bug fixed,

remaining issues in mobilenavbar need pruning, need rewrite the class.

content/Math4111/Exam_reviews/Math4111_E2.md

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