
#1
Feb910, 01:24 AM

P: 197

1. The problem statement, all variables and given/known data
What does it mean to have a union of two subsets? Could someone provide me with an example. Thank you. 



#2
Feb910, 05:13 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi EV33!
It's the subset consisting of everything in either subset. For example, the union of red cars and white cars is all cars which are red or white. And the union of red cars and fast cars is all cars which are red or fast (or both). 



#3
Feb910, 03:24 PM

P: 197

What does this mean geometrically though? I don't see how one subset could be in another subset if they are independent of each other.




#4
Feb910, 03:38 PM

P: 352

union of two subsets
Can you give a little more context? The words "geometrically" and "independent" suggest that you may be thinking of something else.




#5
Feb910, 03:44 PM

Mentor
P: 21,026

If the two subsets have no members in common, their intersection will be empty. For example, the union of O = {1, 3, 5, 7, ...} and E = {0, 2, 4, 6, 8, ...} is the set {0, 1, 2, 3, 4, ...}. They have no members in common.
If A = {0, 4, 8, 12, ...}, A U E = E. In this case, set A is a subset of E, so every member of A is automatically a member of E, but not vice versa; there are members of E that aren't also members of A. Because A is a subset of E, their intersection is not empty. 



#6
Feb910, 07:26 PM

P: 197

Thank you. I think Iknow what you mean now. But just to make sure... Do I have the correct idea?
1. The problem statement, all variables and given/known data So here is my actual problem. Let U and V be the subspaces of R^3, defined by U={x:a^(T)x=0} and V={x:b^(T)x=0} where a= 1 1 0 and b= 0 1 1 Demonstrate that the union of U and V is not a subspace of R^3 2. Relevant equations To be a subspace... 1. it needs to contain the zero vector 2. x+y is in W whenever x and y are in W. 3. ax is in W whenever x is in W and a is any scalar. 3. The attempt at a solution 1. they both have the zero vector because a solution to a^(T)x=0 and b^(T)x=0 is x=0. 2. An arbitrary vector that is in U={x:a^(T)x=0} would be any vector that has two zeros in the first two rows, and an arbitrary vector in V={x:b^(T)x=0}, would be any vector with the bottom two rows as zeros. And because U and V are unioned I can choose one arbitrary vector from each U and V individually and for it to be a subspace it would be in the union of U and V. U= 0 0 S V= T 0 0 If you add these two arbitrary vectors together you get T 0 S which is in neither U or V, therefore the union of U and V is not subspace. 



#7
Feb910, 07:29 PM

P: 352

This is essentially correct. To be concrete, you could exhibit specific elements [tex]u \in U[/tex] and [tex]v \in V[/tex] such that [tex]u + v \notin U \cup V[/tex].



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