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Continuity on a Missing Strip Plane

  1. Jul 31, 2013 #1
    I've seen many definitions of continuous functions. They all describe x in a domain, but there's not really much explanation about the domain considerations beyond examples with "all the reals" and "an interval of the reals."

    I'm trying to figure out what continuity would mean on a missing strip plane. For example, we know that [itex]f(x) = x [/itex] is a continuous function on the reals. But consider a missing strip plane where the interval [itex]x \in (0, 1][/itex] is missing. Is this function still continuous on that domain.

  2. jcsd
  3. Jul 31, 2013 #2
    Sure it is still continuous there. If ##f## is defined on a domain, i.e. ##f:D\rightarrow \mathbb{R}##, then ##f## will remain continuous if we make the domain smaller. So if ##D^\prime## is contained in ##D##, then ##f:D^\prime\rightarrow \mathbb{R}## will be continuous if the original ##f## is.

    However, the following situation might occur, take the following function:


    This is obviously discontinuous in ##0## if the domain is the reals. However, if we take a smaller domain, then the function might become continuous. For example, if we cut out the interval [-1,1], then the function becomes continuous.

    So, in general, if we make the domain smaller then some discontinuous functions might become continuous. But all continuous functions will remain continuous.
  4. Jul 31, 2013 #3
    I attached a picture of my function in the Missing Strip Plane.

    Micromass, I think when you say "make the domain smaller," you are refering to making a subinterval? I have seen that in a lot of textbooks. This is not that type of problem. Rather it is making two infinite intervals and joining them.

    Attached Files:

  5. Jul 31, 2013 #4
    Your domain is the reals without the interval [0,1]. This is a smaller domain than the reals.
    When I said "make the domain smaller", then I meant what I said for any smaller domain, including yours.

    So your function will still be continuous since it's continuous on a larger domain.
  6. Jul 31, 2013 #5


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    All you are doing is restricting a continuous function on ##\mathbb{R}## to the domain ##\mathbb{R}\setminus (0,1]## so yes it will of course still be continuous. This is trivial to show.
  7. Jul 31, 2013 #6
    So, that means that the limit at x =1 in the plane exists and is 1. But in the Missing Strip Plane the left side of that neighborhood immediately contains x=0. Then wouldn't the limit at x=0 approach 1 from the right side?

    I don't understand why you don't collapse the missing strip to make the two sides merge in this case.
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