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Assume S isn't a vector space. Since S isn't a vector space, then V isn't a vector space; however, V is a vector space. By contradiction, S is a subspace.

Correct?

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- Thread starter Dustinsfl
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Assume S isn't a vector space. Since S isn't a vector space, then V isn't a vector space; however, V is a vector space. By contradiction, S is a subspace.

Correct?

- #2

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Since S isn't a vector space, then V isn't a vector space

Why is this true?

- #3

Mark44

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Why is this true?

I said assume S isn't a vector space.

We have [tex]P\rightarrow Q\equiv P\and\sim Q[/tex]

I don't know why the latex is looking messed up but that is supposed to say P and not Q is equivalent to P implies Q

- #5

Mark44

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Which one is A? You started out with S and V.

- #6

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Which one is A? You started out with S and V.

I meant S.

- #7

Mark44

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but you haven't shown that, given that S isn't a subspace, somehow this implies that V is not a subspace. In other words, if S is not a subspace, why does it necessarily follow that V is not a subspace? In fact, you are really concluding the opposite.

I think you are mixing up a proof by the contrapositive with a proof by contradiction.

See post #3.

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That is why I assumed q (S isn't a vector space).

- #9

Mark44

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P: [itex]S\subseteq V[/itex] and V is a vector space

Q: S is a vector space

Forget deMorgan - look at post #3.

- #10

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Assume S isn't a vector space. Since S isn't a vector space, then V isn't a vector space; however, V is a vector space. By contradiction, S is a subspace.

Correct?

I don't know if this way change anything but it should be worded like so:

If [tex]S\subseteq V[/tex] of a vector space V, then S is a vector space.

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- #12

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The answer is supposed to be true so taking a counterexample seems counter-intuitive.

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S is defined as an improper subset of V; so if V is a vector space, S must be as well.

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S is defined as an improper subset of V; so if V is a vector space, S must be as well.

For all we know, it could be a proper subset too. That was just the notation I used.

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For all we know, it could be a proper subset too. That was just the notation I used.

Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.

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Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.

Look at post #3, which has been said a few times now.

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The answer is supposed to be true so taking a counterexample seems counter-intuitive.

The answer was true so I am not to sure about finding a counterexample as being the correct method.

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But a counterexample *does* exist. Therefore, the statement is false (assuming it was meant to be universally true). The statement is true if and only if S also satisfies the closure axioms of a vector space under the operations of addition and multiplication as defined by V (assuming S is a nonempty subset of V).

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- #19

Mark44

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Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = RS is defined as an improper subset of V; so if V is a vector space, S must be as well.

[itex]S~\subseteq V[/itex], and V is a vector space, but S is not a subspace.

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Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = R^{2}. Let S = {(x, y} | y = 1}.

[itex]S~\subseteq V[/itex], and V is a vector space, but S is not a subspace.

I said S was an improper subset.

- #21

HallsofIvy

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S is defined as an improper subset of V; so if V is a vector space, S must be as well.

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S is defined as an improper subset of V; so if V is a vector space, S must be as well.

For all we know, it could be a proper subset too. That was just the notation I used.

Then it is false. There is one vector that S can exclude yet still retain it's subset status. Think of the definition of a vector space.

There you go guys. I said it was improper because of the notation, then I took it back. Sorry to offend you guys by saying something stupid.

- #23

Mark44

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