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Definition: Let $F$ be a real valued function defined on a subset $E$ of $\mathbb{R}$. We say that $F$ is continuous at a point $ x \in E$ iff for each $\epsilon > 0$, there is a $\delta > 0$, such that if $x' \in E$ and $|x'-x|<\delta$, then $|f(x') - f(x)| < \epsilon$.

This definition is taken straight out of Royden-Fitzpatrick Real Analysis.

My question is more related to the intuition behind this definition:

If I take the following function $F$ defined on the natural numbers(which are a subset of $\mathbb{R}$ of course), for which $F(x) = x$, that is, the identity on the natural numbers but that treats them as a subset of $\mathbb{R}$.

Now this function is continous at every point of $\mathbb{N}$. If we take $\epsilon \le 1$, then we can always take some $\delta < 1$. If we take an $\epsilon > 1$, then we can always take a respective $\delta < \epsilon$, but $\delta > $ the largest natural number smaller than epsilon. So, indeed this function is continuous. Is this supposed to happen, and can't we somehow use this definition to show continuity of functions which are intuitively discontinuous at some points.

Is the key element here that we say that the function is continuous/discontinuous at $x$ as a point of a specific set $E$?

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That's exactly right, the point is that both $x$ and $x'$ are constrained to lie within the specified domain $E$. If $E$ is finite or more generally discrete one can always choose $\delta$ small enough so that the condition $x-x'<\delta$ will force $x=x'$ and therefore the condition is vacuous, so the function is trivially continuous at $x$.

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First of all you are right when saying the function $F:\mathbb{N}\subseteq\mathbb{R}\to \mathbb{R}, F(x)=x$ is continuous at every point of $\mathbb{N}$. In fact you may choose for each $\epsilon>0$ the value $\delta=\frac{1}{2}$ to show continuity at a point $x\in\mathbb{N}$. Since $$\{x'\in \mathbb{N}: |x-x'|<1/2\}=\{x\}$$ we obtain \begin{align*} |F(x')-F(x)|=|x'-x|=|x-x|=0<\epsilon \end{align*}

The key element here is that a function is always continuous at isolated points and if we consider the discrete subset $\mathbb{N}\subseteq\mathbb{R}$ each point is isolated.

Somewhat intuitively: Isolated points are points which have no other points near to them. A discontinuity at a point means a jump near to it. So these are two concepts which exclude each other.

Note: This related question might be useful.

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A foremost condition for evaluating the limit of a function $f:E\subseteq \Bbb R\to \Bbb R$ at some point say $x_0\in \Bbb R$ is that $x_0$ must be a limit point of $E$.

In the case where $E=\Bbb N$, the set has no limit points. So forget about testing continuity at some point in $\Bbb N$ we cannot even compute the limits at any point!

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  • $\begingroup$ Understandable, but to be honest, the definition I cited is not based on limit points. Intuitively, it says that if we want to stray no-further than $\epsilon$, we can find a $\delta$ so that wherever we are within the $\delta$ interval, we would be in the corresponding $\epsilon$ interval of the function value. I don't see why we would restrict ourselves to limit points, but I do agree that if we do my intuitive notion of continuity would be satisfied. $\endgroup$ – Georgi Stefanov Jul 12 '17 at 13:30
  • $\begingroup$ Firstly, no matter what your definition of continuity is, all of them are equivalent. Now let me argue in the following manner. Do you agree that a function $f$ is continuous at $x_0$ in the domain $\iff$ the following three conditions are satisfied, (1) $\lim_{x\to x_0}f(x)$ exists (2) $f$ is defined at $x_0$, that is $f(x_0)$ exists [This is obvious if your $x_0$ lies in the domain] (3) $f(x_0)=\lim_{x\to x_0}f(x)$...In the case where the domain is $\Bbb N$ can you tell me what is the value of $\lim_{x\to x_0}f(x)$ ? $\endgroup$ – Naive Jul 12 '17 at 13:54

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