# 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

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

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

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