90 lines
2.7 KiB
Markdown
90 lines
2.7 KiB
Markdown
# Math 4111 Exam 2 review
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$E$ is open if $\forall x\in E$, $x\in E^\circ$ ($E\subset E^\circ$)
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$E$ is closed if $E\supset E'$
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Then $E$ closed $\iff E^c$ open $\iff \forall x\in E^\circ, \exists r>0$ such that $B_r(x)\subset E^c$
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$\forall x\in E^c$, $\forall x\notin E$
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$B_r(x)\subset E^c\iff B_r(x)\cap E=\phi$
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## Past exam questions
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$S,T$ is compact $\implies S\cup T$ is compact
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Proof:
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Suppose $S$ and $T$ are compact, let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $S\cup T$
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(NOT) $\{G_\alpha\}$ is an open cover of $S$, $\{H_\beta\}$ is an open cover of $T$.
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...
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QED
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## K-cells are 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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## Redundant subcover question
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$M$ is compact and $\{G_\alpha\}_{\alpha\in A}$ is a "redundant" subcover of $M$.
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$\exists \{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $M$.
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We define $S$ be the $x\in M$ that is only being covered once.
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$$
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S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)
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$$
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We claim $S$ is a closed set.
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$G_{\alpha_i}\cap G_{\alpha_j}$ is open.
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$\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed
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$S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed.
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So $S$ is compact, we found another finite subcover yeah!
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