voved vo teorijata na mno@estvata i matemati^kata logika
voved vo teorijata na mno@estvata i matemati^kata logika
voved vo teorijata na mno@estvata i matemati^kata logika
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A. ■<br />
13.2 o Sekoj ideal S(a) <strong>na</strong> podredeno mno`est<strong>vo</strong> A e dolen segment <strong>na</strong><br />
Primer<br />
1. Neka A={1,2,3,4,5} e podredeno mno`est<strong>vo</strong> so sledno<strong>vo</strong><br />
podreduvawe:<br />
1 2<br />
3<br />
5<br />
4<br />
6<br />
Toga{: S(5)={6}, S(3)={4,5,6}, S()={3,4,5,6} se ideali <strong>na</strong> A, a {4,6},<br />
{4,5,6}, {5,6} se dolni segmenti <strong>na</strong> A. Pritoa {4,6} i {5,6} ne se ideali <strong>na</strong> A.<br />
13.3 o Dolen segment <strong>vo</strong> dobro podredeno mno`est<strong>vo</strong> A e ili A ili<br />
ideal <strong>na</strong> A.<br />
Dokaz: Ako I e segment <strong>vo</strong> A i I⊂A, toga{ <strong>vo</strong> A\I ima <strong>na</strong>jmal element a.<br />
Da doka`eme deka I=S(a). Ako e x∈I, toga{ e xa 2 >...,<br />
pri {to e {a o ,a 1 ,a 2 ,...}⊆S(a o ). No {a o ,a 1 ,a 2 ,...}≠∅ i nema <strong>na</strong>jmal element, {to<br />
z<strong>na</strong>~i deka S(a o ) ne e dobro podredeno mno`est<strong>vo</strong>. ■<br />
13.5 o Ne postoi biekcija {to go zapazuva podreduvaweto od dobro<br />
podredeno mno`est<strong>vo</strong> A <strong>vo</strong> nekoj ideal <strong>na</strong> A .<br />
Dokaz: Neka e a∈A i f:A→S(a) e biekcija {to zapazuva podreduvawe.<br />
Neka T={x∈A|f(x)
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