Union of subspaces is a subspace iff..

Hi everyone,

I made this post on another site thinking my proof was correct:

http://i1054.photobucket.com/albums/...psb7bca245.jpg

However, I have been told that it isn't correct. I got some help, but I still don't understand why it isn't. Can someone please help me understand why the above proof isn't valid?

Thank-you.

P.S. How can I include LaTeX code in my post?

Re: Union of subspaces is a subspace iff..

Quote:

Originally Posted by

**BryanUrizar** P.S. How can I include LaTeX code in my post?

This subforum will help you with the code.

[TEX]\exists t\in W_1\setminus W_2~\&~ \exists s\in W_2\setminus W_1[/TEX] gives $\displaystyle \exists t\in W_1\setminus W_2~\&~ \exists s\in W_2\setminus W_1 $

If you click on the “go advanced tab” you should see $\displaystyle \boxed{\Sigma} $ on the tool-bar. That gives the [TEX]..[/TEX] wrap. Your LaTeX code goes between them.

For proof by contradiction, make the above assumption.

If $\displaystyle W_1\cup W_2$is a subspace, then $\displaystyle s+t\in W_1\cup W_2~.$

Now what?

Re: Union of subspaces is a subspace iff..

Quote:

Originally Posted by

**Plato** This

subforum will help you with the code.

[TEX]\exists t\in W_1\setminnus W_2~\&~ \exists s\in W_2\setminnus W_1[/TEX] gives $\displaystyle \exists t\in W_1\setminnus W_2~\&~ \exists s\in W_2\setminnus W_1 $

If you click on the “go advanced tab” you should see $\displaystyle \boxed{\Sigma} $ on the tool-bar. That gives the [TEX]..[/TEX] wrap. Your LaTeX code goes between them.

For proof by contradiction, make the above assumption.

If $\displaystyle W_1\cup W_2$is a subspace, then $\displaystyle s+t\in W_1\cup W_2~.$

Now what?

Thanks for the LaTeX link.

I know how to prove it by contradiction, but wanted to know specifically what was wrong with my proof. I finally understand why it isn't correct. The problem is that I haven't shown one definite conclusion for any $\displaystyle x$. As it stands, I have only shown that for any $\displaystyle x$ and any $\displaystyle y \in W_1, W_2$ respectively, that either $\displaystyle x \in W_2$ or $\displaystyle y \in W_1$ which isn't enough to conclude $\displaystyle W_1 \subset W_2$ or $\displaystyle W_2 \subset W_1$. For a direct proof I need to assume $\displaystyle W_2 \not \subset W_1$ and choose $\displaystyle y \in W_2-W_1$ and proceed in a similar fashion.