NoteNextra-origin/pages/Math4111/Math4111_L16.md

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

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.

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 \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.

EOP