# Hey Dr. Spivak, Where’s That Prose? (Part 1)

With the utmost respect to Michael Spivak, his textbook Calculus on Manifolds is commonly referred to (by undergraduates at various universities) as the most difficult analysis book in print. Although elegantly written, it has this stigma precisely due to the lack of common prose. Spivak presents his book in a rigorous Theorem-Proof-Theorem-Proof style, without much breathing room between theorems. While this is wonderful for a second (or third) course on the subject, many students find it difficult to navigate this 137 page book. My goal in these posts is to cover some of the information in Spivak, with additional comments from a student that doesn’t quite think like a professor.

Well first off, Spivak loves to do everything in $\mathbb{R}^n$. So, throw away your fancy metrics and limit points and open balls, because Spivak uses none of these (explicitly, that is). Instead, Spivak proves a set is closed by proving that its complement is open. That is, there’s an open rectangle around each point in the complement completely interior to the set. (Actually, Spivak hasn’t avoided using some sort of metric for his open balls. The open rectangle is the open ball induced by the norm $||x||_\infty$ for $x \in \mathbb{R}^n$. He just never states this. Since this metric is equivalent to the Euclidean Metric, everything turns out ok.)

Great, so Spivak introduces open and closed sets in $\mathbb{R}^n$, open covers $\mathscr{O}$ as collections of open rectangles, and compact sets. Then, bam, Heine-Borel for $\mathbb{R}$. I want to go over this proof a bit, so let’s do that.

Theorem (Heine-Borel): The closed interval $[a,b] \subset \mathbb{R}$ is compact.

Proof (sketch): Okay so here’s the game plan: We’ll take an open cover $\mathscr{O}$ of the interval. We’ll look at all closed interval subsets of $[a,b]$ so that these subsets are covered by finitely many members of $\mathscr{O}$. Then we’ll take the supremum, since this is indeed a bounded non-empty subset of $\mathbb{R}$ and $\mathbb{R}$ is complete! Then, show that this supremum is in the set of finitely covered closed intervals, and finally that this supremum is indeed $b$.

Proof: Let $\mathscr{A} := \{x: x \in [a,b] \, \text{and} \, [a,x] \, \, \text{is covered by finitely many members of} \, \, \mathscr{O} \}$. Clearly this set is non-empty, since $a \in \mathscr{A}$. Let $\alpha = \sup\{\mathscr{A}\}$. To see $\alpha \in \mathscr{A}$, note that $\alpha \in U$ for some $U \in \mathscr{O}$. By definition of supremum (think about the characterization that if we take $\sup\{X\} - \epsilon$ then there exists $x \in X$ so that $\sup\{X\} > x > \sup\{X\} - \epsilon$) there is a $x \in \mathscr{A}$ so that $x \in U$. By definition, $[a,x]$ is covered by finitely many open sets in $\mathscr{O}$, while $[x,\alpha]$ is clearly covered by $U$. Thus, $[a,\alpha]$ is covered by finitely many open sets. That is to say, $\alpha \in \mathscr{A}$.

To see $\alpha = b$, suppose not. Then we must have $\alpha < b$ (i.e. $\alpha = b - \epsilon$ for some $\epsilon$). Then, there is a point $p$ between $\alpha$ and $b$ so that $p \in U$ for the $U \in \mathscr{O}$ that we used to cover $[x,\alpha]$. But then, clearly $p>\alpha$ and $p \in \mathscr{A}$, contradicting the definition of the supremum. Thus, $\alpha = b$. That is, $[a,b]$ is covered by finitely many open sets for any open cover $\mathscr{O}$.

So that’s the first direction of the Heine-Borel. To see a more detailed explanation, see here. Spivak then proves that finite Cartesian Products of compact sets  are compact. (Though, this is easy to see with the notion of a product topology.)

Beyond this, Spivak begins to discuss the notion of functions in $f: \mathbb{R}^n \to \mathbb{R}^m$. A convenient way to represent a function is the “graph” of a function, a subset of $\mathbb{R}^n \times \mathbb{R}^m = \mathbb{R}^{n+m}$. (In fact, the graph of a function is a trivial fiber bundle over $\mathbb{R}^n$.) The simplest way to represent a function is via its $m$ component functions, $(f_1,f_2 \ldots, f_m)$. There are $m$ such functions because we must somehow assign a value for each dimension of $\mathbb{R}^m$. Of remarkable importance in mathematics is the general notion of the projection mapping. (It is used, for example, to define the product topology or the product $\sigma$-algebras in measure theory.) In $\mathbb{R}^n$, we have $\pi_i : \mathbb{R}^n \to \mathbb{R}$ with $\pi_i: (x_1,\ldots,x_i,\ldots,x_n) \mapsto (x_i)$.

When Spivak introduces the notion of a limit, it should be noted that he is doing so in $\mathbb{R}^n$, so that “$|x-a| < \delta$” means such in terms of some metric on $\mathbb{R}^n$. For example, you could interpret this as follows:

Let $x = (x_1,\ldots,x_n)$ and $a = (a_1,\ldots,a_n)$. Then, if $|x_i - a_i| < \frac{\delta}{\sqrt{n}}$, $||x-a||< \delta$ in the Euclidean Metric.

Off of this notion of limits, he defines continuity in an analogous way. He then proceeds to give the usual theorem about continuity implying the inverse image under a continuous function of open subsets are again open subsets…with one subtlety. If the domain is a subset or $\mathbb{R}^n$, say $A$, then “open in $A$” means open in the subspace topology inherited from $\mathbb{R}^n$. More explicitly, this means that:

If $A \subset \mathbb{R}^n$, then $f: A \to \mathbb{R}^m$ is continuous if and only if for every $U \subset \mathbb{R}^m$, there exists an open $V \subset \mathbb{R}^n$ so that $f^{-1}(U) = V \cap A$.

Now Spivak’s proof is more of an intuitive “sketch of proof” than anything. For the reader without a PhD, I’d recommend proving this statement formally in terms of a metric on $\mathbb{R}^n$. After this, Spivak introduces a concept which may be unfamiliar even to those who have taken a full course out of Baby Rudin.

Let $f: A \to \mathbb{R}^m$ be bounded. We can measure “how discontinuous” $f$ is at $a \in A$ as follows:

Define $M(a,f,\delta) := \sup\{f(x) : x \in A \, \, \text{and} \, \, |x-a| < \delta\}$

and $m(a,f,\delta) = \inf\{f(x) : x \in A \, \, \text{and} \, \, |x-a| < \delta\}$

So these are just the “largest” and “smallest” values that $f$ takes in a $\delta$ ball. We can then define the oscillation of $f$ as “how badly behaved” $f$ is as the $\delta$ ball collapses to $0$. Thus, we define:

$o(f,a) := \lim \limits_{\delta \to 0} |M(f,a,\delta) - m(f,a,\delta)|$.

I’ll conclude with a theorem about oscillation:

Theorem 2: Let $f: A \to \mathbb{R}^m$ be bounded. Then, $f$ is continuous at $a \in A$ if and only if $o(f,a) = 0$.

Proof:

I’ll elaborate on Spivak’s proof.

Suppose latex $f$ is continuous. Then for each $\epsilon > 0$, if $|x-a| < \delta$, $|f(x) -f(a)| < \epsilon$ and therefore $|f(x)| < |f(a)| + \epsilon$ so that $\sup \limits_{x \in B_\delta(a)} |f(x)| \leq |f(a)| + \epsilon$ (notice the $\leq$ instead of $<$ by the properties of supremum). By the opposite argument, $\inf \limits_{x \in B_\delta (a)} |f(x)| \geq |f(a)| - \epsilon$ and thus, $|M(f,a,\delta)| - |m(f,a,\delta| \leq |M(f,a,\delta) - m(f,a,\delta)| \leq 2\epsilon$. The proof of the converse is remarkably similar and thus is left to the reader.

I’ll stop here for now, but hopefully this begins to take a little bit of the scare out of reading Spivak. It is really a remarkable book, just terse for first time students of analysis. Thanks for reading; as usual, direct inquiries (say, a request for an article) to erdosninth@gmail.com.

Cheers,

J.T.