# Vector space

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?

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

Why is this true?

Mark44
Mentor
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?

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

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

I meant S.

Mark44
Mentor
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.

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
Mentor
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.

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.

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

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.

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

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.

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.

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.

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.

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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Mark44
Mentor
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.

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
Homework Helper
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"?

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
Mentor
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.