seems working on this small batch
This commit is contained in:
Zheyuan Wu
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# 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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# Math 4111 Exam 3 review
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## Relations between series and topology (compactness, closure, etc.)
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Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\phi\}$
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Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\phi\}$
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$p_n\to p\implies \forall \epsilon>0, \exists N$ such that $\forall n\geq N, p_n\in B_\epsilon(p)$
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### Some interesting results
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#### Lemma
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$p\in \overline{E}\iff \exists (p_n)\subseteq E$ such that $p_n\to p$
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$p\in E'\iff \exists (p_n)\subseteq E\backslash\{p\}$ such that $p_n\to p$ (you cannot choose $p$ in the sequence)
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#### Bolzano-Weierstrass Theorem
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Let $E$ be a compact set and $(p_n)$ be a sequence in $E$. Then $\exists (p_{n_k})\subseteq (p_n)$ such that $p_{n_k}\to p\in E$.
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Rudin Proof:
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Rudin's proof uses a fact from Chapter 2.
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If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\phi$).
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## Examples of Cauchy sequence that does not converge
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> Cauchy sequence in $(X,d),\forall \epsilon>0, \exists N$ such that $\forall m,n\geq N, d(p_m,p_n)<\epsilon$
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Let $X=\mathbb{Q}$ and $(p_q)=\{1,1.4,1.41,1.414,1.4142,1.41421,\dots\}$ The sequence is Cauchy but does not converge in $\mathbb{Q}$.
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This does not hold in $\mathbb{R}$ because compact metric spaces are complete.
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Fact: Every Cauchy sequence is bounded.
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## Proof that $e$ is irrational
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> $e=\sum_{n=0}^\infty \frac{1}{n!}$
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Let $s_n=\sum_{k=0}^n \frac{1}{k!}$
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So $e-s_n=\left(\sum_{k=n+1}^\infty \frac{1}{k!}\right)<\frac{1}{n!n}$
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If $e$ is rational, then $\exists p,q\in\mathbb{Z}$ such that $e=\frac{q}{p}$ and $q!s_q\in\mathbb{Z}$, $q!e=q!\frac{p}{q}\in \mathbb{Z}$, so $q!(e-s_q)\in\mathbb{Z}$
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$0<q!(e-s_q)<\frac{1}{n!n}$ leads to contradiction.
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## $\limsup$ and $\liminf$
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Let $(a_n)=(-1)^n$
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$\limsup a_n=1$ and $\liminf a_n=-1$
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Let $(a_n)\to a$
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$\limsup a_n=\liminf a_n=a$
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### Facts about $\limsup$ and $\liminf$
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#### Convergence of subsequence
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_$\limsup$ is the largest value that subsequence of $a_n$ can approach to._
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_$\liminf$ is the smallest value that subsequence of $a_n$ can approach to._
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#### Elements of sequence
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$\forall x>s^*,\{n:a_n>x\}$ is finite. $\exists N$ such that $\forall n\geq N, a_n\leq x$
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$\forall x<s^*,\{n:a_n>x\}$ is infinite.
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One example is $(a_n)=(-1)^n\frac{n}{n+1}$
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$\limsup a_n=1$ and $\liminf a_n=-1$
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So the size of set of elements of $a_n$ that are greater than any $x<1$ is infinite. and the size of set of elements of $a_n$ that are greater than any $x>1$ is finite.
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#### $\limsup(a_n+b_n)\leq \limsup a_n+\limsup b_n$
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One example for smaller than is $(a_n)=(-1)^n$ and $(b_n)=(-1)^{n+1}$
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$\limsup(a_n+b_n)=0$ and $\limsup a_n+\limsup b_n=2$
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## ($\forall n,s_n\leq t_n$) $\implies \limsup s_n\leq \limsup t_n$
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One example of using this theorem is $(s_n)=\left(\sum_{k=1}^n\frac{1}{k!}\right)$ and $(t_n)=\left(\frac{1}{n}+1\right)^n$
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## Rearrangement of series
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Will not be tested.
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_infinite sum is not similar to finite sum. For infinite sum, the order of terms matters. But for finite sum, the order of terms does not matter, you can rearrange the terms as you want._
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## Ways to prove convergence of series
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### n-th term test (divergence test)
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If $\lim_{n\to\infty}a_n\neq 0$, then $\sum a_n$ diverges.
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### Definition of convergence of series (convergence and divergence test)
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If $\sum a_n$ converges, then $\lim_{n\to\infty}\sum_{k=1}^n a_k=0$.
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Example: Telescoping series and geometric series.
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### Comparison test (convergence and divergence test (absolute convergence))
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Let $(a_n)$ be a sequence in $\mathbb{C}$ and $(c_n)$ be a non-negative sequence in $\mathbb{R}$. Suppose $\forall n, |a_n|\leq c_n$.
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(a) If the series $\sum_{n=1}^{\infty}c_n$ converges, then the series $\sum_{n=1}^{\infty}a_n$ converges.
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(b) If the series $\sum_{n=1}^{\infty}a_n$ diverges, then the series $\sum_{n=1}^{\infty}c_n$ diverges.
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### Ratio test (convergence and divergence test (absolute convergence))
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> $$ \left|\frac{a_{n+1}}{a_n}\right| \leq \alpha \implies |a_n|\leq \alpha^n$$
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Given a series $\sum_{n=0}^{\infty} a_n$, $a_n\in\mathbb{C}\backslash\{0\}$.
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Then
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(a) If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.
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(b) If $\left|\frac{a_{n+1}}{a_n}\right| \geq 1$ for all $n\geq n_0$ for some $n_0\in\mathbb{N}$, then $\sum_{n=0}^{\infty} a_n$ diverges.
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### Root test (convergence and divergence test (absolute convergence))
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> $$ \sqrt[n]{|a_n|} \leq \alpha \implies |a_n|\leq \alpha^n$$
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Given a series $\sum_{n=0}^{\infty} a_n$, put $\alpha = \limsup_{n\to\infty} \sqrt[n]{|a_n|}$.
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Then
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(a) If $\alpha < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.
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(b) If $\alpha > 1$, then $\sum_{n=0}^{\infty} a_n$ diverges.
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(c) If $\alpha = 1$, the test gives no information
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### Cauchy criterion
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### Geometric series
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### P-series
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(a) $\sum_{n=0}^{\infty}\frac{1}{n}$ diverges.
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(b) $\sum_{n=0}^{\infty}\frac{1}{n^2}$ converges.
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### Cauchy condensation test (convergence test)
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Suppose $(a_n)$ is a non-negative sequence. The series $\sum_{n=1}^{\infty}a_n$ converges if and only if the series $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges.
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### Dirichlet test (convergence test)
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Suppose
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(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence.
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(b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
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(c) $\lim_{n\to\infty}b_n=0$.
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Then $\sum a_nb_n$ converges.
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Example: $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converges.
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### Abel's test (convergence test)
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Let $(b_n)^\infty_{n=0}$ be a sequence such that:
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(a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
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(b) $\lim_{n\to\infty}b_n=0$
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Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges.
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# Math 4111 Final Review
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## Weierstrass M-test
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Let $\sum_{n=1}^{\infty} f_n(x)$ be a series of functions.
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The weierstrass M-test goes as follows:
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1. $\exists M_n \geq 0$ such that $\forall x\in E, |f_n(x)| \leq M_n$.
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2. $\sum M_n$ converges.
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Then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly.
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Example:
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### Ver.0
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$\forall x\in [-1,1)$,
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$$
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\sum_{n=1}^{\infty} \frac{x^n}{n}
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$$
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converges. (point-wise convergence on $[-1,1)$)
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$\forall x\in [-1,1)$,
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$$
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\left| \frac{x^n}{n} \right| \leq \frac{1}{n}
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$$
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Since $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we don't know if the series converges uniformly or not using the weierstrass M-test.
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### Ver.1
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However, if we consider the series on $[-1,1]$,
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$$
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\sum_{n=1}^{\infty} \frac{x^n}{n^2}
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$$
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converges uniformly. Let $M_n = \frac{1}{n^2}$. This satisfies the weierstrass M-test. And this series converges uniformly on $[-1,1]$.
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### Ver.2
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$\forall x\in [-\frac{1}{2},\frac{1}{2}]$,
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$$
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\sum_{n=1}^{\infty} \frac{x^n}{n}
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$$
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converges uniformly. Since $\left| \frac{x^n}{n} \right|=\frac{|x|^n}{n}\leq \frac{(1/2)^n}{n}\leq \frac{1}{2^n}=M_n$, by geometric series test, $\sum_{n=1}^{\infty} M_n$ converges.
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M-test still not applicable here.
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$$
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\sum_{n=1}^{\infty} \frac{x^n}{n}
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$$
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converges uniformly on $[-\frac{1}{2},\frac{1}{2}]$.
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> Comparison test:
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>
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> For a series $\sum_{n=1}^{\infty} a_n$, if
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>
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> 1. $\exists M_n$ such that $|a_n|\leq M_n$
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> 2. $\sum_{n=1}^{\infty} M_n$ converges
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>
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> Then $\sum_{n=1}^{\infty} a_n$ converges.
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## Proving continuity of a function
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If $f:E\to Y$ is continuous at $p\in E$, then for any $\epsilon>0$, there exists $\delta>0$ such that for any $x\in E$, if $|x-p|<\delta$, then $|f(x)-f(p)|<\epsilon$.
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Example:
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Let $f(x)=2x+1$. For $p=1$, prove that $f$ is continuous at $p$.
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Let $\epsilon>0$ be given. Let $\delta=\frac{\epsilon}{2}$. Then for any $x\in \mathbb{R}$, if $|x-1|<\delta$, then
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$$
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|f(x)-f(1)|=|2x+1-3|=|2x-2|=2|x-1|<2\delta=\epsilon.
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$$
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Therefore, $f$ is continuous at $p=1$.
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_You can also use smaller $\delta$ and we don't need to find the "optimal" $\delta$._
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## Play of open covers
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Example of non compact set:
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$\mathbb{Q}$ is not compact, we can construct an open cover $G_n=(-\infty,\sqrt{2})\cup (\sqrt{2}+\frac{1}{n},\infty)$.
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Every unbounded set is not compact, we construct an open cover $G_n=(-n,n)$.
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Every k-cell is compact.
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Every finite set is compact.
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Let $p\in A$ and $A$ is compact. Then $A\backslash \{p\}$ is not compact, we can construct an open cover $G_n=(\inf(A)-1,p)\cup (p+\frac{1}{n},\sup(A)+1)$.
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If $K$ is closed in $X$ and $X$ is compact, then $K$ is compact.
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Proof:
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Let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $K$.
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> $A$ is open in $X$, if and only if $X\backslash A$ is closed in $X$.
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Since $X\backslash K$ is opened in $X$, $\{G_\alpha\}_{\alpha\in A}\cup \{X\backslash K\}$ is an open cover of $X$.
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Since $X$ is compact, there exists a finite subcover $\{G_{\alpha_1},\cdots,G_{\alpha_n},X\backslash K\}$ of $X$.
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Since $X\backslash K$ is not in the subcover, $\{G_{\alpha_1},\cdots,G_{\alpha_n}\}$ is a finite subcover of $K$.
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Therefore, $K$ is compact.
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## Cauchy criterion
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### In sequences
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Def: A sequence $\{a_n\}$ is Cauchy if for any $\epsilon>0$, there exists $N$ such that for any $m,n\geq N$, $|a_m-a_n|<\epsilon$.
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Theorem: In $\mathbb{R}$, every sequence is Cauchy if and only if it is convergent.
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### In series
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Let $s_n=\sum_{k=1}^{n} a_k$.
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Def: A series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $\{s_n\}$ converges.
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$\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
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$$
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|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
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$$
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## Comparison test
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If $|a_n|\leq b_n$ and $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty} a_n$ converges.
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Proof:
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Since $\sum_{n=1}^{\infty} b_n$ converges, $\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
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$$
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\left|\sum_{k=n+1}^{m} b_k\right|<\epsilon.
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$$
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By triangle inequality,
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$$
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\left|\sum_{k=n}^{m}a_k\right|\leq \sum_{k=n+1}^{m} |a_k|\leq \sum_{k=n+1}^{m} b_k<\epsilon.
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$$
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Therefore, $\forall \epsilon>0$, there exists $N$ such that for any $m>n\geq N$,
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$$
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|s_m-s_n|=\left|\sum_{k=n+1}^{m} a_k\right|<\epsilon.
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$$
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Therefore, $\{s_n\}$ is Cauchy, and $\sum_{n=1}^{\infty} a_n$ converges.
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