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Let $L \in \mathbb{R}$ and let $f: S \rightarrow \mathbb{R}$ be a function. Suppose $c \in S$ is not a cluster point and let $\varepsilon>0 .$ Then there exists $\delta>0$ such that if $x \in S \backslash\{c\}$ and $|x-c|<\delta,$ then $|f(x)-L|<\varepsilon$

I am just a few chapters into an intro to real analysis course. This is a prove or disprove.

So, this is just the definition of a function limit, but we have that $c$ is NOT a cluster point. I was told this is a situation where it is vacuously true, but I am not sure if that is correct. If it is, how do I actually go about proving it?

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  • $\begingroup$ Let $A$ be some set. Is the following true? For all $a\in A$, $a$ is an orange. If such statement is not true then its negation is true: there exists some $a\in A$ that is not an orange. If $A=\emptyset$ then the negated statement makes no sense since we need to exhibit some element of $A$. So, if $A=\emptyset$ then the (oranges) statement is always true. $\endgroup$ – EBO Mar 6 at 0:41
  • $\begingroup$ I think I see what you are saying but I am having a hard time applying it to my proof. $\endgroup$ – user756226 Mar 6 at 1:39
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If $c$ is not a cluster point of $S$, then it exist a $\delta>0$ such that there are no $x\in S\backslash\{c\}$ such that $|x-c|<\delta$. This is the delta that we are going to use in the definition of continuity.

Let $\epsilon>0$, then the following statement is true

if $x\in S\backslash\{c\}$ such that $|x-c|<\delta$ (there are no such $x$), then $\lvert f(x)-f(c)\rvert<\epsilon$.

If it is not true, then there exist a $x\in S\backslash\{c\}$ such that $|x-c|<\delta$ which is impossible since $c$ is a not a cluster point.

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  • $\begingroup$ To see if my thinking is correct, the statement is vacuously true because it says "if this x exists..." but that x can't exist (based on your above reasoning)? $\endgroup$ – user756226 Mar 6 at 3:08
  • $\begingroup$ Exactly. Since there are no $x$, the statement is always true. $\endgroup$ – Alain Remillard Mar 6 at 3:40

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