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

[MDolphins]

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

 

[WannabeNewton]

What have you tried?

 

[MDolphins]

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

 

[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?

 

[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.

 

[MDolphins]

would i use contradiction

 

[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?

 

[micromass]

Would you use the property that a W is a subspace of V if and only if W is closed under vector addition and scalar mulitplication in V?

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 $w\in V$.

 

[micromass]

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?

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

 

[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.

 

[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?

 

[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}?

 

[micromass]

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}?

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

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

 

[MDolphins]

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

 

[MDolphins]

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

 

[Dick]

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

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?

 

[MDolphins]

True

 

[Dick]

True

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

 

[MDolphins]

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

 

[Dick]

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

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.

 

[MDolphins]

It is not a sunspace because it is not closed under multiplication.

 

[Dick]

It is not a sunspace because it is not closed under multiplication.

I'd go back to saying 'subspace' instead of 'sunspace'. But yes, a subspace would need to be closed under scalar multiplication, which means you should be able to take any element of the subspace and multiply by any real and get an element of the subspace. (-2,2) doesn't work.

 

[MDolphins]

So going back to the proof, the real numbers and 0 are the only subspaces of R bc they seethe only sets that hold under. Scalar multiplication.

 

[Dick]

So going back to the proof, the real numbers and 0 are the only subspaces of R bc they seethe only sets that hold under. Scalar multiplication.

Yes, if a set contains any nonzero number then it must contain all real numbers.