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.