di(x,y)) - (p(d, 2(x,y))\ < <p(d 2(x,y)) + d*(d ud 2) because Mi(*,»)-<«.(*, y)l < d* {dud2 ). (l + </i(®,Ji))(l + d2(a:,s)) Taking supremum in the previous inequality we obtain f(d\) < f(d 2) + d*(di,d 2), therefore f(d\) — f(d 2) < d*(d\,d 2). From symetrics we have f(d 2) - f(d x) < d*{di,d 2) and \f(di) - f(d 2)\ < d*(d ud 2). From this we see that the function / is uniformly continuous on A4. Obviously for x, y £ X Id l(x,y) - d 2(x,y)\ > di(x,y) - d 2(x,y) > g(d x) - d 2(x,y). Then (1) g{di)-g(d 2)< inf |di(®,y) - d 2(x,y)\ < d k(d ud 3). x , y fc X According the inequality \d\ (x, y) - d 2(x, y)\ > d 2(x,y) — d\(x,y), similarly to the previous we get (2) g(d 2)-g(d l)<d^d ud 2).
