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Math Help - Algebra

  1. #1
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    Algebra

    Hi all
    Last edited by glen-cullen; April 2nd 2008 at 07:06 AM.
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  2. #2
    Junior Member teuthid's Avatar
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    It's been a long time since I did anything like this, but I'll give it a try. I'm confident in the basic approach of the proof, but I feel like there might be something missing between the first and second step...



    Let (x_0,y_0) be a limit point of gr(f).

    \rightarrow y_0=\lim_{x\rightarrow x_0} f(x)

    \rightarrow y_0=f(x_0) because f(x) is continuous.

    \rightarrow (x_0, y_0)\  \epsilon\  gr(f)

    \rightarrow gr(f) is a closed subset of \Re^2
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  3. #3
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    Opalg's Avatar
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  4. #4
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    algebra

    Thank you again for your help
    Last edited by glen-cullen; April 2nd 2008 at 07:07 AM.
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  5. #5
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    Suppose that \left( {a,b} \right) \notin gr(f) that implies f(a) \ne b.
    Let \varepsilon  = \left| {f(a) - b} \right| > 0. From continuity we get \left( {\exists \delta  > 0} \right)\left[ {\,\delta  < \frac{\varepsilon }{4}\& \,\left| {x - a} \right| < \delta  \Rightarrow \left| {f(a) - f(x)} \right| < \frac{\varepsilon }{4}} \right]
    As a sort of lemma, any point in the ball centered at (a,b), \left( {x,y} \right) \in B_\delta  \left[ {(a,b)} \right]\quad  \Rightarrow \quad \left| {a - x} \right| < \delta \,\& \,\left| {b - y} \right| < \delta .
    Can you show that? Hint: \left| {x - a} \right| = \sqrt {\left( {x - a} \right)^2 }  \le \sqrt {\left( {x - a} \right)^2  + \left( {y - b} \right)^2 } .

    So suppose some point of the graph is such that \left( {c,f(c)} \right) \in B_\delta  \left[ {(a,b)} \right].
    Can you explain the following contradiction?
    \varepsilon  = \left| {f(a) - b} \right| \le \left| {f(a) - f(c)} \right| + \left| {f(c) - b} \right| < \frac{\varepsilon }{4} + \frac{\varepsilon }{4} = \frac{\varepsilon }{2}

    That shows that the complement of the gr(f) is open.
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  6. #6
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    algebra

    Thank you Plato so much for your help. I understand that method much better. Thanks again
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