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By Stanislaw Saks

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Implies Z (this · the distance between the points z1 and in agreement with the definitions in the Introduction, §§ 8, 13. ,Zn are F on the set Z closed, then every finite and = Z1 x Z2 x ... x Zn is uniformly continuous on this set. Moreover, if the furiction then at a certain point of the set Z con­ bounded and F is real, it attains the upper bound of its · values on this set. F is not {zJk=i,2, ... in Pr o o f. Let us assume that the function on Z. such that F(z)-oo. tracted from the bounded Z {z(k1)} be a convergent subsequence ex· sequence {zJ.

4. i r e) a set of tke jif'Bt categOf'JI. Prove that . in a complete metric space the complement of every set of the first category is an everywhere dense set in the space (BaiTe's tkeOTem ); therefore no complete metric space can be represented as the sum of a se­ quence of sets nowhere dense in this space. Since every closed set in a complete metric space can· also be considered as a complete metric space, deduce from this that every perfect set in a com­ plete metric space is non-denumerable (generalization of theorem the indispensability of t�e assumption that the space is complete.

A) KnCKn--1 for n>l, (b) the centre of every circle Kn (c) the closed circle Kn belongs to the set A, does not contain the point K1 With this in view, let us select as an. an arbitrary circle whose centre belongs to the set A and which does not contain the point a1 in the interior or on the boundary. Next, let us assume that there have been defined r circles K K 1, 2, • • • , K,. in such a way that the conditions (a), (b), and (c) are satisfied for n�r. Since the circle Kr has a centre at a point belonging to the perfect set A, it contains an infinite number of points of this set.

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