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Differentitation - Why open set?

  1. Sep 21, 2008 #1
    Hey guys,

    why is it that when we define [tex]f:U\subset R^n \rightharpoonup R^m[/tex], why does U have to be open?
    what happens if it's closed?

    Thanks
     
  2. jcsd
  3. Sep 21, 2008 #2

    HallsofIvy

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    I am a bit confused. Your title is "Differentiation- Why open set?" but you don't mention differentiation in the body.

    If you are asking about mappings in general, no U does not have to be open. But if you are asking about points at which you can define the derivative of f, that's a different story.

    Think about what happens in the "Calculus I", R1 case. The derivative of f(x), at x= a, is defined by
    [tex]\lim_{h\rightarrow 0} \frac{f(a+h)- f(a)}{h}[/tex]
    In order for that to exist, f(a+ h) has to be defined for all h close to 0- positive or negative. And that means that a has to be in the interior of some interval on which f is defined. If a is on the boundary, we can only define the "one sided derivative", using a one sided limit.

    Now, in the more general, n dimensional, case, we have to be able to define f(x) for all x close to a. In other words, a must be an interior point of the domain of f. The set of all interior points of a set is, of course, an open set.
     
  4. Sep 21, 2008 #3
    oops... my bad.
    Was trying to find an answer in my textbook as I wrote that question.

    Nothing about it in there, but thanks a lot :)
     
    Last edited: Sep 21, 2008
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