Is Every Set in ℝ3 a Subspace?

[MarcL]

Homework Statement

A. {(x,y,z)ι x<y<z }
B.{(x,y,z)ι -4x+2y=0, -5x-7z=0 }
C.{(x,y,z)ι -9x-3y+8z=7}
D.{-7x-8y,9x+6y,3x-6y ι x,y arbitrary numbers}
E.{(x,y,z)ι x+y+z=0 }
F.{(x,x+4,x-2 }

Homework Equations

So it is a subspace therefore I need the additive axiom: u+v = v+u, the one where u + o = u and the additive inverse where u + (-u) = (-u)+u=0

The Attempt at a Solution

I thought all of them would be a subspace of ℝ³ because they all have 3 components and ℝ³ includes all numbers in the subspace, or am I seeing this wrong?

sorry for the picture, edit done!

 

[LCKurtz]

I'm not seeing it at all. Posting unreadable images doesn't help anyone. Type your questions.

 

[Mark44]

Yes, type your problems. See Guidelines for Students and Helpers, #5: Do not simply post images of the problem statement or your work

 

[MarcL]

I am sorry , I couldn't write them down, was on my chromebook, was really hard to do, ill edit right now.

 

[MarcL]

Edit done, sorry about that!

 

[LCKurtz]

Homework Statement

A. {(x,y,z)ι x<y<z }
B.{(x,y,z)ι -4x+2y=0, -5x-7z=0 }
C.{(x,y,z)ι -9x-3y+8z=7}
D.{-7x-8y,9x+6y,3x-6y ι x,y arbitrary numbers}
E.{(x,y,z)ι x+y+z=0 }
F.{(x,x+4,x-2 }

Homework Equations

So it is a subspace therefore I need the additive axiom: u+v = v+u, the one where u + o = u and the additive inverse where u + (-u) = (-u)+u=0

The Attempt at a Solution

I thought all of them would be a subspace of ℝ³ because they all have 3 components and ℝ³ includes all numbers in the subspace, or am I seeing this wrong?

sorry for the picture, edit done!

Yes you are seeing it wrong. You have to check the axioms. Just to get you started, in part A is (0,0,0) in the described set?

 

[MarcL]

forgot the statement. Which of the following sets are subspaces of

was the problem statement. However because of the restriction it wouldn't be right?

 

[LCKurtz]

forgot the statement. Which of the following sets are subspaces of

was the problem statement. However because of the restriction it wouldn't be right?

You need to quote what you are replying to, otherwise it becomes impossible to follow the thread. Because of what restriction what wouldn't be right and why? Explanation required. Maybe even some math statements.

 

[MarcL]

You need to quote what you are replying to, otherwise it becomes impossible to follow the thread. Because of what restriction what wouldn't be right and why? Explanation required. Maybe even some math statements.

This would mean that my 4^th axiom would not hold because u + o = u, my "0" vector would be (0,0,0) to satisfy the axiom, but it would not satisfy my restriction where x<y<z, right? ( if that is what you were trying to get to?)

 

[LCKurtz]

This would mean that my 4^th axiom would not hold because u + o = u, my "0" vector would be (0,0,0) to satisfy the axiom, but it would not satisfy my restriction where x<y<z, right? ( if that is what you were trying to get to?)

Yes, that is what I was getting at. So your first example is not a subspace and that is the reason. It is easy to show something isn't a subspace when it isn't because you only have to find one axiom that doesn't work. There may be some others in your problem that aren't subspaces and you have to show why. What's worse is that some of them are subspaces and you can't just say that I see they are. You have to check that all the axioms work, which can be lots of busy work.

Also, even though your problem image is now readable, the reason posting problems in images is frowned on is that we can't copy, for example, part B to talk about it when we are discussing that problem.

 

[MarcL]

Thank you, I'll give a shot to the rest of them and post if I have another question ,and I understand about the picture. won't happen again :)

 

[Mark44]

You don't need to confirm that all 10 (or 11 or 12, whatever) vector space axioms are satisfied, because all of your sets already belong to a vector space, $\mathbb{R^3}$.

All you need to check are these properties:
1. The 0 element belongs to the set.
2. The set is closed under vector addition. IOW, if u and v are in the set, then u + v is also in the set.
3. The set is closed under scalar multiplication. IOW, if u is in the set and k is a scalar, the ku is also in the set.

If a particular set satisfies all three of these properties, the set is a subspace of the larger vector space.