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I started to study the topic "Limit of function" in the book "Introduction to real analysis" by G. Bartle, R. Sherbert. The definition of Limit of function is:

Let $A \subseteq \mathbb R$, and let $c$ be a cluster point of $A$. For a function $f:A \to \mathbb R$ a real number $L$ is said to be a limit point of $f$ at $c$ if, given any $\epsilon>0$, there exists a $\delta>0$ such that if $x \in A$ and $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.

My question is: Do we really need $c$ to be a cluster point of $A$ in this defition?

My answer is : I think yes, but I'm not sure. Because if $c$ is cluster point, then $x \neq c$ because of definition of cluster point and it can be useful if fuction is not defined at $c$.

Can somebody explain me whether I am right or not? If not, where do I have missunderstanding?

mathguruu
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    If $c$ is not a cluster point then, for small enough $\delta$, the set of values $x\in A$ with $0<|x-c|<\delta$ will be empty so the claim would be vacuously true no matter what $L$ you used. – lulu Dec 29 '22 at 16:24
  • Note: you meant to write $|f(x)-L|<\epsilon$. Without the absolute value, you could just take any sufficiently large $L$. – lulu Dec 29 '22 at 16:25
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    The reason you want cluster points is that if $c$ were an isolated point, then the definition would be met for any value of $L$ by taking a $\delta$ for which no points in the domain different from $c$ lie within $\delta$ of $c$. – Arturo Magidin Dec 29 '22 at 16:25
  • @lulu thank you, sure It has to be with absolute value. I edited) – mathguruu Dec 29 '22 at 16:26
  • @lulu so I understand, why set of values $x \in A$ with $0<|x-c|<\delta$ will be empty. But I don't understand, why would the claim be vacuously true no matter what L I used? Could you explain, please? – mathguruu Dec 29 '22 at 16:40
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    If there are no such $x$, then it is certainly true that $|f(x)-L|<\epsilon$ no matter what $L$ and $\epsilon$ are as there no possible counterexamples. That's what "vacuous truth" is like. The claim "every integer between $1.1$ and $1.2$ is prime" is vacuously true as there are no counterexamples to the claim. – lulu Dec 29 '22 at 16:52
  • @lulu thank you. I understood it. It's about logic. I got it) – mathguruu Dec 29 '22 at 16:54

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If $c$ was not a cluster point of $A$, then there would exist some $\delta_0>0$ such that the set $\{x\in A : 0<\lvert x-c\rvert<\delta_0\}$ is empty. This implies that whatever $L\in\mathbb{R}$ you want would be a limit of $f$ at $c$. Indeed: let $\epsilon>0$ be arbitrary, take $\delta=\delta_0$. Then for any $x\in A$, $0<\lvert x-c\rvert<\delta=\delta_0$ (no such $x$ exists) we have that $\lvert f(x)-L\rvert<\epsilon$ (vacuously true: the claim $\forall x\in\emptyset, P(x)$ is always true regardless of what $P$ represents, check here). This is, of course, not convenient.

Samuel M. A. Luque
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