# Monthly Archives: October 2013

teaching

## 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?

teaching

## 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.

teaching

## 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.