# Tag Archives: M32AH

## Compactness and Open Covers

Let $$S \subseteq \mathbf{R}^n$$ and suppose $$\mathcal{A} = \{A_i | i \in I\}$$ is a family of open subsets of $$\mathbf{R}^n$$ such that
$S \subseteq \bigcup_{i \in I} A_i.$
We call such a family $$\mathcal{A}$$ an open cover of $$S$$. We call a subset $$K \subseteq \mathbf{R}^n$$ topologically compact if every open cover $$\mathcal{A}$$ of $$K$$ admits a finite subcover. That is for every open cover $$\mathcal{A}$$, there exists $$k \in \mathbf{N}$$ such that there exist $$A_1, A_2, \ldots, A_k \in \mathcal{A}$$ with
$K \subseteq \bigcup_{i = 1}^k A_i.$

Theorem. If $$K \subseteq \mathbf{R}^n$$ is topologically compact, then it is closed and bounded.

Proof. We first show that $$K$$ is bounded. To this end, consider the family of open sets given by $$A_i = \{x \in \mathbf{R}^n \big| |x| < i\}$$. Clearly, we have $K \subseteq \mathbf{R}^n = \bigcup_{i = 1}^\infty A_i.$ Since $$K$$ admits a finite subcover, there exist $$i_1 < i_2 < \cdots < i_k$$ such that $K \subseteq A_{i_1} \cup A_{i_2} \cup \cdots \cup A_{i_k}.$ Since we have $$A_i \subseteq A_j$$ for $$i < j$$, we in fact have $$K \subseteq A_{i_k}$$, hence $$K$$ is bounded. We will now argue that $$K$$ is closed. Suppose toward a contradiction that $$K$$ is not closed. Then there exists some limit point $$x$$ of $$K$$ which is not contained in $$K$$. Now consider the family of sets $A_i = \{y \in \mathbf{R}^n \big| |x - y| > 1/i\}.$
Note that $$\bigcup A_i = \mathrm{R} \setminus\{x\}$$, thus (since $$x \notin K$$) $$\{A_i\}$$ is an open cover of $$K$$. Since this family admits an open subcover and the $$A_i$$ are nested ($$i < j$$ implies $$A_i \subseteq A_j$$) we have, in fact, $$K \subseteq A_k$$ for some $$k \in \mathbf{N}$$. But this implies that $$B(x, 1/k) \subseteq K^c$$, contradicting the assumption that $$x$$ is a limit point of $$K$$.∎ The converse of the theorem above is also true. The equivalence of being topologically compact and being closed and bounded (in $$\mathbf{R}^n$$) is known as the Heine-Borel Theorem.

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## Math 32A/H Final Review Materials

I have started compiling some review materials for math 32A and 32AH for this quarter. They consist of

I will post solutions to the exercises before the review session on Thursday. As always, let me know in the comments below if anything is unclear or incorrect.

Update I have posted the solutions to the practice problems. Also, a student pointed out a typo in statement of the final problem in the review. Spherical coordinates should be given by
$$x = \rho \cos \theta \sin \varphi, \quad y = \rho \sin \theta \sin \varphi, \quad z = \rho \cos \varphi.$$
These changes have been made in the version of the problems and solutions online.

Update 2 There was a mistake in the solution of problem 10 involving curvature. This has (hopefully) been fixed.

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## Math 32AH Final Review

I have written up an overview of much of the material covered in Math 32AH this quarter. Hopefully this review will serve as a good study guide for the course. The bulk of the review is a statement of the definitions of most of the terms defined in class. I didn’t write out any detailed proofs in the review, but left the proofs of many statements as exercises. The exercises therefore tend to be rather abstract and conceptual. Later this week, I will post more review questions (with solutions) of a more computational nature that will appeal both to 32A and 32AH.

Over the next week I will update the review with a bit more material to be covered before the final.

At any rate, here is a link to the review. As usual, let me know in the comments below if anything is unclear or incorrect.

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## 32A Midterm 2 Review Materials

I have written up some review materials for the second midterm for Math 32A. They include:

I think the questions and solutions cover the majority of the conceptual material likely to appear on the midterm. I’ve taken some of the problems/solutions from previous times I’ve taught this course, so some of the notation may not agree with what you are used to — so be warned.

As usual, please let me know in the comments below if anything is unclear or incorrect.

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## Unions and Intersections of Closed Sets

Let $$S \subset \mathbf{R}^n$$. We call $$x \in \mathbf{R}^n$$ a boundary point of $$S$$ if for every $$\varepsilon > 0$$ the ball $$B(x, \varepsilon)$$ centered at $$x$$ of radius $$\varepsilon$$ contains at least one point $$y \in S$$ and at least one point $$z \notin S$$. The set of all boundary points of $$S$$ is called the boundary of $$S$$ denoted $$\partial S$$. We say that $$S$$ is closed if $$\partial S \subset S$$.

Now let $$\{S_\alpha\}$$ be a collection of closed sets indexed by $$A$$ (that is $$\alpha$$ ranges over all possible values in $$A$$). We would like to show that the set
$S = \bigcap_{\alpha \in A} S_\alpha$
is closed. (Recall that $$x \in \bigcap_{\alpha \in A} S_\alpha$$ if and only if $$x \in S_\alpha$$ for all $$\alpha \in A$$.) To this end, we must show that every $$x \in \partial S$$ is also in $$S$$. To see this is the case, let $$x \in \partial S$$. Since $$x \in \partial S$$, for every $$\varepsilon > 0$$, there exist $$y \in S$$ and $$z \notin S$$ such that $$y, z \in B(x, \varepsilon)$$. Since $$y \in S$$, from the definition of $$S$$, we must have $$y \in S_\alpha$$ for all $$\alpha$$. Therefore, $$x \in S_\alpha \cup \partial S_\alpha$$ for each $$\alpha$$, hence $$x \in S_\alpha$$ for each $$\alpha$$ as $$\partial S_\alpha \subset S_\alpha$$. Thus we may conclude that
$x \in \bigcap_{\alpha \in A} S_\alpha = S$
which is what we needed to show.

Now we move on to consider unions of closed sets. First let $$S = \bigcup_{i = 1}^k S_i$$ be a finite union of closed sets. Suppose $$x \in \partial S$$. Again, for every $$\varepsilon > 0$$ there exists $$y \in S$$ and $$z \notin S$$ with $$y, z \in B(x, \varepsilon)$$. We would like to show that $$x \in S$$, hence $$S$$ is closed. Consider a sequence $$y_1, y_2, \ldots \in S$$ with $$|x – y_i| < 1/i$$. Such a sequence exists by choosing $$\varepsilon = 1, 1/2, 1/3, \ldots$$. Since each $$y_i \in S$$, $$y_i \in S_j$$ for some $$j$$. Since there are infinitely many $$y_i$$ and only finitely many $$S_j$$, some $$S_j$$ must contain infinitely many of the $$y_i$$. I will leave it to my diligent readers to verify that for this $$S_j$$ we have $$x \in \partial S_j \subset S_j$$, and therefore $$x \in S$$. This shows that $$S$$ is closed. Notice that the previous argument relies crucially on the fact that $$S$$ is a union of finitely many closed sets. What if $$S = \bigcup_{i = 1}^\infty S_i$$ is an infinite union? Can we still conclude that $$S$$ is closed? To see this is not the case, consider the sets $S_i = [1/i, 1 - 1/i]$ where $$i$$ is $$2, 3, 4, \ldots$$. Is the union of these sets closed?

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## Basic Logic

I have written up a short essay that gives a brief and informal overview of the logical syntax used in mathematics. It covers the basic logical connectives and quantifiers, and gives a few examples. In particular, the final example (negation of the ε-δ definition of continuity) should be helpful in solving problem 2 in this week’s homework for Math 32AH.

Here is a link to the essay: Basic Logic. As usual, let me know in the comments below if anything in the essay is incorrect or unclear.

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## Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is among the most useful inequalities in all of mathematics. Suppose $$V$$ is a (real) vector space and $$\langle \cdot, \cdot \rangle$$ be an inner product on $$V$$. That is, for all $$\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$$ and $$a, b \in \mathbf{R}$$, the inner product $$\langle \cdot, \cdot \rangle : V \times V \to \mathbf{R}$$ satisfies the following properties:

1. $$\langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v} ,\mathbf{u} \rangle$$
2. $$\langle a \mathbf{u} + b \mathbf{v}, \mathbf{w} \rangle = a \langle \mathbf{u}, \mathbf{w} \rangle + b \langle \mathbf{v}, \mathbf{w} \rangle$$
3. $$\langle \mathbf{u}, \mathbf{u} \rangle \geq 0$$ with $$\langle \mathbf{u}, \mathbf{u} \rangle = 0$$ if and only if $$\mathbf{u} = \mathbf{0}$$

Under these conditions, the Cauchy-Schwarz inequality states that
$|\langle \mathbf{u}, \mathbf{v} \rangle| \leq |\mathbf{u}| |\mathbf{v}|.$
Here, the norm of a vector is given by $$|\mathbf{v}|^2 = \langle \mathbf{v}, \mathbf{v} \rangle$$.

To prove the inequality, we employ the following trick: For any $$\lambda \in \mathbf{R}$$, we have
$\langle \mathbf{u} + \lambda \mathbf{v}, \mathbf{u} + \lambda \mathbf{v} \rangle \geq 0$
by property 3. The Cauchy-Schwarz inequality follows from choosing a suitable value for $$\lambda$$. Using properties 1 and 2, we can expand
$\langle \mathbf{u} + \lambda \mathbf{v}, \mathbf{u} + \lambda \mathbf{v} \rangle = \langle \mathbf{u}, \mathbf{u} \rangle + 2 \lambda \langle \mathbf{u}, \mathbf{v} \rangle + \lambda^2 \langle \mathbf{v}, \mathbf{v} \rangle \geq 0.$
Notice that this is a quadratic equation in the variable $$\lambda$$. Recall that the quadratic equation $$a \lambda^2 + b \lambda + c$$ obtains its minimum (for $$a > 0$$) when $$\lambda = \frac{- b}{2 a}$$. Therefore, we take
$\lambda = – \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle}.$
Plugging this into the previous expression, we obtain
$\langle \mathbf{u}, \mathbf{u} \rangle – 2 \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle} \langle \mathbf{u}, \mathbf{v} \rangle + \frac{\langle \mathbf{u}, \mathbf{v} \rangle^2}{\langle \mathbf{v}, \mathbf{v} \rangle^2} \langle \mathbf{v}, \mathbf{v} \rangle \geq 0.$
Multiplying both sides by $$\langle \mathbf{v}, \mathbf{v}\rangle$$ and rearranging gives
$\langle \mathbf{u}, \mathbf{v} \rangle^2 \leq \langle \mathbf{v}, \mathbf{v} \rangle \langle \mathbf{u}, \mathbf{u} \rangle$
whence the Cauchy-Schwarz inequality immediately follows.

## Math 32AH Final Review Problems

I am almost finished writing up solutions to the practice problems for Math 32AH (available here). I will update the file as I finish the write-up.

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## Math 32A Final Review

I just uploaded a review for the final exam for Math 32A (multivariable differential calculus) which is available here. If you find typos or think that some portion needs clarification, please let me know in the comments below. The review sheet contains an overview of the topics covered in the course, as well as many examples worked out in full detail.

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## Midterm II Review

Here are some examples that I will go over in the midterm review. If you notice any typos or errors, feel free to let me know in the comments. You can add math symbols to your comments using LaTeX syntax surrounded by dollar signs. For example, the line

D\lim_{x \to \infty} f(x) = \frac{1}{2}D

with the D’s replaced by dollar signs becomes

$\lim_{x \to \infty} f(x) = \frac{1}{2}$.