Demostraciones de álgebra
Mis apuntes de Matemática Discreta.
El lema 1 de reticulados nos dice:
Sea (A,+,.) perteneciente a la clase de los reticulados, entonces (A,<=) es un conjunto ordenado en el que todo par de elementos tiene infimo y supremo, donde el infimo es x.y y el supremo x+y, para todo x,y perteneciente a A.
Demostracion:
Sea (A.+,.) un reticulado, veamos que (A,<=) es un conjunto ordenado. Para ello probemos que
a<=b sssi a.b=a sssi a+b=b
es un orden parcial.
Reflex)
Para todo a perteneciente a A se verifica que a<=a. Para todo a perteneciente a A se verifica que a+a=a. Por idempotencia, a<=a.
Antisim)
Para todo a,b perteneciente a A se verifica que si a<=b y b<=a entonces a=b.
Para todo a,b,c perteneciente a A se verifia que si a<=b y b<=c entonces a<=c.
si a<=b y b<=c entonces a+b=b y b+c=c
a+c=a+(b+c)=(a+b)+c=b+c=c entonces a<=c
Por lo tanto, <= es una relacion de orden.
Debo probar que x<=x+y e y<=x+y
x+(x+y)=(x+x)+y=x+y entonces x<=x+y
Analogamente se prueba que y<=x+y
Por lo tanto, x+y es cota superior de x e y.
Ahora, debo probar que x+y es la menor cota superior.
Sea u cota superior de {x,y}, veamos que x+y<=0:
Como u es cota superior de {x,y} entonces x<=u y y<=u. Por esto, x+u=u y y+u=u
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