Problem 1-23 and 1-24 from Spivak's Calculus on Manifolds

[psie]

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.

 

[FactChecker]

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.

If $a$ is not an accumulation point, there is no way for $x$ to approach $a$. So the point is moot.

 

[psie]

If $a$ is not an accumulation point, there is no way for $x$ to approach $a$. So the point is moot.

Ok, but all function are continuous at non-accumulation points, right? Since for any neighborhood $V$ around $f(a)$, there is always a neighborhood $U$ around $a$ whose image $f(U)\subseteq V$ (this is one of equivalent definitions of continuity at a point). By negation of accumulation point, $U$ is simply $\{a\}$.

If this is true, then I feel like the statement in the problem must be provable somehow at the non-accumulation points.

 

[psie]

Having thought some more about this, I think there's nothing to prove when $a$ is an isolated point, i.e. not an accumulation point. The statement "$f$ is continuous at $a$ iff all of its components are continuous at $a$" is just trivially true. There is nothing to prove really, since the hypothesis and conclusion are always true in either direction, as the function and its components share the same domain and thus are always continuous at $a$.

 

[pasmith]

Use the actual definition of continuity: f is continuous at a iff for every \epsilon &gt; 0 there exists a \delta &gt; 0 such that if \|x - a\| &lt; \delta then \|f(x) - f(a)\| &lt; \epsilon. If a is isolated, we can always take 0 &lt; \delta &lt; \inf \{ \|x - a\| : x \neq a\} so that \|x - a\| &lt; \delta \Rightarrow \|f(x) - f(a)\| = 0 &lt; \epsilon.

 

[WWGD]

For one side of 1-24, use the fact that projections$\pi_i$( into the $x_i$) are continuous. EDIT: And the composition of continuous functions is continuous.

 

[WWGD]

I believe Spivak refers to accumulation points as points that aren't isolated points. A point $p$ is an isolated point of$S$ if there is a neighborhood of $p$ that doesn't intersect $S$.