# Prove the function limit definition still works if c is not a cluster point.

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?

• 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. – EBO Mar 6 at 0:41
• I think I see what you are saying but I am having a hard time applying it to my proof. – user756226 Mar 6 at 1:39

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.
• Exactly. Since there are no $x$, the statement is always true. – Alain Remillard Mar 6 at 3:40