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I am trying to understand a part of the following proof.

Prove that $(1)$ and $(2)$ are equivalent:
$(1)$ $\lim_{x \to c}f(x)=f(c)$
$(2)$ $f$ is continuous at $c$.

I understood the proof of $(1) \Rightarrow (2)$. My question is about a part of the proof of $(2) \Rightarrow (1)$.

Definition

Definition of continuity I am using is as follows.

A function $f:E \rightarrow \mathbb{R}$ is continuous at $c$ if for all $\varepsilon \gt 0 $, there exists $\delta \gt 0$ such that if $\left | x-c \right| \lt \delta$ and $x \in E$, then $\left | f(x)-f(c) \right| \lt \varepsilon$.

Proof

Suppose that $\lim_{x \to c}f(x)=f(c)$. For any $\varepsilon \gt 0$, there exists $\delta \gt 0$, such that for all $x \in (a,b)$ satisfying $0 \lt \left | x-c \right| \lt \delta$, we have $ \left | f(x)-f(c) \right| \lt \varepsilon$.

If $\left | x-c \right| = 0$, then $x=c$. In this case $ \left | f(c)-f(c) \right| = 0 \lt \varepsilon$.

Hence for all $x \in (a,b)$ with $\left | x-c \right| \lt \delta$, we have $ \left | f(x)-f(c) \right| \lt \varepsilon$. Thus $f$ is continuous at $c$.

$Q.E.D$

Question

Why do we need the part of the proof which is blocked? And why does it eliminate the need for $0 \lt \left | x-c \right|$ ?

Thank you for your time.

user4167
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  • what is the definition of continuity that you are using? (reason I'm asking is that most commonly, $f$ is continuous at $c$ is defined to mean that $\lim _{x\to c}f(x)=f(c)$). – Ittay Weiss Apr 12 '13 at 19:18
  • Hi, I have added the definition above. – user4167 Apr 12 '13 at 19:27
  • @euler : whoever foisted that definition on you was cruel. It is more common and sensible to use (1) as the definition of (2). – Stefan Smith Apr 12 '13 at 23:38

1 Answers1

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$\lim_{x\to c}f(x)=f(c)$ means that for all $\epsilon>0$ there is some $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ whenever $x\in\text{dom} f$ such that $0<|x-c|<\delta$. To show that $f$ is continuous at $c$, we must show that for all $\epsilon>0$ there is some $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ whenever $x\in\text{dom} f$ such that $|x-c|<\delta$. The blocked part simply takes care of the only remaining case: $|x-c|=0$.

Added: One pivotal distinction not made explicit in your definition is that $c$ must be an element of $E,$ and cannot be isolated in $E$. This is essential, as if $c\notin E,$ then $f(c)$ makes no sense, and if $c$ is isolated in $E$, then $\lim_{x\to c}f(x)$ makes no sense.

Cameron Buie
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