By von Neumann J.

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**Extra resources for Collected works. Vol.4 Continuous geometry and other topics**

**Sample text**

Now we deﬁne a multiplication in M by ^ a b :D f˛ ˇ j ˛, ˇ 2 Z ^ ˛ 2 a ^ ˇ 2 bg. a,b2M In order not to leave the set Z while executing the product, we replace the real and/or imaginary part of the product ˛ ˇ by r whenever the former and/or the latter do in fact exceed r . Then fM , g is a groupoid with the neutral element f1g. Now set r D 5 and consider two special elements a, b 2 M . 12 (a). 1 C i/g. 2 Here a, b are elements of S M . a b/. When the groupoids are ordered, additional properties can be derived for them.

S (resp. 4 : M ! S) the monotone downwardly (resp. upwardly) directed rounding. For each element a 2 M let I :D Œ 5 a, 4a and let I1 and I2 with I1 < I2 be subsets2 of M which partition I :D I1 [ I2 . Then the mapping : M ! S is a monotone rounding if and only if ( ^ ^ 5 a for all a 2 I1 aDa ^ aD . 4a for all a 2 I2 a2SÂM a2M nS Proof. (a) It is clear that every such mapping is a monotone rounding. (b) We still have to show that every monotone rounding has the property stated in the theorem. Now let I1 :D fa 2 I j a D 5 ag and let I2 :D fa 2 I j a D 4ag.

A ı b/. a,b2S (a) If has the property (Ri), i D 1, 2, 3 deﬁned for roundings, then fS, property (RGi), i D 1, 2, 3, respectively. ı g has the (b) If the groupoid fM , ıg has a right neutral element e and e 2 S, then (RG1), (RG2), and (RG3) imply (RG). Proof. (a) We omit the proof of this property since it is straightforward. (b) We give the proof in the case of a lower screen. a ı b/. a ı b/ Ä a ı b. , e is also a right neutral element in fS, ı g. a ı b/ Ä a ı b. 2) 38 Chapter 1 First concepts (RG3) yields a (R1) and (R2) ı b Ä a ı b.