Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say we have a convergent sequence $(x_n)$ where $x_n \in E$ for all $n \in \mathbb{N}$ and $E$ is a subset of a metric space $(X,d)$.

With this setup, we usually define it's limit as a point $x \in X$ such that for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $n > N$ implies $d(x_n,x) < \epsilon$.

In some sense, I think that the above definition of the limit effectively mean that given any "$\epsilon > 0$, there is some kind of $N$ that we can use to get the sequence within $\epsilon$ of the limit $x$.

I am wondering whether this can be reformulated as follows: there is a function $f: \mathbb{R} \rightarrow \mathbb{N}$, so that $f(\epsilon) = N$ and $n > N$ implies $d(x_n,x) < \epsilon$. If so, the function $f$ would have some nice properties (it would be onto, and monotonically decreasing in $\epsilon$ for instance).

Is there any use to thinking about functions in this way / has it been introduced in this way?

share|cite|improve this question
Note that your $f$ need not be onto. – Hagen von Eitzen Dec 19 '12 at 21:14
up vote 1 down vote accepted

This concept is often known as the modulus of convergence, though for recursion-theoretic purposes it's often given as a function $g(n): \mathbb{N}\to\mathbb{N}$, where (in terms of your definition for $f()$) $g(n)$ would be defined as $g(n) = f(\frac{1}{n})$ or $g(n) = f(2^{-n})$. The concept is important in recursion theory because, for instance, we can compute the digits of the limit recursively in $g$ (i.e., using a Turing machine for $g$). Note that the function $g$ doesn't have to be computable even if the series converges; see, for instance, .

share|cite|improve this answer

To obtain a well-defined function you might define $f(\epsilon)$ to be the smallest $N$ such that $d(x_n,x)<\epsilon$ for all $n>N$. Convergence of the sequence $x_n$ guarantees existence of such $N$'s and one of them has to be the smallest.

Conversely, if there exists a function $f$ such that, for every $\epsilon$, $d(x,x_n)<\epsilon$ for all $n>f(\epsilon)$, then $x_n$ is convergent with limit $x$, because you may take $N=f(\epsilon)$.

These are nothing but quite tautological reformulations of each other, though.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.