# Theorem 4.6 from baby Rudin

Theorem 4.6. In the situation given in Definition 4.5, assume also that $p$ is a limit point of $E$. Then $f$ continuous at $p$ if and only if $\lim \limits_{x\to p}f(x)=f(p)$.

Proof: Let $f$ continuous at $p$. Then for any $\varepsilon>0$ exists $\delta>0$ such that $d_Y(f(x),f(p))<\varepsilon$ for any $x\in E$ such that $d_X(x,p)<\delta$. Hence $d_Y(f(x),f(p))<\varepsilon$ is also true for any $x\in E$ such that $0<d_X(x,p)<\delta$ and $p$ is also limit point. Hence $\lim \limits_{x\to p}f(x)=f(p)$.

Let $\lim \limits_{x\to p}f(x)=f(p)$. Then for any $\varepsilon>0$ exists $\delta>0$ such that $d_Y(f(x),f(p))<\varepsilon$ for any $x\in E$ such that $0<d_X(x,p)<\delta$. But I can replace the last inequality to $d_X(x,p)<\delta$ because if $d_X(x,p)=0$ then $x=p$ and $p\in E$ and $d_Y(f(p),f(p))=0<\varepsilon$.

Is my proof true?

What would be if we consider that $p$ is isolated point of $E$?

## 1 Answer

Yes, your proof is correct.

What would be if we consider that p is isolated point of E?

Rudin already considers this right before Theorem 4.6.

"If $p$ is an isolated point of $E$, then our definition implies that every function $f$ which has $E$ as its domain of definition is continuous at $p$. For, no matter which $\epsilon > 0$ we choose, we can pick $\delta > 0$ so that the only point $x \in E$ for which $d_X(x,p) < \delta$ is $x = p$; then $d_Y(f(x),f(p)) = 0 < \epsilon$. "

• Yes I read this and also I understand it. But why if function continuous at $p$ then $f(x)\to f(p)$. Can you explain it? Rudin wrote about continuous but nothing about $f(x)\to f(p)$ – ZFR Sep 9 '15 at 6:43
• in limits of function $p$ must be a limit point but in our case it's isolated point. – ZFR Sep 9 '15 at 7:03
• @Pacman, yes it must be a limit point in order for the limit of a function to exist, and that's why Theorem 4.6 demands that it must be a limit point. Hence if it is an isolated point, Theorem 4.6 doesn't apply. – ignoramus Sep 9 '15 at 7:15