- #1

psie

- 182

- 18

- Homework Statement
- I'm working two problems in Spivak's Calculus on Manifolds. I'm getting a stuck at an annoying technicality. See the problems in the quoted text below.

- Relevant Equations
- A function is continuous at an accumulation point ##a## if ##\lim_{x\to a}f(x)=f(a)##.

1-23: If ##f: A \rightarrow \mathbb{R}^m## and ##a \in A## [is an accumulation point], show that ##\lim _{x \rightarrow a} f(x)=b## if and only if ##\lim _{x \rightarrow a }f^i(a)=b^i## for ##i=1, \ldots, m##.

Proof: Suppose that ##\lim _{x \rightarrow a }f^i(x)=b^i## for each i. Let ##\epsilon>0##. Choose for each ##i##, a positive ##\delta_i## such that for every ##x \in A\setminus\{a\}## with ##|x-a|<\delta_i##, one has ##\left|f^i(x)-b^i\right|<\epsilon / \sqrt{n}##. Let ##\delta=\min \left(\delta_1, \ldots, \delta_n\right)>0##. Then, if ##x \in A\setminus\{a\}## satisfies ##|x-a|<\delta##, then ##|f(x)-b|<\sqrt{\sum_{i=1}^n \epsilon^2 / n}=\epsilon##. So, ##\lim _{x \rightarrow a} f(x)=b##.

Conversely, suppose that ##\lim _{x \rightarrow a} f(x)=b, \epsilon>0##, and ##\delta## is chosen as in the definition of ##\lim _{x \rightarrow a} f(x)=b##. Then, for each ##i##, if ##x## is in ##A\setminus\{a\}## and satisfies ##|x-a|<\delta##, then ##\left|f^i(x)-b^i\right| \leq|f(x)-b|<\epsilon##. So ##\lim _{x \rightarrow a} f^i(x)=b^i##.

1-24: Prove that ##f: A \rightarrow \mathbb{R}^m## is continuous at ##a## if and only if each ##f^i## is.

Proof: A function ##f:A\to\mathbb R^m## is continuous at an accumulation point ##a## if ##\lim_{x\to a}f(x)=f(a)##, so in that case this is a simple consequence of the previous exercise.

Now, what troubles me is that I don't know how to prove the statement at the non-accumulation points. After all, ##\lim_{x\to a}f(x)=f(a)## is only the definition of continuity at an accumulation point ##a##. I'm a bit bewildered about this.