Is the Union and Intersection of a Null Collection Valid in Set Theory?

[JasonRox]

The book that I'm reading is saying...

If C is the null collection of subsets of S then,

(Union) C = Null

(Disjoint) C = S

Is this true?

 

[Enuma_Elish]

How does your book define a null collection of subsets?

 

[matt grime]

The book that I'm reading is saying...
If C is the null collection of subsets of S then,
(Union) C = Null
(Disjoint) C = S
Is this true?

Take it as a definition, or read this, for example:

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2004;task=show_msg;msg=0896.0001

since i presume by (disjoint) you actually mean intersection.

incidentally i got that answer by insertingf the words empty intersection into google and clicking the first link.

empty union requires you to follow the third (non indented) link.

you might want to remember that the next time you struggle to check a definition,

 

[JasonRox]

The truth is that I searched and searched. Then I thought and thought, then searched again.

Using the definition it is true, and I see that, but I was skeptical about it.

Thanks, for the link.

 

[mathwonk]

imagine a list of rules for a game which has no rules at all!

the intersection of a null collection of sets, corresponds to those moves which satisfy all the rules, hence any move at all, i.e. S.etc...you do the other case

 

[inquire4more]

Wow, the exact problem I was having a couple of months or so ago...
This might be of some help, maybe...

https://www.physicsforums.com/showthread.php?t=94777

 

[HallsofIvy]

The book that I'm reading is saying...
If C is the null collection of subsets of S then,
(Union) C = Null
(Disjoint) C = S
Is this true?

if x is in (Union)C, then it must be in at least one of the members of C. But C has no members so that is always false. Yes, (Union) C= Null set.

By (Disjoint) C do you mean the intersection[\b] of all the members of C?

Let x be any member of S. If x is NOT in (intersection) C, then there must be some member of C such that x is NOT in it. But that's NEVER true because C has no members! Therefore every member of S is in (Intersection) C.