1.6. Five Big Theorems
Compact Set. A nonempty subset $C \subset \R^n$ is compact if it is closed and bounded.
BolzanoWeierstrass Theorem
Theorem. A compact set $C \subset \R^n$ contains a sequence $i \mapsto x_i$, ifandonlyif that sequence has a convergent subsequence $j \mapsto (x_i)_j$ whose limit is in $C$.
If $i \mapsto x_i$ is a sequence in $C$, then it is bounded so it has a convergent subsequence, and it is closed so the limit is in $C$.
If $C$ is unbounded, it contains a sequence $i \mapsto x_i \to \infty$, and that sequence has no convergent subsequence. There exists a sequence $i \mapsto x_i$ which converges in $\R^n$ to a point not in $C$.
Existence of Maxima and Minima
Theorem. Let $X \subset \R^n$ be compact, and $f: X \to \R$ be continuous. There exists $x_\circ \in X$ such that $\forall x \in X, \space f(x) \leq f(x_\circ)$. In particular, $f$ is bounded.
Proof. We use proof by contradiction.
 Say that $f$ is unbounded. There exists a sequence $n \mapsto x_n \in X$ such that $f(x_n) > n$.
 There exists a subsequence $(x_n)_i$ which converges to a point $y$ of $X$.

Since $f$ is continuous at $y$, $\forall \varepsilon > 0, \exists \space \delta > 0$ such that $ zy < \delta \implies f(z)  f(y) < \varepsilon$. 
Take $\varepsilon = 1$. Find such a $\delta$. $ zy < \delta \implies f(z) < 1 + f(y)$. Since $n \mapsto x_n$ converges to $y$, for $i$ sufficiently large, $ (x_n)_i  y < \delta$. If $n_i > f(y) + 1$, then this is a contradiction. It is both smaller and bigger than $f(y) + 1$.  Now that we know $f$ is bounded, let $M = \sup f(x)$. There must exist a sequence $n \mapsto x_n$ such that $\lim_{n \to \infty}f(x_n) = M$
 Since $X$ is compact, there is a convergent subsequence $i \mapsto (x_n)_i$ converging to $y \in X$.

$\forall \varepsilon > 0, \space \exists \delta > 0$ such that $ zy < \delta \implies f(z)  f(y) < \varepsilon$. $\exists I$ such that $i > I \implies (x_n)_iy < \delta \implies f((x_n)_i)  f(y) < \varepsilon$.  $\lim_{i \to \infty} f((x_n)_i) =f(y)= M$ because $f$ is continuous.
Mean Value Theorem
Theorem. If $f: [a,b] \to \R$ is differentiable, then there exists $c \in (a,b)$ such that $f’(c) = \frac{f(b)  f(a)}{ba}$.
Example:
\[f(x) = \begin{cases}x^2 \sin\frac{1}{x} & \text{if }x \neq 0 \\ 0 & \text{if }x = 0 \end{cases}\]Taking the derivative,
\[f'(x) = \begin{cases}2x\sin\frac{1}{x} \cos\frac{1}{x} & \text{if } x \neq 0\\0 & \text{if } x =0\end{cases}\]This is terribly discontinuous.
If a derivative of a differentiable singlevariable function is discontinuous, it will always look erratically sinusoidal. Point discontinuities will never exist.
Proof.
Lemma. If $f$ is differentiable at $x \in \R$ and has a local maximum or minimum at $x$, then $f’(x) = 0$.
Proof of Lemma.
\[f'(x) = \lim_{h \to 0}\frac{f(x+h)  f(h)}{h} \text{ if } h \neq 0\]If $h > 0$, then
\[\lim_{h \to 0}\frac{f(x+h)  f(h)}{h} \leq 0\]If $h < 0$, then
\[\lim_{h \to 0}\frac{f(x+h)  f(h)}{h} \geq 0\]The limit can be a limit of both positive and negative numbers only if $f’(x) = 0$.
Let $g(x) = f(x)  \frac{f(b)  f(a)}{ba}(xa)f(a)$. Because $f$ is continuous, $g$ is continuous. $g(a) = g(b) =0$. $g$ achieves its maximum or minimum at a point of $c \in (a,b)$. By the lemma, $g’(c) = 0 = f’(c)  \frac{f(b)f(a)}{ba}$. The theorem is proved.
The Fundamental Theorem of Algebra
Theorem. If $p(z)$ is a polynomial of degree $d \geq 1$, then there exists $z_\circ \in \Complex$ such that $P(z_\circ)=0$.
D ‘Alembert, Lagrange, Gauss all tried to prove it. There exists several proofs, all of which use topology somewhere. The general idea is that if $  z  $ is big, big powers dominate small powers. If $  z  $ is small, small powers dominate big powers. $10^3 » 10$ and $\frac{1}{10}^3 « \frac{1}{10}$. 
Proof. $z\mapsto  p(z)  $ is continuous because $p$ is a polynomial. 

There exists $R$ such that $\inf_{ z \leq R} p(z) = \inf_{z\in \Complex} p(z) $. In particular, since ${z \space \space z \leq R}$ is compact, there exists $z_0$ such that $\forall z \in \Complex, p(z_0) \leq {p(z)} $. This is also true for real polynomials. 
For some $R$, suppose $ z > R$, and $R > 1$. By the triangle inequality,
If $(z) > R =  a_{n=1}  +\dots  a_1  +  a_0  $, then $  p(z)  > a_0 = p(0)$. So $\inf_{  z  > R} p(z) > p(0) > \inf_{  z  \leq R}  p(z)  $. 
Therefore, there exists $z_0 \in \Complex$, with $  z_0  \leq R$ such that $\forall z \in \Complex,  p(z_0) \leq  p(z)  $. 
The polynomial achieves its global minimum on a compact set ${z\in \Complex \space  \space  z  \leq R}$. 
Dog on a Leash
$p(z_0) = 0$. $p(z_0+u) = q(u)$. $q$ is a polynomial of degree $n$.
For example, suppose
\[p(z) = z^2 + z + 1, \space z_0 =1.\\\space p(z_0+u) = (1+u)^2 + 1 + u + 1 = \underbrace{u^2 +3u + 3}_{q(u)}\]We set up our analogy as follows,
\[q(u) = \underbrace{\underbrace{u^k+\dots}_{\text{leash}}+\underbrace{+b_{j+1}u^{j+1}+b_ju^j+\underbrace{b_0}_{\text{flag pole}}}_{\text{man}}}_{\text{dog}}\]Here, $b_0 = p(z_0)$. We need to show that $b_0 = 0$. By contradiction, suppose $b_0 \neq 0$, I will find $u$ such that $  q(u)  <  b_0  $. When $u$ goes in a circle, $q(u)$ also moves in a circle. The man takes a walk of radius $\rho =  u  $ around $b_0$. At some point during this walk, the man is exactly between the origin and the flag pole. 
If I choose $\rho>0$ sufficiently small, then there is an instance when the leash is shorter than the distance ($  b_j  \rho^j$) from the man to the flagpole. 
Proof. Let $\rho =  u  $. Set $B = \max{  b_{j+1}  ,\dots,b  _k  }$ and choose $\rho$ satisfying 
If $\rho < 1$,
\[\leq b_{k+1}\rho^{k+1}+b_{k+2}\rho^{k+1} + \dots + \rho^{k+1}\\ \leq \rho^{k+1}(b_{k+1}+b_{k+2}+\dots1)\\ \stackrel{?}{<}\rho^kb_k\\ b_k > \rho(b_{k+1}+\dots + 1)\\ \rho<\frac{b_k}{b_{k+1}+\dots + 1}\]Second Part of the Fundamental Theorem of Algebra
For $p(z)$, there exists $x_1,\dots, x_n \in \Complex$ such that $p(z) = (zx_1)(zx_2)\dots (zx_n)$.
Suppose $p(x_1) = 0$.
\[p(z) = q(z)(zx_1)+r(z)\text{, and }\deg r < 1\\ 0=p(x_1)=q(x_1)(0)+r(x_1) \text{ so } r= 0\\ \deg q(z) = n1 \space \space \space q(z)=(zx_2)\dots(zx_n)\]Inductively, we can find each root of $p(z)$.
Gauss’s Proof
Complex numbers were not accepted at the time! Every real polynomial can be factored as a product of linear and quadratic factors.
If $p$ is a real polynomial and $p(a+ib) = 0$, then $p(aib) = 0$, so $p(z)$ is divisible by $(z(a+ib))(z(aib)) = z^22az+(a^2+b^2)$.