Prove that ℝ has no subspaces except ℝ and {0}.

1. Mar 19, 2013

MDolphins

Prove that ℝ has no subspaces except ℝ and {0}.

Last edited by a moderator: Mar 20, 2013
2. Mar 19, 2013

WannabeNewton

What have you tried?

3. Mar 19, 2013

MDolphins

Ive tried using a nontrivial subspace of R and showing that it equals R but I am having touhg time doing that

4. Mar 19, 2013

WannabeNewton

Ok that's a good way to do it. Let's call this non-zero subspace $V$ and let $v\in V$ be a non-zero vector. Can you think of a property of vector spaces that we can use here to give us a critical result?

5. Mar 19, 2013

5hassay

Assuming we are dealing in the realm of undergraduate linear algebra...

Continuing on what WannabeNewton said, think about why $\mathbb{R}$ is a subspace of itself, and then consider some nonempty set that is not $\mathbb{R}$ or the set consisting of just $0$.

Recall that a nonempty set is a subspace of another set if and only if it is a subset of it and it is closed under both addition and "scalar" multiplication.

6. Mar 20, 2013

MDolphins

7. Mar 20, 2013

MDolphins

Well, if we were to look at a subspace that is not in ℝ, it would not be closed under the same addition or multiplication that is in ℝ. And additionally, from the theorem "if a subset S of a vector space V does not contain the zero vector 0 of V, then S is not a subspace of V". From this, the 0 vector of the ℝ is the set 0 itself. THerefore, the subspace must be ℝ and {0}.

DOes this seem right?

8. Mar 20, 2013

micromass

Staff Emeritus
Yes.

So you have a nontrivial vector space $V\subseteq \mathbb{R}$. You have a nonzero vector $v$.

You wish to prove somehow that $V=\mathbb{R}$. So given any $w\in \mathbb{R}$, you wish to prove that $w\in V$.

9. Mar 20, 2013

micromass

Staff Emeritus
No, this is completely incorrect. I would wish I could tell you what was wrong, but I don't understand your proof at all.

10. Mar 20, 2013

MDolphins

So to show that w is in the vector space V, would I assume that it is not. Then, go on to show that if it is not it is not a part of the real numbers. However, under the assumption I said w is apart of the real numebrs, therefore a contradiciton.

11. Mar 20, 2013

WannabeNewton

You don't need contradiction here mate. Did you try what I said? Let me be a bit more direct: let $V\subseteq \mathbb{R}$ be a non zero subspace of $\mathbb{R}$ over the field $\mathbb{R}$ (This is crucial! I'm assuming your assignment was to prove this assuming the field was $\mathbb{R}$ correct?). Let $v\in V$ be non zero. Remember that $v$ is still technically a non zero real number. How can you use the closure of a vector space under scalar multiplication to help you finish the proof?

12. Mar 20, 2013

MDolphins

Well, the closure property holds for v, because since v is a part of the reals, any real number times v will still be apart of the reals. Therefore, the only subspace of the reals are the reals and {0}?

13. Mar 20, 2013

micromass

Staff Emeritus
Could you expand more? Why does the "therefore" hold??

For example, can you tell me why $(-2,2)$ is not a subspace of $\mathbb{R}$?

14. Mar 20, 2013

MDolphins

I'm confused as to why (-2,2) is not a subspace. I am lost.

15. Mar 20, 2013

MDolphins

Is (-2,2) not a sunspace because all the scalers are of the natural numbers?

16. Mar 20, 2013

Dick

You are talking gibberish. The scalars are all real numbers. Since 1 is in your 'sunspace' the k*1 should be in it for all real k. That's the scalar product. True or false?

17. Mar 20, 2013

MDolphins

True

18. Mar 20, 2013

Dick

Explain why you think so. If k=3 then 3*1=3. 3 is not in (-2,2), is it?

19. Mar 20, 2013

MDolphins

No 3 is not. The sunspace (-2,2) does not contain all of the real numbers of R. For example 3.

20. Mar 20, 2013

Dick

I hope you are saying that (-2,2) is not a subspace, and I hope you know why. Can you explain why? Don't just say "it doesn't contain all real numbers". Give a good reason why a subspace that's not {0} would have to.

Last edited: Mar 20, 2013