What is single point continuity?

Question

According to my textbook ,

Function should be continuous only at one point and to be defined everywhere.

For example ,

$$F(x)=\begin{cases} x\ \text{if} \ x\in Q \\-x\ \text{if} \ x\notin Q \end{cases}$$

The above example is continuous at $x = 0$ and defined everywhere.

My question is what does this means to be defined everywhere and how a function can be continuous at one point?

Answer 1

Let's look at definition of continuity using neighborhood conception (John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), case for single variable on page 43, multiple - page 186 ):

$f$ is continuous at some point $a$ iif $f$ is defined in $a$ and holds $\varepsilon-\delta$ definition for points from $a$'s neighborhood in which function is defined. I.e. definition puts restriction on intersection of domain of $f$ and $a$'s neighborhood.

Using such definition continuity for function can be defined, together with definition in limit point of domain, also in isolated point of domain - in such cases function is automatically continuous and we can say, that it is continuous at single point.

Addition.

How it became known from the comments. We are talking about the book AMIT M Agarwal - A Textbook of Differential Calculus, Chapter 6, page 470. Under the title "Function continuous only at one point and defined everywhere. (Single Point Continuity)". eg,(ii) is function brought in OP.

1. It is defined everywhere, obviously, because $\mathbb{Q} \cup (\mathbb{R}\setminus \mathbb{Q})=\mathbb{R}$.
2. It's continuous at $x=0$, because $\lim\limits_{x\to 0}f(x)=f(0)=0$. I'll brought proof by definition: we need $\forall \varepsilon \gt 0, \exists \delta \gt 0, |x| \lt \delta \Rightarrow |f(x)-f(0)|=|f(x)| = |x| \lt \varepsilon$ , which is obvious, if we take $\delta=\varepsilon$.
3. It's discontinous for any $x_0 \ne 0$, because we can find sequence of rational numbers $x_n \to x_0$ and sequence of irrational numbers $y_n \to x_0$. For first will be $f(x_n)=x_n \to x_0$ and for second $f(y_n)=-y_n \to -x_0$, which makes continuity impossible.

Answer 2

Continuity is DEFINED pointwise, i.e. a function is continuous at a point $a$ in its domain if and only if $\displaystyle \lim_{x\to a}f(x)$ exists, $f(a)$ is defined, and the two are equal, i.e. $$\lim_{x\to a}f(x)=f(a)$$

We extend from this the notion of a function being continuous on a subset of the domain (Typically intervals in the terms of real functions, though not always) to mean that the function is continuous at every point in that subset. (Or the whole domain if it works for the entire domain)

A function is defined on its domain, that's the set of inputs. Here the author was showing you how you could have a function that is continuous only at 1 point. Typically we visualize continuity as "being able to draw a graph without picking up your pen", but that fails on weirder stuff.

For example, the function $f(x)=0$ if $x$ is irrational and $f(x)=\frac 1 q$ if $x=\frac p q$ where $\gcd (p,q)=1$ is a weird one that is continuous at every irrational and discontinuous at every rational. The reason why for that is you can get close enough to any irrational such that all the denominators of fractions near it are as large as you like, so you can make the function as small as you want.