Fixed points of mappings of metric spaces by Ivanov

By Ivanov

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By virtue of b) one can ~>0 and ~0 such that if#>~g0, then T~('ut~(A))=~{ CA) . Since ~ ~(~K, ~)= 0, one has ~r ~(~K,~T)= 0 and under our assumption about the sequence ( K)K=O one can find a K~ such that for K~K0, and consequently, But then for ~ ~(~K)~ ~{ (~)), which contradicts the condition ~ Now we consider the numbers 4 - ~ , with which ~ ~Ke[~(A) j~=-G~ F=-G, s $~ K = and ~ o , ~(~(~K)~ @(~)) = 0. ,T~)~ ~ * ( ~ , ~ ) . For the number indicated here, (i) and (2) are satisfied and it remains only to verify that (3) is satisfied.

4,T=~. ~) - ~(T~,TCT~))'~ ~, which contradicts the assumption. Thus, it is proved that ~{~ ~(~n,Tx~)= 0. Since I ( ~ , . 4)== 0. Now we consider the double sequence ( ~ ( z ~ . m ) ) a n d such that for some arbitrarily large for ~ according to (2). numbers ~ and ~ , we assume that one can find an ~>0 ~(~K,~,~) ~ ~s Obviously, one can assume that ~ we have ~ ( m ~ , T ~ ) ~ f , d'< & . Let cF~0 be a number found For sufficiently large so one can find arbitrarily large numbers m and ~ for which *Condition of Meir and Keeler [43].

Since As the co- 0 is a countable set, one can find a line passing through 0 which does not contain any other points of ~ . On this line we fix an arbitrary direction and we take the axis so obtained as the m axis. Then the ~ axis will be uniquely determined and one can consider the map T: ~ - - ~ d e f i n e d in Paragraph 4. general c a s e , T ( X ) r X, so one is led to seek another map T ~ : ~ - - - L , Tf(X)c~. ~ ~X except for such that 3 [ ~ ~ , taking as ~ ~0, we consider some closed for ~<~, B ~ we take the map T itself.

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