Why is an open interval needed in this definition? (definition of a limit of a function)

Question

Here's a part of the definition Ross' Elementary Analysis states for limits of a function:

20.3 Definition
(a) For $a\in\mathbb R$ and a function $f$ we write $\lim_{x\to a} f(x)=L$ provided $\lim_{x\to a^S} f(x)=L$ for some set $S=J\setminus\{a\}$ where $J$ is an open interval containing $a$. $\lim_{x\to a} f(x)$ is called the [two-sided] limit of $f$ at $a$. Note $f$ need not be defined at $a$ and, even if $f$ is defined at $a$, the value $f(a)$ need not equal $\lim_{x\to a} f(x)$. In fact, $f(a)=\lim_{x\to a} f(x)$ if and only if $f$ is defined at $a$ and $f$ is continuous at $a$.
(b) For $a\in\mathbb R$ and a function $f$ we write $\lim_{x\to a^+} f(x)=L$ provided $\lim_{x\to a^S} f(x)=L$ for some open interval $S=(a,b)$. $\lim_{x\to a^+} f(x)$ is the right-hand limit of $f$ at $a$. Again $f$ need not be defined at $a$.

In both parts of the definition, why are open intervals needed? Would it fail if it were a closed interval instead?

Answer 1

Here are my opinions (Not accepted facts from math society):
Definitions usually should be short and provide good intuitive for incoming definitions.

1. Open intervals guaranty that some points in two side of point $a$ exist. But if $J=[a,c]$ or $J=[d,a]$ then $\lim_{x\to a^S}f(x)$ will be one sided. So it needs to use some extra words explaining that $a$ shouldn't be at start or end of which isn't short any more.
   
2. Open intervals will have good relevance to their generalized things in topology (open set) which needs for definition of continuous functions.

Answer 2

An open interval is not needed and is stronger than necessary.

Usually, in other books, this definition will only require that $a$ be a limit point.†

I think perhaps for simplicity, Ross's Elementary Analysis tries to avoid having to talk about limit points. So, it uses the stronger condition that $a$ is in an open interval (where $f$ is defined on this interval except possibly at $a$). It can be shown that this condition implies that $a$ is a limit point of the domain of $f$.

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A closed interval would fail. I consider only part (a) of the definition:

Define $f: \mathbb R \to \mathbb R$ by $f(x)=-1$ for $x<0$ and $f(x)=1$ for $x\ge 0$. Let $a=0$.

Intuitively, we "want" $\lim_{x\to a} f(x)$ to be undefined.

But now if we allow $J$ to be the closed interval $[a,1]=[0,1]$, so that $S=J\setminus \{a\}=(0,1]$, by the definition, we'd have $\lim_{x\to a} f(x)=1$, which isn't what we "want".

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†As for why this is required, see e.g. Limit of function at non limit point.