153 lines
5.1 KiB
Markdown
153 lines
5.1 KiB
Markdown
# 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 $t$.
|
|
|
|
QED
|
|
|
|
## 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 $p_{n_i}\to q$.
|
|
|
|
QED
|
|
|
|
### 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
|
|
|
|
$$
|
|
3,3.1,3.14,3.141,3.1415,\cdots
|
|
$$
|
|
|
|
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
|
|
|
|
$$
|
|
p_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}
|
|
$$
|
|
|
|
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.*
|
|
|
|
QED
|
|
|
|
#### 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 $\forall n\in \mathbb{N}$, $p_n\in B_r(p_N)$.
|
|
|
|
QED
|
|
|
|
> 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**, $\bigcap_{n=1}^{\infty}K_n\neq \phi$. 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.
|
|
|
|
QED
|
|
|