<Home>Teaching><MATH 131B (Winter 2022)>Week 3 Discussion Notes

Week 3 Discussion Notes

Homework Comments
- Limits
- Problem 2
Cauchy Sequences and Completeness
Homework Hints

Homework Comments

Limits

I noticed that a lot of students are very "limit-happy," i.e., many of you like to move limits around without justification. Starting with Homework 3, I will require that you justify any instances where you move a limit around. This is especially important if you want to swap $\lim$ with $\max$ , $\min$ , $\sup$ , or $\inf$ . (In fact, in many instances, that won't be legal.)

For example, here is a model solution:

Example 1.

Assume that $\lim_{n\to\infty} x_n = x$ and $\lim_{n\to\infty} y_n = y$ . Show that

\lim_{n\to\infty} \p{x_n + y_n}^2 = \p{x + y}^2.

Solution.

\begin{aligned} \lim_{n\to\infty} \p{x_n + y_n}^2 &= \p{\lim_{n\to\infty} \p{x_n + y_n}}^2 &&\p{t^2 \text{ is continuous}} \\ &= \p{x + y}^2 &&\p{\text{limit law}} \end{aligned}

Problem 2

While grading this one, I noticed that a lot of you guys wrote something like this:

$\lim_{n\to\infty} \abs{x_i^{\p{n}}} = 0 \text{ for all } i \implies \lim_{n\to\infty} \max_{1 \leq i \leq k}{\abs{x_i^{\p{n}}}} = 0$

Many of you also wrote the equivalent statement

$\max_{1 \leq i \leq k} \lim_{n\to\infty} \abs{x_i^{\p{n}}} = 0 \implies \lim_{n\to\infty} \max_{1 \leq i \leq k}{\abs{x_i^{\p{n}}}} = 0$

While it turns out this is true for this problem, it's a non-trivial fact, meaning you have to prove it. What's special in this problem is that $i$ ranges over a finite set, $i \in \set{1, \ldots, k}$ . When you take a maximum or supremum over an infinite set, you can no longer switch $\lim$ and $\max$ in general:

Example 2.

Let

a_i^{\p{n}} = \begin{cases} 1 & \text{if } i = n, \\ 0 & \text{otherwise}. \end{cases}

In other words, $a^{\p{n}}$ is a sequence which is all $0$ 's except at the $n$ -th term. For example,

\begin{aligned} a^{\p{1}} &= \p{1, 0, 0, 0, \ldots} \\ a^{\p{2}} &= \p{0, 1, 0, 0, \ldots} \\ a^{\p{3}} &= \p{0, 0, 1, 0, \ldots} \\ &\,\,\:\vdots \end{aligned}

Then

\lim_{n\to\infty} \abs{x_i^{\p{n}}} = 0 \text{ for all } i \iff \max_i \lim_{n\to\infty} \abs{x_i^{\p{n}}} = 0,

but for each $n$ , the maximum coordinate of $x^{\p{n}}$ is $1$ , so you also have

\lim_{n\to\infty} \max_i{\abs{x_i^{\p{n}}}} = \lim_{n\to\infty} 1 = 1.

Moral: Pointwise convergence (i.e., convergence in each coordinate) does not imply uniform convergence (i.e., convergence of the maximum or supremum) in general.

We'll see more examples of this phenomenon later.

Cauchy Sequences and Completeness

If you remember the definition of a Cauchy sequence in $\R$ , then this definition should be no surprise:

Definition

Let $\p{X, d}$ be a metric space. We say a sequence $\set{x_n}_n$ is Cauchy if for all $\epsilon > 0$ , there exists $N$ such that $d\p{x_n, x_m} < \epsilon$ whenever $n, m \geq N$ .

Morally, Cauchy sequences "should" converge. In $\p{\R, \abs{\:\cdot\:}}$ , a sequence was convergent if and only if it is Cauchy (this is the Bolzano-Weierstrass theorem from MATH 131A). However, this is not true for general metric spaces, which is why we have an additional definition:

Definition

Let $\p{X, d}$ be a metric space. If every Cauchy sequence converges in $X$ , then we say that $X$ is complete.

Example 3.

$\R$ is complete with respect to the Euclidean metric..
$\p{0, 1}$ is not complete with respect to the Euclidean metric.
$\br{0, 1}$ is complete with respect to the Euclidean metric..

Example 4.

Show that $\p{\R^k, d_{\ell^2}}$ is complete.

Solution.

To show this, we need to show that any Cauchy sequence $\set{x^{\p{n}}}_n$ in $\R^n$ converges. Each element in our sequence is a vector:

x^{\p{n}} = \p{x^{\p{n}}_1, \ldots, x^{\p{n}}_k}

As you can imagine, we're going to need to use completeness of $\R$ to do this.

Let $\epsilon > 0$ . Since $\set{x^{\p{n}}}_n$ is Cauchy, there exists an $N$ such that if $n, m \geq N$ , then

\p{\sum_{i=1}^k \abs{x^{\p{n}}_i - x^{\p{m}}_i}^2}^{1/2} = d_{\ell^2}\p{x^{\p{n}}, x^{\p{m}}} < \epsilon.

Notice that

\abs{x^{\p{n}}_j - x^{\p{m}}_j} \leq \p{\sum_{i=1}^k \abs{x^{\p{n}}_i - x^{\p{m}}_i}^2}^{1/2} < \epsilon

if $n, m \geq N$ , i.e., the components $\set{x^{\p{n}}_j}_n$ is Cauchy in $\R$ . Thus, by completeness, there exists $x_j \in \R$ such that $\lim_{n\to\infty} x^{\p{n}}_j = x_j$ for each $1 \leq j \leq k$ . The vector $x = \p{x_1, \ldots, x_k}$ will be our candidate for the limit of the sequence. We just need to prove that $\set{x^{\p{n}}}_n$ actually converges to $x$ with respect to $d_{\ell^2}$ .

This is easy, though. The sum in $d_{\ell^2}$ is a finite sum, so we can apply our limit laws:

\begin{aligned} \lim_{n\to\infty} d_{\ell^2}\p{x^{\p{n}}, x} &= \lim_{n\to\infty} \p{\sum_{i=1}^k \abs{x^{\p{n}}_i - x_i}^2}^{1/2} \\ &= \p{\lim_{n\to\infty} \sum_{i=1}^k \abs{x^{\p{n}}_i - x_i}^2}^{1/2} && \p{t^{1/2} \text{ is continuous}} \\ &= \p{\sum_{i=1}^k \lim_{n\to\infty} \abs{x^{\p{n}}_i - x_i}^2}^{1/2} && \p{\text{limit law; sum is finite}} \\ &= 0 && \p{t^2 \text{ is continuous; } \lim_{n\to\infty} x^{\p{n}}_i = x_i} \end{aligned}

Thus, $\set{x^{\p{n}}}_n$ converges to $x$ in $d_{\ell^2}$ , so $\p{\R^k, d_{\ell^2}}$ is complete.

Example 5.

Show that $\p{C\p{\br{0,1}}, d}$ is complete, where $d$ is the $\sup$ -norm.

Solution.

As a reminder:

\begin{gathered} C\p{\br{0,1}} = \set{\func{f}{\br{0,1}}{\R} \st f \text{ is continuous}} \\ d\p{f, g} = \sup_{x \in \br{0,1}} \abs{f\p{x} - g\p{x}} \end{gathered}

Let $\set{f_n}_n$ be a Cauchy sequence. Like in the previous example, we'll first want to find a candidate for the limit of this sequence. Notice that for a fixed $x \in \br{0, 1}$ ,

\abs{f_n\p{x} - f_m\p{x}} \leq \sup_{x \in \br{0,1}} \abs{f_n\p{x} - f_m\p{x}} = d\p{f_n, f_m}.

Like before, this estimate implies that $\set{f_n\p{x}}_n$ is Cauchy for each $x$ , so by completeness of $\R$ , there exists a number $f\p{x} \in \R$ such that $\lim_{n\to\infty} f_n\p{x} = f\p{x}$ .

There are two more things to show: (i) that $\set{f_n}_n$ converges to $f$ with respect to $d$ , (ii) and that $f$ is actually in $C\p{\br{0,1}}$ , i.e., that $f$ is continuous.

To show (i), let $\epsilon > 0$ . Because $\set{f_n}_n$ is Cauchy, there exists $N$ such that for any $x \in \br{0, 1}$ ,

\begin{aligned} \abs{f_n\p{x} - f_m\p{x}} &\leq \sup_{x \in \br{0,1}} \abs{f_n\p{x} - f_m\p{x}} \\ &= d\p{f, g} \\ &< \frac{\epsilon}{2} \end{aligned}

whenever $n, m \geq N$ . If we fix $n \geq N$ and send $m \to \infty$ , we get

\begin{gathered} \abs{f_n\p{x} - f\p{x}} = \lim_{m\to\infty} \abs{f_n\p{x} - f_m\p{x}} \leq \frac{\epsilon}{2} \\ \implies d\p{f_n, f} = \sup_{x \in \br{0,1}} \abs{f_n\p{x} - f\p{x}} \leq \frac{\epsilon}{2} < \epsilon, \end{gathered}

and this inequality is true for any $n \geq N$ . In other words, $\set{f_n}_n$ converges to $f$ with respect to $d$ .

Finally, we need to show (ii). But this is immediate: convergence with respect to $d$ is exactly uniform convergence. Thus, $\set{f_n}_n$ is a sequence of continuous functions which converges uniformly to $f$ , so $f$ itself must be continuous (this is a fact from MATH 131A).

Remark.

When proving (i), you cannot write the following:

$\begin{aligned} \sup_{x \in \br{0,1}} \abs{f_n\p{x} - f\p{x}} &= \sup_{x \in \br{0,1}} \lim_{m\to\infty} \abs{f_n\p{x} - f_m\p{x}} \\ &= \lim_{m\to\infty} \sup_{x \in \br{0,1}} \abs{f_n\p{x} - f_m\p{x}} \\ &< \epsilon. \end{aligned}$

The second equality is not true in general: Consider $f_n\p{x} = x^n$ on $\pco{0, 1}$ . Then

\lim_{n\to\infty} \sup_{x \in \pco{0,1}}{x^n} = 1 \quad\text{but}\quad \sup_{x \in \pco{0,1}} \lim_{n\to\infty} x^n = 0.

This means that you cannot swap $\sup$ and $\lim$ without additional justification.

Homework Hints

Here's a useful proposition (that's listed as an exercise in Terry's book) for Problem 2, 3(a), and 5:

Proposition (1.2.5(h))

If $E$ is any subset of $X$ , then $\Int{E}$ is open, and given any other open set $V \subseteq E$ , we have $V \subseteq \Int{E}$ . Similarly, $\cl{E}$ is closed, and given any other closed set $K \supseteq E$ , we have $K \supseteq \cl{E}$ .

Proof.

There are a number of things to prove:

$\Int{E}$ is open:

Let $x \in \Int{E}$ . By definition, this means there exists $r > 0$ such that $B\p{x, r} \subseteq E$ . We'll show that $B\p{x, r} \subseteq \Int{E}$ also, which shows by definition that $\Int{E}$ is open.

$B\p{x, r}$ is an open set, so given any $y \in B\p{x, r}$ , there exists $r' > 0$ such that $B\p{y, r'} \subseteq B\p{x, r} \subseteq E$ . In other words, $y \in \Int{E}$ , so $B\p{x, r} \subseteq \Int{E}$ .

$\Int{E}$ is the largest open set in $E$ :

Assume $V \subseteq E$ and that $V$ is open. Given $x \in V$ , this means there exists $r > 0$ such that $B\p{x, r} \subseteq V \subseteq E$ . Thus, $x \in \Int{E}$ by definition, so $V \subseteq \Int{E}$ .

$\cl{E}$ is closed:

This was proven in lecture.

$\cl{E}$ is the smallest closed set containing $E$ :

Assume $K \supseteq E$ and $K$ is closed. Given $x \in \cl{E}$ and $r > 0$ ,

\emptyset \neq B\p{x, r} \cap E \subseteq B\p{x, r} \cap K.

Moreover, $K$ is closed, so $x \in \cl{K} = K$ , which shows that $\cl{E} \subseteq K$ . $\square$

You can use this proposition freely as long as you cite it properly (i.e., write down the proposition number every time you use it).

Week 3 Discussion Notes

Table of Contents

Homework Comments

Limits

Example 1.

Solution.

Problem 2

Example 2.

Cauchy Sequences and Completeness

Definition

Definition

Example 3.

Example 4.

Solution.

Example 5.

Solution.

Remark.

Homework Hints

Proposition (1.2.5(h))

Proof.