Homework Help: Find a Subset that is not a Subspace

1. Jul 8, 2009

1. The problem statement, all variables and given/known data
Give an Example of a subset U of R2 that is closed under addition and under taking additive inverses (i.e., -u in U whenever u in U), but is not a subspace of R2

Okay, I know that this problem is not hard, but I just need a hint. I don't want to just start arbitrarily guessing conditions to impose on my subset.

From the last example that I did, it is pretty clear that subsets are closed when one element is simply a multiple of another or when there elements sum to 0.

So given {$U=(x_1,x_2)\in\mathbf{R}^2 :\, \dots$}

2. Jul 8, 2009

What about imposing the condition that x1=x2+b for all b > 0 ? Is that too restrictive a condition to impose?

If x=(x1, x2) and y=(y1+y2) then x+y=(x2+y2+2b, x2+y2)

is that in U since I have a 2b ?

3. Jul 8, 2009

CoCoA

I would first think about what condition of being a subspace is going to be violated by the subset. It is closed under addition by your problem statement, so your subset must not be closed under scalar multiplication.

4. Jul 8, 2009

Or the additive identity, right? Which is what I have tried to do with the above.
However, I do not think that my subset is closed under addition

So I am looking for a subset that is closed under addition but not under scalar...

5. Jul 8, 2009

CoCoA

Pick a vector v in U. By the additive closure, we know rv in U for each integer r. What about xv for some non-integer scalar x?

6. Jul 8, 2009

I am not sure why the definition of scalar multiplication conflicts with non-integers?

a(x1,x2) = (a*x1,a*x2) for all a in R.... non-integers are certainly real, right?

7. Jul 8, 2009

I actually think that this works. If b is in R, then 2*b is also in R. Thus it is closed under addition, but the additive ID does not exist in U.

And -(x1,x2) = (-x2-b, x2) is in U

8. Jul 8, 2009

snipez90

CoCoA meant that U = Z^2 (pairs of integers) satisfies the problem conditions. Remember, U is a subspace, so when we are talking about scalar multiplication, we are referring to the field F of the original vector space (in this case, R). Clearly, an element of Z^2, upon multiplication by a non-integer element of R, is no longer an element of Z^2.

9. Jul 8, 2009

I am sorry, I do not follow Why do you say that the elements of U are "pairs integers?"

Where does it say that? Sorry. I am now lost.

10. Jul 8, 2009

I am sorry, I do not follow Why do you say that the elements of U are "pairs integers?"

Where does it say that they have to be integers? Sorry. I am now lost.

11. Jul 8, 2009

CoCoA

I'm not sure without your definition of your set U. On the face of what you wrote, I am concerned that you give (-x2-b,-x2) in U; here we have x1=-x2-b which is not equal to -x2+b as in your condition on elements of U. In fact, if v and -v are in U, then 0 is in U by additive closure. Thus, since your problem statement gives every additive inverse in U, you cannot violate 0 in U in this case.

On the other hand, if you think about the condition that I give, you can problably come up with a specific example where not every scalar multiple of a vector in U is also in U; you do know that all the integer multiples are in U by the problem statement, so I was just hinting at what type of subset you need to consider to find an element with not every scalar multiple in the subset as well.

12. Jul 8, 2009

There is no definition of U. We are trying to define it. That is the problem statement.

"Define U such that it is closed under addition; the additive inverse is in U, and is not a subspace."

CoCoA:

I see your point about additive closure and the inverse.... but, I still don't understand what the problem statement has to do with integers.

I am new to all of this terminology. But to my understanding if R^2 is a vector space over R then its elements are all possible ordered pairs that are comprised of real numbers.

If U is a subset of R^2 , then isn't U just a set of more ordered pairs that come from R^2 ?

If yes, then why do they have to be integers?
If no, then where is my interpretation of the definition of a subset failing?

13. Jul 8, 2009