# Re: Set operations - CORRECTED - please look at this one instead

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by T.Gracken on October 02, 2002 at 07:57:31:

In Reply to: Re: Set operations - CORRECTED - please look at this one instead posted by Joel on October 01, 2002 at 19:14:45:

: : using the following symbols:
: : u = union
: : e = "is member of"
: : v = disjunction (THIS IS THE CORRECTION)
: : | = such that
: : c= = "is subset of"

: : is this a valid proof that
: : (A u B) c= (A u B u C) ("A union B is a subset of A union B union C")?

: : given an X | X e (A u B)
: : then (X e A) v (X e B)
: : then (X e A) v (X e B) v (X e C) [this is the step that worries me, since no information is given regarding membership or non-membership in C, but there has to be some way to introduce C into the picture]

: : then X e (A u B u C)
: : therefore any (X | X e (A u B)) e (A u B u B)
: : so by the definition of subset, (A u B) c= (A u B u C)

: : The whole question seems so obvious and trivial, and this argument seems so simple that it is hard to believe that this is a legitimate proof. Also, I'm bothered by the fact that "C" seems to materialize out of thin air, but there has to be some "legal" way to get C into the argument. Is this it?

There is no proof needed. By definition, if x is an element of set A, then x is an element of any union of sets in which A is included.

Name:
E-Mail:

Subject: