Proof: S is a Subspace of Vector Space V

[Dustinsfl]

If $$S\subseteq V$$ and V is a vector space, then S is a vector space.

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?

 

[VeeEight]

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

Why is this true?

 

[Mark44]

Suppose V = R³ and S = {(x, y, z)| x + y - 2z = 3}. S is clearly a subset of V, but is S a subspace of V?

 

[Dustinsfl]

Why is this true?

I said assume S isn't a vector space.

We have $$P\rightarrow Q\equiv P\and\sim Q$$

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

 

[Mark44]

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

 

[Dustinsfl]

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

I meant S.

 

[Mark44]

P ==> Q <===> ~P ==> ~Q
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.

 

[Dustinsfl]

I was using de morgan's laws which says p implies q is equiv to ~p or q which is equiv to p and ~ q.

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

 

[Mark44]

Based on your original post, here are P and Q.
P: $S\subseteq V$ and V is a vector space
Q: S is a vector space

Forget deMorgan - look at post #3.

 

[Dustinsfl]

If $$S\subseteq V$$ and V is a vector space, then S is a vector space.

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 $$S\subseteq V$$ of a vector space V, then S is a vector space.

 

[VeeEight]

Take Mark's counterexample (or some other counterexample you can think of), that is the best approach.

 

[Dustinsfl]

Take Mark's counterexample (or some other counterexample you can think of), that is the best approach.

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

 

[Squeezebox]

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

 

[Dustinsfl]

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.

 

[Squeezebox]

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.

 

[Squeezebox]

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.

 

[Dustinsfl]

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.

 

[Subdot]

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).

 

[Mark44]

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

Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = R². Let S = {(x, y} | y = 1}.
$S~\subseteq V$, and V is a vector space, but S is not a subspace.

 

[Squeezebox]

Not true. Not all subsets of vector spaces are themselves subspaces. Here's another example, with V = R². Let S = {(x, y} | y = 1}.
$S~\subseteq V$, and V is a vector space, but S is not a subspace.

I said S was an improper subset.

 

[HallsofIvy]

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

Where was "S defined as an improper subset of V"?

 

[Squeezebox]

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.

 

[Mark44]

No offense taken, just trying to clear up incorrect information. In this problem it doesn't matter whether it said $S \subset V$ or $S \subsetex V$. The latter notation means "S is a subset of V or is equal to V." S can still be a proper subset of V without contradicting this statement.