Definition of Continuity of Real Valued Functions

Question

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 $E$?

Answer 1

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!

Answer 2

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.

Answer 3

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 $x$.