Theoretical groups proof 1,1

1. Oct 26, 2009

lom

V is a vectoric space.

$$W_1,W_2\subseteq V\\$$
$$W_1\nsubseteq W_2\\$$
$$W_2\nsubseteq W_1\\$$
prove that $$W_1 \cup W_2$$ is not a vectoric subspace of V.

i dont ave the shread of idea on how to tackle it

i only know to prove that some stuff is subspace

but constant mutiplication

and by sum of two coppies

this question here differs alot

Last edited: Oct 26, 2009
2. Oct 26, 2009

Staff: Mentor

Differs a lot from what? Is there any other information given in your problem? For example, you have that
$$W_1\nsubseteq W_2$$
and
$$W_2\nsubseteq W_1$$

but are you given anything about
$$W_1 \bigcap W_2$$
?

BTW, the term is "vector space" not vectoric space. No such word as vectoric.

3. Oct 26, 2009

lom

it differs by its pure theoretic way

i am used to prove that f(x)=0 is a subspace
by the constant multiplication and sum of two copies law
i only know that
$$W_1,W_2\subseteq V\\$$

i have written all the given stuff

4. Oct 26, 2009

Tedjn

Since you are only given conditions on two subspaces, the only thing you can do next is look at the individual elements. The two conditions $W_1 \not\subseteq W_2$ and $W_2 \not\subseteq W_1$ imply the existence of what elements in these sets?

5. Oct 26, 2009

lom

W1 and W2 are foreign to each other
there is no intersection between them

6. Oct 26, 2009

Tedjn

Are you sure that there is no intersection between them? Let's take a step back. What is the actual condition that $W_1 \not\subseteq W_2$?

7. Oct 26, 2009

lom

its not only
$W_1 \not\subseteq W_2$
its both
$W_1 \not\subseteq W_2$
and $W_2 \not\subseteq W_1$

as for what you say:
$W_1 \not\subseteq W_2$
means that all the members of W1 are not a part W2 group

8. Oct 26, 2009

Tedjn

No, it does not mean that all the members of W1 are not in W2. It means that there exists a member of W1 that is not in W2. Does that make sense? If so, where can you go from there?

9. Oct 26, 2009

Staff: Mentor

You can't conclude that from these two statements:
$$W_1\nsubseteq W_2\\$$
$$W_2\nsubseteq W_1\\$$

It's very possible that W1 contains some elements that are in W2, but other elements that aren't in W2. Same thing for the other statement. That's why I asked if you were given that these two sets are disjoint. You said you weren't given that information, and now here you're saying that they are.