Date modified: Sun 2023-07-02 . 02:22 PM
Date created: Sat 2023-10-07 . 09:02 PM
# Compactness, and FTA. Compactness is an important property in analysis and topology. It is a sort of generalization of finiteness for infinite sets. Quite often there are mathematical statements that are either (1) trivially true for finite sets, (2) true for compact sets, and (3) false for noncompact sets. For example extreme value theorem. A paraphrasing of Weyl on compactness: "If a city is compact it can be guarded by a finite number of arbitrarily near-sighted policemen." This is a brief description of the topological definition of compactness: We say a subset $K$ is compact if every open cover $\{U_\lambda\}_{\lambda\in\Lambda}$ for $K$, there exists a finite subcover. Namely, there are some finitely many $U_{\lambda_1},\ldots,U_{\lambda_k}$ from the collection that will cover $K$. ![[---images/---assets/---icons/question-icon.svg]] Show if $f:K\to X$ is a continuous function from a compact set to any topological space $X$, then the image $f(K)$ is compact. ![[---images/---assets/---icons/question-icon.svg]] Show that a compact set in any metric space is closed and bounded. ![[---images/---assets/---icons/question-icon.svg]] Show that there exists a metric space where a set being closed and bounded need not be compact. ![[---images/---assets/---icons/question-icon.svg]] In Euclidean space $\mathbb{R}^n$, we have **Heine-Borel theorem**, which asserts the equivalence of a set being compact and a set being closed and bounded. Try recalling its proof for $\mathbb{R}$. ![[---images/---assets/---icons/question-icon.svg]] Show that if $f:K\to\mathbb{R}$ is a continuous real-valued function on a compact set $K$, then $f$ attains a minimum in $K$, as well as a maximum in $K$. This is **extreme value theorem**. ![[---images/---assets/---icons/question-icon.svg]] Show that real or complex polynomial $p$ attains a global minimum on some closed disk $D$ centered at $0$ with radius $R > 0$, for some large enough $R$. ![[---images/---assets/---icons/question-icon.svg]] Show the function $f:T\to \mathbb{R}$ given by $f(x,y)=x^2 + y^2 + \frac{1}{|x-y|}$ attains a global minimum somewhere on the set $T=\{(x,y):x\neq y\}$. Hint: Cut out big behaviors parts. When d'Alembert produced his proof of the fundamental theorem of algebra, it was essentially correct, however at the time it was not known rigorously why a polynomial would necessarily attain a minimum on a large enough compact disk. But now that we know this, we can prove FTA in a first principle manner! Here is the sketch of the idea: Given polynomial $p(z)$, we know the global minimum of $|p(z)|$ would occur in some large enough closed disk $D_R$ of radius $R > 0$. Say this occurs at some $z_0$. Now suppose to the contrary that $p(z_0) \neq 0$, we shall produce some $z$ such that $|p(z)| < |p(z_0)|$, contradicting minimality of $|p(z_0)|$. To see this, note, by centering the polynomial at $z_0$, we can rewrite $p(z)=a_n(z-z_0)^n+\cdots+a_k(z-z_0)^k + a_0$, where $a_k,a_0$ both not zero, and $k$ is the least positive integer such that $a_k \neq 0$. This term $a_k(z-z_0)^k$ would dominator the behavior of $p$ when we are near $z_0$. So we should try to wiggle $z$ in such a way that $|p(z)| < |p(z_0)|=|a_0|$ ! Here are the details. ![[---images/---assets/---icons/question-icon.svg]] Recall by de Moivre theorem, for every nonzero complex number $w = r(\cos(t)+i\sin(t))$, we can find its $k$-th root, given by $$ \sqrt[k]{r}\left( \cos\left( \frac{t+2\pi m}{k} \right) +i\sin\left( \frac{t+2\pi m}{k} \right) \right) $$where $m=0,1,2,\ldots,k-1$. ![[---images/---assets/---icons/question-icon.svg]] Consider complex polynomial $p(z)$ such that $|p(z)|$ attains global minimum at $z_0$, Suppose that $p(z_0)=a\neq0$. Show the transformed polynomial $\tilde p(z) = \frac{1}{a} p(z+z_0)$ such that $|\tilde p(z)|$ attains minimum at $z=0$, with $\tilde p(0)=1$. Furthermore, if we write $\tilde p(z)=\cdots+b z^k+ 1$ where $k$ is the least positive integer such that the term $z^k$ has nonzero coefficient $b\neq0$, we can let $\lambda$ to be any $k$-th root of $-\frac{1}{b}$ and consider the polynomial $q(z)=\tilde p(\lambda z)=\frac{1}{a}p(\lambda z +z_0) = \cdots - z^k + 1$. Show this $q(z)$ also have $|q(z)|$ attaining a global minimum at $z=0$ with $q(0)=1$. ![[---images/---assets/---icons/question-icon.svg]] By considering small positive real inputs $z$, show you can make $|q(z)| < 1$. Hence proving fundamental theorem of algebra!