By Stanislaw Saks

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**Example text**

1. Since these semi (x0±r,y0), C(z0;r), is also a continuum circumferences have common points, namely the points their sum, i. e. the circumference (cf. 2). We can deduce further from this the connectedness of every P(z0;r1,r1), where we can assume again (cf. § 8, p. 20) that z0 =t= oo. Let z1 and z 1 be arbitrary points of this annulus and, for brevity, let R 1= lz 1 Z0 ! 2) is therefore also a continuum; as w� perceive at once,· it con z 1 and z 1, and is itself contained in the annulus P(z0;r1,r1).

Z.. are closed sets, and F(t,z1' z2, • • • , Zn) a finite atid continuous function on the set T x Z 1 X Z2 X .

3. The set of irrational points of every interval not reducing to one point, is non-denumerable and of the same power as the set of all real numbers. 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.