Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Open interval (set) end points and differentiability.

  1. Jul 30, 2016 #1
    When we talk about differentiability on a
    Set X, the set has to be open.

    And if a set X is open there exists epsilon> 0 where epsilon is in R.

    Then if x is in X, y=x+ or - epsilon and y is also in X

    But this contradicts to what i was taught in highschool; end points are excluded in the open interval.

    Could anyone clarify this for me?

    Also, since epsilon is arbitrary number, if set it to be infinite then would X be R?
    (Entire Real number set)
     
  2. jcsd
  3. Jul 31, 2016 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    That isn't a grammatically correct statement and it isn't the correct definition for "set X is open".

    A variable representing a single number can't be "set to be infinite" and it can't be set equal to a set of numbers.
     
  4. Jul 31, 2016 #3

    pwsnafu

    User Avatar
    Science Advisor

    Let ##X \subset \mathbb{R}##. We say ##X## is an open set if and only if for all ##x \in X## there exists ##\epsilon > 0## such that ##(x-\epsilon, x+\epsilon) \subset X##.

    The order of the ##\epsilon## and ##x## is important: ##\epsilon## depends on ##x##. This means it is not arbitrary.
     
  5. Jul 31, 2016 #4

    fresh_42

    Staff: Mentor

    Why? What about ##f: [0,1] \rightarrow \mathbb{R}## with ##f(x) = x## or ##f(x) = \frac{1}{x}##? Why shouldn't we talk about differentiability here?
     
  6. Jul 31, 2016 #5

    Stephen Tashi

    User Avatar
    Science Advisor

    It's also clearer to say "for each ##x \in X##" because we aren't insisting that there is a single ##\epsilon## that works for all ##x##.
     
  7. Jul 31, 2016 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    It is not usually done in analysis books (at least not introductory ones). The question is how we should define differentiability at the boundary. You probably have quite a good idea how to do that, but it does complicate matters somewhat. Furthermore, differentiability at the boundary is almost never needed for elementary analysis results.

    Also, when we generalize analysis to ##\mathbb{R}^n## it becomes even more difficult to describe differentiability at the boundary. It's not impossible to do it, but it takes some effort which is usually spent in other places.

    In fact, I know of two good generalizations of differentiability at the boundary, and I'm not sure whether they even are equivalent. I should think a bit about this.
     
  8. Jul 31, 2016 #7

    fresh_42

    Staff: Mentor

    I basically wanted to animate the OP to think about why an open neighborhood is "needed", and to sharpen his argumentation, because "the set has to be open" isn't correct in this generality. A reflex, if you like. I thought it would help to understand differentials as linear approximations and what approximation really means in this context.
    (I wonder if any derivation on any ring could be considered a differentiation .... with no metric in sight ...)
     
  9. Jul 31, 2016 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You mean in the sense of differential algebra? https://en.wikipedia.org/wiki/Differential_algebra
     
  10. Jul 31, 2016 #9

    fresh_42

    Staff: Mentor

  11. Aug 5, 2016 #10

    Thanks. And I am sorry for the confusing statements and wording.

    I sort of understand why we need an "open set" when we define differentiability on a set X. As far as I know, we'd need to define right or left differentiabilty if the set
    is closed.

    But here I have two questions

    1. I learned in highschool that open intervals exclude end points. But in an open set X, it would include end points and its neighborhood.
    Are they different in terms of definition?

    2. If we consider the boundary points of X, intuitively half of (x-eps,x+eps) would include the interval outside the set X. Do we define (x-eps, x+eps) to be in X because epsilon is small?
     
  12. Aug 5, 2016 #11

    fresh_42

    Staff: Mentor

    That's correct.
    No. ##(a,b) = \{x \in \mathbb{R} \, | \, a < x < b \}## is an open set, ##(a,b] = \{x \in \mathbb{R} \, | \, a < x \leq b \}## is neither open nor closed, and ##[a,b] = \{x \in \mathbb{R} \, | \, a \leq x \leq b \}## is a closed interval.
    What do you mean by "half of ... outside"?

    We consider an open interval ##(a,b)## in ##X = \mathbb{R}##, a point ##x \in (a,b) \subset X## and a neighborhood ##(x-\epsilon,x+\epsilon) \subset (a,b)## which is entirely within our open set / interval. We then speak of differentiability at (or in) ##x##. The fact that ##(a,b)## is open and ##x \in (a,b)## that is ##a < x < b## guarantees us, that we can always find some neighborhood left and right of ##x## that is still in ##(a,b)## so we must not deal with any boundaries, where e.g. there is no limit from the left if it is the left end point. And we can get as close to ##x## as we like - from both sides.
     
  13. Aug 5, 2016 #12
    Oh shoot. I see. Somehow I thought the open set would include the end points.. Thanks. It is so weird. where I got confused of this..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Open interval (set) end points and differentiability.
  1. Open Intervals (Replies: 5)

Loading...