NoteNextra-origin/pages/Math4111/Math4111_L12.md

Lecture 12

Review Questions

For a metric space (X,d), we say a subset S\subset X si bounded if there exists p\in X and r>0 such that S\subset B_r(p).

Consider the following statement: If a set S\subset X is compact, the its is bounded.

1. Will the proof of this statement involve an arbitrary open cover (one that you, the prover, do not get to choose) or a specific open cover (one that you can choose)? We should choose a specific cover so that we can construct cover that have a set that is a superset of S.
2. Give a proof of the statement. [Suggestion: If you prefer, you could try proving the contrapositive. Both a direct proof and a proof by contrapositive are roughly of the same difficulty.]

Continue on compact sets

Lemma

If S\in X is compact, then S is bounded.

Proof:

Fix p\in X, then \{B_n(p)\}_{n\in \mathbb{N}} (specific open cover) is an open cover of S (Since \bigcup_{n\in \mathbb{N}}=X). Since S is compact, then \exists a finite subcover {n\in \mathbb{N}}_{i=1}^k=S, let r=max(n_1,...n_k), Then S\subset B_r(p)

EOP

Definition k-cell

A 2-cell is a set of the form [a_1,b_1]\times[a_2,b_2]

Theorem 2.39 (K-dimension of Theorem)

Theorem 2.38 replace with "closed and bounded intervals" to "k-cells".

Ideas of Proof:

Apply the Theorem to each dimension separately.

Theorem 2.40

Every k-cell is compact.

We'll prove the case k=1 and I=[0,1] (This is to simplify notation. This same ideas are used in the general case)

Proof:

That [0,1] is compact.

(Key idea, divide and conquer)

Suppose for contradiction that \exists open cover \{G_a\}_{\alpha\in A} of [0,1] with no finite subcovers of [0,1]

Step1. Divide [0,1] in half. [0,\frac{1}{2}] and [\frac{1}{2},1] and at least one of the subintervals cannot be covered by a finite subcollection of \{G_\alpha\}_{\alpha\in A}

(If both of them could be, combine the two finite subcollections to get a finite subcover of [0,1])

Let I_1 be a subinterval without a finite subcover.

Step2. Divide I_1 in half. Let I_2 be one of these two subintervals of I_1 without a finite subcover.

Step3. etc.

We obtain a seg of intervals I_1\subset I_2\subset \dots such that

(a) [0,1]\supset I_1\supset I_2\supset \dots
(b) \forall n\in \mathbb{N}, I_n is not covered by a finite subcollection of \{G_\alpha\}_{\alpha\in A}
(c) The length of I_n is \frac{1}{2^n}

By (a) and Theorem 2.38, \exists x^*\in \bigcap^{\infty}_{n=1} I_n.

Since x^*\in [0,1], \exists \alpha_0 such that x^*\in G_{\alpha_0}

Since G_{\alpha_0} is open, \exist r>0 such that B_r(x^*)\subset G_{\alpha_0}

Let n\in \mathbb{N} be such that \frac{1}{2^n}<r. Then by (c), I(n)\subset B_r(x^*)\subset G_{\alpha_0}

Then \{G_{\alpha_0}\} is a cover of I_n which contradicts with (b)

EOP

Theorem 2.41

If a set E in \mathbb{R}^k (only in \mathbb{R}^k, not for general topological space or metric spaces.) has one of the following three properties, then it has the other two:

(a) E is closed and bounded.
(b) E is compact.
(c) Every infinite subset of E has a limit point in E.

We'll prove (a)\implies (b)\implies (c)\implies (a)

Proof:

(a)\implies (b)

Suppose (a) holds, then \exists k-cell I such that E\in I.By Theorem 2.40, I is compact. By Theorem 2.35, E is compact.

(b)\implies (c)

Follows from Theorem 2.37

(c)\implies (a)

We will proceed by contrapositive, which says that \neg (a)\implies \neg (c)

\neg (a): E is not closed or not bounded.

\neg (c): \exists infinite subcover S\subset E such that S'\cup E=\phi

Suppose (a) does not hold. There are two cases to consider

Case 1: E is not bounded. Then \forall v\in \mathbb{N},\exists x_n\in E such that |x_n|\geq n

Let S=\{x_n,...,n\in\mathbb{N}\}, then S'=\phi (by Theorem 2.20)

Case 2: E is not closed. Then z\in E'\backslash E.

Since z\in E', \forall n\in \mathbb{N},\exists x_n\in E such that |x_n-z|<\frac{1}{n}

Let S=\{x_n:n\in \mathbb{N}\}, we claim S'\subset \{z\}. (In fact s'\in\{z\}, but we don't need this in the proof)

We'll show if y\neq z, then y\notin S'

\forall w\in B_r(y)

\begin{aligned}
    d(w,z)&\geq d(y,z)-d(y,w)\\
    &>d(y,z)-r\\
    &=d(y,z)-\frac{1}{2}d(y,z)\\
    &=\frac{1}{2}d(y,z)
\end{aligned}

So B_r(y)\cap S is finite. By Theorem 2.20, y\notin S, this proves the claim so S'\cap E=\phi

EOP