Lecture 16

Review

Let $(s_n)$ be a sequence in $\mathbb{R}$ satisfying the following properties:

1. It is bounded ($\exists M>0$ such that $\forall n\in \mathbb{N}, |s_n|\leq M$)
2. It is monotonic increasing ($\forall n\in \mathbb{N}, s_n\leq s_{n+1}$)

Let $E=\{s_n:n\in \mathbb{N}\}$ and $t=sup E$. Prove that $s_n\to t$. [Hint: The proof begins with "Let $\epsilon>0$ be arbitrary." What do we know about $t-\epsilon$?]

Proof:

Let $\epsilon>0$ be arbitrary. Then since $t-\epsilon$ is not an upper bound of $E$, $\exists N$ such that $t-\epsilon<s_N$.

Let $n\geq N$. Since $(s_n)$ is monotonic increasing, $s_n\geq s_N>t-\epsilon$. Since $t$ is an upper bound of $E$, $s_n<t$. Therefore, $|s_n-t|<\epsilon$.

So $(s_n)$ converges to $t$.

EOP

New materials

Subsequences

Theorem 3.7

Let $X$ be a metric space. If $(p_n)$ is a sequence in $X$. $E^*=\{p\in X:\exists \textup{ subsequence } (p_{n_i}) \textup{ such that } p_{n_i}\to p\}$. Then $E^*$ is closed.

Proof:

Let $q\in (E^*)'$. We will show that $q\in E^*$.

Step 1: Since $q\in (E^*)'$, $\exists x_1\in B_1(q)\cap E^*$.

Since $x_1\in E^*$, $\exists n_1\in \mathbb{N}$ such that $p_{n_1}\in B_1(x_1)$.

By triangle inequality, $d(p_{n_1},q)\leq d(p_{n_1},x_1)+d(x_1,q)<1+1=2$.

Step 2: Since $q\in (E^*)'$, $\exists x_2\in B_{1/2}(q)\cap E^*$ and $n_2>n_1$ (by definition of $E^*$. If $x_2\in E^*$, then there are infinitely many $p\in \mathbb{N}$ such that $p_n\in B_{1/2}(x_2)$).

Since $x_2\in E^*$, $\exists n_2\in \mathbb{N}$ such that $p_{n_2}\in B_{1/2}(x_2)$.

By triangle inequality, $d(p_{n_2},q)\leq d(p_{n_2},x_2)+d(x_2,q)<\frac{1}{2}+\frac{1}{2}=1$.

Step 3: By induction, we can get a sequence $n_1,n_2,\cdots$ such that $\forall i\in \mathbb{N}, d(p_{n_i},q)<\frac{2}{i}$.

Then $(p_{n_i})$ is a subsequence of $(p_n)$ and $p_{n_i}\to q$.

EOP

Cauchy Sequences

Definition 3.8

A sequence $(p_n)$ in a metric space $X$ is called a Cauchy sequence if for every $\epsilon>0$, there exists $N\in \mathbb{N}$ such that $\forall m,n\geq N$, $d(p_m,p_n)<\epsilon$.

The terms are getting closer to each other.

Example:

$X=\mathbb{Q}$ with the usual metric. Let $(p_n)$ be a sequence

If $m\leq n$, $|p_m-p_n|<\frac{1}{10^{m}}$.

Then $(p_n)$ is a Cauchy sequence. Let $\epsilon>0$ be arbitrary. Choose $N$ such that $\frac{1}{10^{N}}>\epsilon$. Then if $m,n\geq N$, then $|p_m-p_n|\leq \frac{1}{10^{m}}<\epsilon$.

This sequence does not converge in $\mathbb{Q}$.

$X=\mathbb{R}$ with the usual metric. Let $(p_n)$ be a sequence

This sequence is not bounded above. (by Theorem 3.28), so (as we will prove) it is not a Cauchy sequence.

The fact that $p_{n+1}-p_n=\frac{1}{n+1}\to 0$ is not relevant to determining whether $(p_n)$ is a Cauchy sequence.

Theorem 3.11 (a)

$(p_n)$ converges $\implies$ $(p_n)$ is a Cauchy sequence.

Proof:

Since $(p_n)$ converges, $\exists p\in X$ such that $p_n\to p$. Let $\epsilon>0$ be arbitrary. Then $\exists N\in \mathbb{N}$ such that $\forall n\geq N$, $d(p_n,p)<\epsilon$.

If $m,n\geq N$, then $d(p_m,p_n)\leq d(p_m,p)+d(p,p_n)<\epsilon+\epsilon=2\epsilon$.

You can also use $\frac{\epsilon}{2}$ instead of $\epsilon$ in the above proof, just for fun.

EOP

Lemma 3.11 (b)

If $(p_n)$ is a Cauchy sequence, then $(p_n)$ is bounded above.

Proof:

Since $(p_n)$ is a Cauchy sequence, $\exists N\in \mathbb{N}$ such that $\forall m,n\geq N$, $d(p_m,p_n)<1$.

Let $r=max\{d(p_i,p_j);1\leq i,j\leq N\}+1$.

Then $\forall n\in \mathbb{N}$, $p_n\in B_r(p_N)$.

EOP

Note: This proof is nearly identical to the proof of convergent sequences implies bounded.

Definition 3.9

Let $E$ be a nonempty subset of a metric space $X$, and let $S$ be the set of all real numbers of the form $d(p,q)$ for $p,q\in E$. The diameter of $E$, denoted by $diam E$, is defined to be the supremum of $S$.

Exercise:

Prove that $(p_n)$ is a Cauchy sequence if and only if $\lim_{N\to \infty}diam\{(p_n):n\geq N\}=0$.

Theorem 3.10

(a) $diam E=diam(\overline{E})$
(b) If $K_n$ is a sequence of nonempty compact sets and $K_1\supset K_2\supset \cdots$, then $\bigcap_{n=1}^{\infty}K_n$ has exactly one point.

Proof:

(a) The idea is still, triangle inequality.

Since $E\subset \overline{E}$, $diam E\leq diam(\overline{E})$.

Now we want to show that $diam(\overline{E})\leq diam E$.

Claim: $\forall \epsilon>0$, $2\epsilon+diam E$ is an upper bound of $\{d(p,q):p,q\in \overline{E}\}$.

Let $p,q\in \overline{E}$.

Since $p\in \overline{E}$, $\exists p'\in E\cap B_\epsilon(p)$.

Since $q\in \overline{E}$, $\exists q'\in E\cap B_\epsilon(q)$.

Then $d(p,q)\leq d(p,p')+d(p',q')+d(q',q)<\epsilon+diam E+\epsilon=diam E+2\epsilon$.

This proves the claim.

By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\overline{E})\leq 2\epsilon+diam E$. So $diam(\overline{E})\leq diam E$.

(b) By Theorem 2.36, $\bigcap_{n=1}^{\infty}K_n\neq \emptyset$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0.

EOP