135 lines
4.4 KiB
Markdown
135 lines
4.4 KiB
Markdown
# Lecture 12
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## Review Questions
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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 $S\subset B_r(p)$.
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Consider the following statement: If a set $S\subset X$ is compact, the its is bounded.
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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)?
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We should choose a specific cover so that we can construct cover that have a set that is a superset of $S$.
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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.]
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### Continue on compact sets
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#### Lemma
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If $S\in X$ is compact, then $S$ is bounded.
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Proof:
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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 $S\subset B_r(p)$
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QED
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#### Definition k-cell
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A 2-cell is a set of the form $[a_1,b_1]\times[a_2,b_2]$
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#### Theorem 2.39 (K-dimension of Theorem)
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Theorem 2.38 replace with "closed and bounded intervals" to "k-cells".
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Ideas of Proof:
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Apply the Theorem to each dimension separately.
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#### Theorem 2.40
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Every k-cell is compact.
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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)
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Proof:
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That $[0,1]$ is compact.
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(Key idea, divide and conquer)
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Suppose for contradiction that $\exists$ open cover $\{G_a\}_{\alpha\in A}$ of $[0,1]$ with no finite subcovers of $[0,1]$
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**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 $\{G_\alpha\}_{\alpha\in A}$
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(If both of them could be, combine the two finite subcollections to get a finite subcover of $[0,1]$)
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Let $I_1$ be a subinterval without a finite subcover.
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**Step2.** Divide $I_1$ in half. Let $I_2$ be one of these two subintervals of $I_1$ without a finite subcover.
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**Step3.** etc.
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We obtain a seg of intervals $I_1\subset I_2\subset \dots$ such that
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(a) $[0,1]\supset I_1\supset I_2\supset \dots$
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(b) $\forall n\in \mathbb{N}$, $I_n$ is not covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$
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(c) The length of $I_n$ is $\frac{1}{2^n}$
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By (a) and **Theorem 2.38**, $\exists x^*\in \bigcap^{\infty}_{n=1} I_n$.
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Since $x^*\in [0,1]$, $\exists \alpha_0$ such that $x^*\in G_{\alpha_0}$
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Since $G_{\alpha_0}$ is open, $\exist r>0$ such that $B_r(x^*)\subset G_{\alpha_0}$
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Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subset B_r(x^*)\subset G_{\alpha_0}$
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Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
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QED
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#### Theorem 2.41
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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:
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(a) $E$ is closed and bounded.
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(b) $E$ is compact.
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(c) Every infinite subset of $E$ has a limit point in $E$.
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We'll prove $(a)\implies (b)\implies (c)\implies (a)$
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Proof:
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$(a)\implies (b)$
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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.
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$(b)\implies (c)$
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Follows from **Theorem 2.37**
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$(c)\implies (a)$
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We will proceed by contrapositive, which says that $\neg (a)\implies \neg (c)$
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$\neg (a)$: $E$ is not closed or not bounded.
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$\neg (c)$: $\exists$ infinite subcover $S\subset E$ such that $S'\cup E=\phi$
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Suppose $(a)$ does not hold. There are two cases to consider
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Case 1: $E$ is not bounded. Then $\forall v\in \mathbb{N},\exists x_n\in E$ such that $|x_n|\geq n$
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Let $S=\{x_n,...,n\in\mathbb{N}\}$, then $S'=\phi$ (by **Theorem 2.20**)
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Case 2: $E$ is not closed. Then $z\in E'\backslash E$.
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Since $z\in E'$, $\forall n\in \mathbb{N},\exists x_n\in E$ such that $|x_n-z|<\frac{1}{n}$
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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)
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We'll show if $y\neq z$, then $y\notin S'$
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$\forall w\in B_r(y)$
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$$
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\begin{aligned}
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d(w,z)&\geq d(y,z)-d(y,w)\\
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&>d(y,z)-r\\
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&=d(y,z)-\frac{1}{2}d(y,z)\\
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&=\frac{1}{2}d(y,z)
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\end{aligned}
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$$
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So $B_r(y)\cap S$ is finite. By **Theorem 2.20**, $y\notin S$, this proves the claim so $S'\cap E=\phi$
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QED
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