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Math Help - Smooth = continuous?

  1. #1
    Senior Member slevvio's Avatar
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    Smooth = continuous?

    Is a smooth (infinitely differentiable) map always continuous?

    I am not sure about this. Here is my problem:

    I have a smooth map \phi:W \rightarrow \mathbb{R}^3, where W\subseteq\mathbb{R}^3 is open. For p\in W, ive got an open neighbourhood of \phi(p) \in \mathbb{R}^3, say V.

    It is claimed that  \phi^{-1}(V) is an open neighbourhood of p. Surely \phi must be continuous for this to be the case? Any help with this would be appreciated!
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  2. #2
    MHF Contributor

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    In order to be once differentiable, a function must be continuous.
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  3. #3
    Senior Member slevvio's Avatar
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    Thanks!
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