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Continuity in the metric spaces

  1. Aug 11, 2011 #1
    I'm currently reading Ross's Elementary Analysis, which presents the definition of continuity as such: (not verbatim)

    Let x be a point in the domain of f. If every sequence (xn) in the domain of f that converges to x has the property that:

    lim f(xn) = f(x)

    then we say that f is continuous at x.

    For most functions, which are defined on uncountable sets, this presents no problems. But note that the sequences in question fall into two categories: those which converge "from the outside" (sequences in dom(f)\{x}) and those sequences which are eventually equal to x (and yes, I know that this does not actually form a partition of those sequences, but that won't really matter.)

    So consider a function defined on some isolated point x. It may or may not be defined elsewhere, and the other places it is defined may or may not be isolated points, as long as none of them intersect some neighborhood of x. We've eliminated one of the two "types of convergent sequence." Every sequence in the domain that converges to x must eventually be equal to x, so f(xn) eventually equals f(x), and the function is continuous at x, even though it's not even defined in the neighborhood of x.

    The (equivalent) epsilon-delta formulation appears to allow for the same thing (obviously, or else it wouldn't be equivalent). Am I interpreting this correctly? Is it generally the case that we say such a sequence is continuous? Would it not make more sense to require the sequence in question to be in dom(f)\{x}? (Then again we rarely speak about continuity at a point anyways. All the interesting results concern continuity on a set, so I suppose it shouldn't affect much.)
  2. jcsd
  3. Aug 11, 2011 #2
    Yes, you are correct. If I have interpreted your post correctly, here's a formal way of describing what you said:

    If, for some [itex]a,n,m[/itex], we can define [itex]f[/itex] to be
    [itex] f(x) = \begin{cases}
    undefined, & \mbox{if } a<x<m \\
    n, & \mbox{if } x = m \\
    undefined, & \mbox{if } m<x<a \\
    g(x), & \mbox{otherwise }\\
    then [itex]f[/itex] is continuous at [itex]m[/itex].

    This is true, both from your definition as provided as well as the typical limit definition of continuity.

    I should also mention that this may not "jive" with the common notion of continuity, that being a line drawn without picking up the pencil from the paper, however in the spirit of reducing all required restrictions of the definition to make it as simple as possible it is always defined this way. I do not know of any useful situations where it is actually a benefit (perhaps a more experienced member can suggest some), however as you pointed out most interesting theorems involving continuity require the entire set to be continuous anyway.
    Last edited: Aug 11, 2011
  4. Oct 23, 2011 #3
    Hey -
    I think if you define continuity to be:

    For any positive number ε>0 there exists a δ>0 such that for

    l x - a l < δ → l f(x) - f(a) l < ε

    You are correct. But that's only provided we don't restrict i x - a l from being greater than zero, which is what I've most often come across.
  5. Oct 23, 2011 #4


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    Yes, AFAIK, the function is defined to be automatically continuous at isolated points. This is why any function defined on a discrete space is automatically continuous; every point in the domain is an isolated point.
  6. Oct 23, 2011 #5
    MarlyK, the definition you gave is equivalent to the one I gave. You're noting that if we allow δ=0, we enormously weaken the definition of continuity. (In fact, we make it into a tautology.) Including δ=0 in the ε-δ definition, however, is not the same as including sequences inside the point in the sequence-limit definition.

    The difference is that ε-δ states that there is some delta, so allowing an additional value in the domain of δ is weakening. The quantification in the sequence definition is universal, though, so including additional sequences makes the definition stronger.

    In fact, since I posted this (a while ago) I recognized that not including sequences that converge "from the inside" does not change the way the definition is applied to isolated points. If we exclude those sequences, then there are no sequences at all converging to our isolated point, so we end up with a vacuous implication (quantification over the empty set) and the function is still vacuously continuous.
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