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Math Help - Union of subspaces is a subspace iff..

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    Union of subspaces is a subspace iff..

    Hi everyone,

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



    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?
    Last edited by BryanUrizar; May 10th 2013 at 01:48 AM.
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    Re: Union of subspaces is a subspace iff..

    Quote Originally Posted by BryanUrizar View Post
    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  \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 \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 W_1\cup W_2is a subspace, then s+t\in W_1\cup W_2~.

    Now what?
    Last edited by Plato; May 10th 2013 at 07:13 AM.
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    Newbie BryanUrizar's Avatar
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    Re: Union of subspaces is a subspace iff..

    Quote Originally Posted by Plato View Post
    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  \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 \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 W_1\cup W_2is a subspace, then 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 x. As it stands, I have only shown that for any x and any y \in W_1, W_2 respectively, that either x \in W_2 or y \in W_1 which isn't enough to conclude W_1 \subset W_2 or W_2 \subset W_1. For a direct proof I need to assume W_2 \not \subset W_1 and choose y \in W_2-W_1 and proceed in a similar fashion.
    Last edited by BryanUrizar; May 10th 2013 at 07:16 AM.
    Thanks from emakarov
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