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

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- Thread starter MDolphins
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- #1

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

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- #2

WannabeNewton

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What have you tried?

- #3

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- #4

WannabeNewton

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

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

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would i use contradiction

- #7

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DOes this seem right?

- #8

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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##.

- #9

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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.

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- #11

WannabeNewton

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- #13

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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}##?

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

- #15

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

- #16

Dick

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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?

- #17

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True

- #18

Dick

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True

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

- #19

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

- #20

Dick

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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.

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- #21

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

- #22

Dick

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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.

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- #24

Dick

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

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