Vector Spaces, Subsets, and Subspaces

[mrroboto]

Homework Statement

What is an example of a subset of R^2 which is closed under vector addition and taking additive inverses which is not a subspace of R^2?

R, in this question, is the real numbers.

Homework Equations

I know that, for example, V={(0,0)} is a subset for R^2 that is also a subspace, but I can't figure out how something can be a subset and not a subspace.

The Attempt at a Solution

Does this have anything to do with scalar multiplication being closed on the vector space?

 

[Dick]

Think about the set of all (x,y) where x and y are both integers.

 

[mrroboto]

So for example, if we let the subset = (a,b) s.t. a,b are elements of Z. Then it is closed under addition but not under scalar multiplication. i.e. Let (a,b) = (1,3) and multiply by 1/2 for example (which is the example we used to figure it out). Then you get (1/2, 3/2), neither of which are in Z.

 

[Dick]

Sure, but it does have additive inverses.

 

[andytoh]

Think about the set of all (x,y) where x and y are both integers.

or both rational numbers

 

[HallsofIvy]

Homework Statement

What is an example of a subset of R^2 which is closed under vector addition and taking additive inverses which is not a subspace of R^2?

R, in this question, is the real numbers.

Homework Equations

I know that, for example, V={(0,0)} is a subset for R^2 that is also a subspace, but I can't figure out how something can be a subset and not a subspace.

It's very easy to be a subset without being a subspace! Just look at any subset that does not satisfy the requirements for a subpace- what about { (1, 0)}?

The Attempt at a Solution

Does this have anything to do with scalar multiplication being closed on the vector space?

In order for a subset to be a subspace, it must be closed under addition, have additive inverses, and be closed under scalar multiplication. Since you are asked about a subset that IS closed under addition and has additive inverses, looks like there is only one thing left!