continuity at isolated point

Question

In our booklet it is written that :

A function is continuous at every isolated point.

MY doubt:-

Let us consider an example: let $f:\mathbb N \rightarrow \mathbb{R} $ such that $f(x)=x$. As $\ \mathbb N =\{1,2,3,...\}$ and $1,2,3,..$ are all isolated points i.e $1,2,3... $ are not limit points of $\mathbb N $.

Now according to the above statement the function is continuous at every isolated point. But according to the definition of continuity, the continuity at point $a$ is $$\lim_{x\to a}f(x)=f(a)$$

Now take any number from set $\mathbb N $ , for example take $ 2$ then $f(2)=2$ and $$\lim_{x\to 2}f(x)=\text{not possible to determine or cannot be evaluate }$$

More precisely $2$ is not a limit point, so limit at $2$ cannot be calculated. So, the function is not continuous at all isolated points in this example.

How the function can be continuous at all isolated points? Can anyone tell me?

Answer 1

This is because the main difference between the definition of limit in a point and continuity in a point is the inclusion (in the last case) of the distance zero in the domain, i.e. $|x-c|<\delta$ for continuity and $0<|x-c|<\delta$ for limit.

The definition of the limit of a function at a point (to exist a limit the point must be a limit point):

$$\forall\varepsilon>0,\exists\delta>0,\forall x\in\mathcal D:0<|x-c|<\delta\implies|f(x)-L|<\varepsilon$$

where $\mathcal D$ is the domain of the function. Notice that $c$ doesn't need to belong to the domain of $f$. Now the definition of continuity of a function at a point is:

$$\forall\varepsilon>0,\exists\delta>0,\forall x\in\mathcal D:|x-c|<\delta\implies|f(x)-f(c)|<\varepsilon$$

Notice that here we need that $c\in\mathcal D$ and for $x=c$ the definition is trivially true, that is what happen in an isolated point.

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Trying to answer the comment. We can define the limit of a function in some point using sequences, this is named the sequential characterization of the functional limit: if for any sequence $(x_n)_n$ in the domain of the function that converges to some point $c$ (maybe in the domain or not) the sequence $(f(x_n))_n$ converge to some point $L$ in the codomain (maybe not in the range of the function) then we says that $L$ is the limit of the function $f$ at $c$.

Symbolically if

$$\big(\forall (x_n)_n\in\mathcal D^{\mathbb N },\forall j\in\Bbb N: (x_n)_n\to c\land x_j\neq c\implies (f(x_n))_n\to L\big) \iff\lim_{x\to c}f(x)=L$$

Notice that if $(x_n)_n\to c$ and there is some finite number of $x_j=c$ then we can quit these points of the sequence and produce a subsequence $(x'_n)_n\to c$ such that $x'_n\neq c$ for all $n\in \mathbb N$, then this subsequence hold the condition $|x'_n-c|>0$ that is required in the $\delta,\varepsilon$-definition of the functional limit.

Answer 2

By definition, a function $f$ is continuous at a point $p$ if, as you get near $p$, $f($points near p$)$ approaches $f(p)$.

For isolated points, there are no points near $p$, so the statement is trivially true!

It's like saying: if there were unicorns, I would be green. The statement is always true if there are no unicorns, as the precondition is never satisfied.

Answer 3

Well, your mistake is really conceptual. It all comes back to the following:

$$p\implies q$$

is true, if $p$ is false.

So, if there are no sequence $x_n\in N$ converging to $2$, then, $$x_n\to 2\implies f(x_n)\to 2$$ is still correct.

Also, you're missing a fact. Although $2$ is not a limit point, there are sequences that converge to $2$. For example, $2,2,2,2,2,\ldots$ converges to $2$. More precisely, any sequence that constantly becomes $2$ after some point converges to $2$, and these are the only sequences in that are converging to $2$ in $N$.

Answer 4

One can interpret $$\lim_{x\to a}f(x)=f(a)$$

to be true vacuously, since $f$ is not defined in a sufficiently small deleted neighbourhood of $a$. Alternatively, you could consider the standard $\epsilon-\delta$ definition of continuity and note that it is satisfied.