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A very basic question about continuous function

  1. Jan 18, 2012 #1
    Hello everyone,

    This is probably a really newbie question and I apologise for it.

    So a continuous function is one that is differentiable with respect to its input parameters. What happens when the input parameters can only take discrete values? So, I guess the function can, of course, not be continuous everywhere.

    However, what about the corresponding function values at those discrete points? I am guessing that the function cannot be continuous for those inputs as well because of the discreteness of the input variables but am able to convince myself.

    Could someone be kind enough to shed some light on this?

    Thanks,
    Luca
     
    Last edited by a moderator: Jan 18, 2012
  2. jcsd
  3. Jan 18, 2012 #2
    Well, it turns out that a function is always continuous at an "isolated point" (an intuitively pleasing term which can easily be made rigorous). I'll assume that you're interested in function from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R}[/itex]. If in fact you want the whole shebang for metric spaces, it's basically the same, with open intervals replaced by open balls.

    Let [itex]A\subset\mathbb{R}[/itex] and [itex]x\in A[/itex]. The point [itex]x[/itex] is an isolated point of [itex]A[/itex] if there is a [itex]\delta>0[/itex] such that [itex](x-\delta, x+\delta)[/itex] does not contain any other points of [itex]A[/itex]. For example, [itex]0[/itex] is an isolated point of [itex]\{0\}\cup (1,2)[/itex], since [itex](-1,1)[/itex] is an open interval about [itex]0[/itex] which does not contain any points of [itex](1,2)[/itex] other than [itex]0[/itex] itself.

    Recall that a function [itex]f\colon A\to\mathbb{R}[/itex] is continuous at [itex]a\in A[/itex] if for every [itex]\epsilon>0[/itex] there exists [itex]\delta>0[/itex] such that [itex]|x-a|<\delta[/itex] implies [itex]|f(x)-f(a)|<\epsilon[/itex].

    Now consider what happens when [itex]a[/itex] is an isolated point of [itex]A[/itex]. Then there is a [itex]\delta>0[/itex] such that [itex](a-\delta, a+\delta)\cap A = a[/itex]. So no matter what [itex]\epsilon>0[/itex] you choose, if [itex]x\in (a-\delta, a+\delta)\cap A[/itex] can only happen if [itex]x=a[/itex], so that [itex]|f(x)-f(a)|=0<\epsilon[/itex]. All this really takes is some rewording of definitions.

    Here's another way to put it: for [itex]f[/itex] to be continuous at [itex]a[/itex], you have to be able to make the "error" ([itex]\epsilon[/itex]) from [itex]f(a)[/itex] as small as you want by accordingly adjusting the "distance" ([itex]\delta[/itex]) from [itex]a[/itex]. As long as you are in the "empty space" around the isolated point, the function doesn't even have any values, so it doesn't even make sense to talk about the "error," i.e. this is vacuously true.
     
  4. Jan 18, 2012 #3

    chiro

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    Hey pamparana.

    It turns out that not all continuous functions are differentiable.

    In fact if you ever do any kind of stochastic calculus, you'll see that for many applications including physics and finance, we deal with functions that are everywhere un-differentiable but still continuous where the function is defined by a random variable.

    But even if we were not dealing with random variables, the Weierstrauss function is an example of a deterministic function that is continuous but not differentiable.

    Differentiable functions over some domain are continuous over that domain if the derivative exists over that domain, but the other way around is not always true.
     
  5. Jan 18, 2012 #4

    pwsnafu

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    Fixed for you :tongue:
    There are ways to make that statement precise, but it'll be too much for pamparana.
     
  6. Jan 19, 2012 #5

    chiro

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    Cheers dude :)
     
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