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

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

Posted by Joel on October 01, 2002 at 19:14:45:

In Reply to: Set operations - is this a valid proof? posted by Joel on October 01, 2002 at 19:11:12:

: 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?

Follow Ups:

Re: Set operations - CORRECTED - please look at this one instead T.Gracken 07:57:31 10/02/02 (3)
- Re: Set operations - CORRECTED - please look at this one instead Joel 10:06:30 10/02/02 (2)
  - not so... T.Gracken 11:55:25 10/02/02 (1)
    - You make it sound so easy. Thanks. Joel 22:50:31 10/02/02 (0)

Post a Followup

Name:
E-Mail:

Subject:

Comments:
: : 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?

Optional Link URL:
Link Title:
Optional Image URL:

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