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Thread: Continuity definiton problem

  1. #1
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    Continuity definiton problem

    Show that $\displaystyle f: A \rightarrow R^{m} $ is continuous at $\displaystyle x_0 $ iff for every $\displaystyle \epsilon > 0 $, there exist $\displaystyle \delta > 0 $ such that $\displaystyle || x-x_0 || \leq \delta $, we have $\displaystyle ||f(x)-f(x_0)|| \leq \epsilon $

    So I can replace the definition < with $\displaystyle \leq $?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Show that $\displaystyle f: A \rightarrow R^{m} $ is continuous at $\displaystyle x_0 $ iff for every $\displaystyle \epsilon > 0 $, there exist $\displaystyle \delta > 0 $ such that $\displaystyle || x-x_0 || \leq \delta $, we have $\displaystyle ||f(x)-f(x_0)|| \leq \epsilon $
    So I can replace the definition < with $\displaystyle \leq $?
    I am not sure that I follow your question.
    Do note that the tradition definition is for <.
    However, it is also $\displaystyle \left( {\forall \varepsilon > 0} \right)\left[ {\frac{\varepsilon }
    {2} < \varepsilon } \right]$, that is for all.
    So $\displaystyle \left\| {f(x) - f(x_0 )} \right\| \leqslant \frac{\varepsilon }
    {2} \Rightarrow \quad \left\| {f(x) - f(x_0 )} \right\| < \varepsilon$
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  3. #3
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    I asked the professor today and he said this problem is suppose to show that the two definitions for continuity are equiv.

    Well, from the last post I understand how to go from $\displaystyle \leq $ to <

    How to do this the other way around is killing me...

    Any hints? Thanks.
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