- #1
EV33
- 192
- 0
Homework Statement
What does it mean to have a union of two subsets?
Could someone provide me with an example.
Thank you.
Click For Summary
Homework Help Overview
The discussion revolves around the concept of the union of two subsets, particularly in the context of vector spaces in R^3. The original poster seeks clarification on the definition and an example of this union, as well as its implications for subspaces.
Discussion Character
- Conceptual clarification, Problem interpretation
Approaches and Questions Raised
- Participants explore the definition of the union of subsets, with examples provided to illustrate the concept. Questions arise regarding the geometric interpretation of subsets and their independence. The original poster presents a specific problem involving subspaces U and V in R^3 and attempts to demonstrate that their union is not a subspace.
Discussion Status
The discussion is active, with participants providing examples and clarifications. Some guidance has been offered regarding the properties of subspaces and the implications of the union of U and V. The original poster is seeking confirmation of their understanding and the correctness of their reasoning.
Contextual Notes
The original poster's problem involves specific definitions of subspaces in R^3, and they reference conditions that must be satisfied for a set to be considered a subspace. There is an emphasis on the intersection of subsets and the implications of having no common members.
What does it mean to have a union of two subsets?
Could someone provide me with an example.
Thank you.
- #2
tiny-tim
Science Advisor
Homework Helper
- 25,837
- 258
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).
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
EV33
- 192
- 0
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
ystael
- 352
- 0
Can you give a little more context? The words "geometrically" and "independent" suggest that you may be thinking of something else.
- #5
Mark44
Mentor
- 38,107
- 10,661
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.
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
EV33
- 192
- 0
Thank you. I think Iknow what you mean now. But just to make sure... Do I have the correct idea?
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
To be a sub-space...
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.
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.
Homework Statement
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
Homework Equations
To be a sub-space...
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.
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
ystael
- 352
- 0
This is essentially correct. To be concrete, you could exhibit specific elements u \in U and v \in V such that u + v \notin U \cup V.
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