A linguist would be shocked to learn that if a set is not closed this does not
mean that it is open, or again that “E is dense in E” does not mean the
same thing as “E is dense in itself.”

Understanding these basic definition and statement~

  • We say $A \subset \mathbb{R}$ is open if for every $x \in A$ there exists $\varepsilon>0$ such that $(x-\varepsilon, x+\varepsilon) \subseteq A . A$ is closed if its complement $A^{c}$ is open. Similarly a set $A$ in a metric space $(M, d)$ is called open if for each $x \in A$, there exists an $\varepsilon>0$ such that $B(x ; \varepsilon) \subset A$. Here,
    B(x ; \varepsilon)={y \in M: d(x, y)<\varepsilon}
    is the $\varepsilon$-ball (also called $\varepsilon$-neighborhood) around $x$.
  • Let $A$ be a subset of a metric space $(M, d)$ and $x \in M$. We say $x$ is an accumulation point
    of $A$ if every open set $U$ containing $x$ contains some point $y \in A$ with $y \neq x$.
  • Let $A \subset(M, d)$. We say $x \in M$ is a limit point of a set $A$ provided $U \cap A \neq \emptyset$ for every neighborhood $U$ of $x$.
  • A set $A$ is closed in a metric space $(M, d)$ if and only if the accumulation points of $A$ belong to $A$ and we set $\bar{A}:=A \cup{x \in M: x$ is an accumulation point of $A}$.

A subset $A$ in a metric space $M$ is called compact if one of the following equivalent conditions is satisfied:

a) Every open cover of $A$ has a finite subcover.
b) Every sequence in $A$ has a convergent subsequence converging to a point in $A$ (sequential compactness).

Furthermore, if $A \in \mathbb{R}^{n}$, the above two conditions are equivalent to saying that $A$ is closed and bounded (Heine-Borel Theorem).

  • A metric space $(M, d)$ is called totally bounded if for each $\varepsilon>0$ there is a finite set $\left{x_{1}, x_{2}, \cdot, x_{k}\right}$ in $M$ such that
    A \subset \bigcup_{i=1}^{k} B\left(x_{i} ; \epsilon\right)
  • A metric space $(M, d)$ is compact if and only if $M$ is complete and totally bounded.
    A subset $A$ of a metric space $(M, d)$ is called nol connecled if there are disjoint open sets $U$ and $V$ such that $A \subseteq U \cup V$ and $A \cap U \neq \emptyset \neq A \cap V .$ Otherwise, the set $A$ is said to be connected.
  • A subset $A$ of a metric space $(M, d)$ is said to be path connected if for each pair of points $x$ and $y$ in $A$, there is a path in $A$ connecting $x$ to $y$, i.e., there is a continuous function $\psi:[0,1] \rightarrow A$ such that $\psi(0)=x$ and $\psi(1)=y$
  • Let $M$ be a metric space and $A$ be a subset of $M$, and $f: A \rightarrow \mathbb{R}$ be a continuous function. Suppose $B \subset A$ is connected and $x, y \in B$. Then for every real number $c$ such that $f(x)<$ $c<f(y)$, there exists a point $z \in B$ such that $f(z)=c$ Notice that, since intervals (open or closed) are connected, the above statement is a generalized version of the Intermediate Value Theorem given for intervals.

Try the following problem to test yourself!

Problem 1.

Let $S_{1}=\left[0, \frac{1}{3}\right] \cup\left[\frac{2}{3}, 1\right]$ be obtained from $[0,1]$ by removing the middle third $\left(\frac{1}{3}, \frac{2}{3}\right) .$ Repeat the process to obtain $S_{2}=\left[0, \frac{1}{9}\right] \cup\left[\frac{2}{9}, \frac{1}{3}\right] \cup\left[\frac{2}{3}, \frac{7}{9}\right] \cup\left[\frac{8}{9}, 1\right] .$ In general, $S_{n+1}$ is obtained
from $S_{n}$ by removing the middle third of each interval in $S_{n} .$ Let $C=\bigcap_{n \geq 1} S_{n}$, also known as the Cantor set. Prove that

  1. $C$ is compact.
  2. $\operatorname{int}(C)=\emptyset$.
  3. $C$ has infinitely many points.
  4. The total length of the intervals removed is equal to 1 .

Proof .

  1. Notice that $C \subset[0,1]$, so it is bounded. $C=\bigcap_{n \geq 1} S_{n}$, where each $S_{n}$ is a union of finitely many closed intervals and so is closed. Thus, the set $C$ is an intersection of closed sets and so $C$ is closed as well. Therefore, $C$ is a closed and bounded subset of $\mathbb{R}$, so $C$ is compact.
  2. The length of each of the subintervals making up the set $S_{n}$ is $\frac{1}{3^{n}}$, so the intersection can contain no interval longer than this. Since $\frac{1}{3^{n}} \rightarrow 0$ as $n \rightarrow \infty$, the intersection can contain no interval of positive length. If $a \in C$ is an interior point, we should be able to find an interval around $a$ with positive length $s .$ There are no such intervals, so int $(C)=\emptyset$.
  3. Begin by noting that $C$ contains the endpoints of all the intervals for each $S_{n}$ and that each $S_{n}$ has a total of $2^{n+1}$ endpoints. Since the number of endpoints in $S_{n}$ goes to $\infty$ as $n \rightarrow \infty$, we have that $C$ has infinitely many points.
  4. By summing up the length of the deleted intervals, we are able to obtain:
    \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\cdots=\frac{1}{3}\left[1+\frac{2}{3}+\left(\frac{2}{3}\right)^{2}+\cdots\right]=\frac{1}{3} \cdot \frac{1}{1-\frac{2}{3}}=1

Problem 2.

Given a metric space $(X, d)$ and a nonempty bounded subset $A$, the real number $\delta(A):=\sup {d(x, y): x, y \in A}$ iProblem $\mathbf{1 0 . 5 3}$ Let $C$ be a closed subset of $\mathbb{R}$. Show that $C=P \cup F$, where $P$ is perfect, $F$ is countable, and $P \cap F=\emptyset$. This is known as the Cantor-Bendixson Theorem.

Proof .

Let $(X, d)$ be a discrete metric space where $X$ has more than one element. Then
S\left(x_{0} ; 1\right)=X \backslash\left{x_{0}\right} \quad \text { and } \quad \delta\left(S\left(x_{0} ; 1\right)\right)=1<2 \text {. } $$ For $r>0$ and $r \neq 1, S\left(x_{0} ; r\right)=\emptyset$ and $\delta\left(S\left(x_{0} ; r\right)\right)$ is not dLet $P$ be the set of all condensed points of $C .$ Set $F=C \backslash P .$ Then $C=P \cup F$. We have $P \cap F=\emptyset$. According to Problem $10.52, P$ is perfect, and $F$ is countable.

实分析代写|Fundamentals of Topology请认准UpriviateTA

复分析代写,数学代写Riemann surface请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

Fourier analysis代写



Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析