Lecture 13

Review

Consider the metric space $X=\mathbb{R}$ (with the usual metric $d(x,y)=|x-y|$). Let $E=(0,1)$.

1. Find several examples of sets $Y\subset \mathbb{R}$ such that $E\subset Y$ and $E$ is closed in $Y$.
   Example:

   1. $Y=E$, $E$ is closed in $Y$.
      We can prove this using normal ways, or Theorem 2.23 $E$ is closed in $Y\iff E^c$ is open in $Y$
      $Y\iff E^c=\phi$ and it's open.
   2. $Y=\mathbb{R}\backslash\{0,1\}$
      $Y\backslash E=(-\infty,0)\cup (1,\infty)$ Theorem 2.30 $E\subset Y\subset X$, $E$ is open in $Y\iff$ $\exists G$ open in $X$ such that $G\cap Y=E$
      $G\cap Y=Y\backslash E$ And we know $Y\backslash E$ is open in $Y$. By Theorem 2.23 $E$ is closed in $Y$.

2. If $Y$ is as in part 1, we can conclude that $E$ is closed and bounded in $y$. Part of Theorem 2.41 says: "If a set is closed and bounded, then it is compact." Why doesn't that theorem apply here.
   The set is not closed in $\mathbb{R}^k$.

New stuffs

Connected sets

Definition 2.45

$A,B\subset X$, we say $A$ and $B$ are separated in $X$ if $A\cup \overline{B}=\phi$ and $\overline{A}\cup B=\phi$

• $E\subset X$ disconnected in $X$ if $\exists$ nonempty separated $A,B\subset X$ such that $E=A\cup B$
• $E\subset X$ is connected in $X$ if it is not disconnected.

Example 2.46

$(0,1),(1,2)$ are separated [so $(0,1)\cup (1,2)$ is disconnected]

$[0,1],(1,2)$ are not separated [so $[0,1]\cup (1,2)=\{1\}$] So this doesn't tell us where $[0,1]\cup (1,2)$ is connected or not.

Theorem 2.47

Suppose $E\subset \mathbb{R}$

$E$ is connected $\iff \forall (x,y,z)$ with $x,y\in E,x<z<y$ such that $z\in E$.

By negating, this is equivalent to

$E$ is disconnected $\iff \exists (x,y,z)$ with $x,y\in E,x<z<y$ such that $z\notin E$.

Proof:

$\impliedby$

Suppose $\exists (x,y,z)$ with $x,y\in E,x<z<y$ such that $z\notin E$.

Let $A=(-\infty,z)\cap E,B=(z,\infty)\cap E$

Lemma: $E\subset F\implies \overline{E}\subset \overline{F}$

Since $A\subset (-\infty,z), \overline{A}\subset (-\infty,z]$

Since $\overline{A}\cap B=\phi$, similarly, $A\cap \overline{B}=\phi$. So $A,B$ are separated. Also they are non empty ($x\in A,y\in B$) and $E=A\cap B$. So $E$ is disconnected.

$\implies$

Suppose $\exists$ nonempty separated $A,B\subset X$ such that $E=A\cup B$.

Our goal is to find $x,y\in E,x<z<y$ such that $z\notin E$.

$A\neq \phi,B\neq \phi\implies \exists x\in A,y\in B$

Without loss of generality, assume $x<y$.

Let $w=sup(A\cup [x,y])$, $w\in \overline{A\cup [x,y]}$ (by Theorem 2.28) This implies $w\subset \overline{A}$ and $w\in [x,y]$

Since $A,B$ are separated, $w\notin B$ ($\overline{A}\cup B=\phi$).

Since $y\in B$, $w\in [x,y)$

Consider 2 cases,

Case 1. $w\notin A$.

let $z=w$, and $x,y,z$ satisfy the desired properties

Case 2. $w\in A$

Since $A,B$ are separated, $w\notin \overline{B}$ ($A\cup \overline{B}=\phi$).

Thus, $\exists r>0$ such that $(w-r,w+r)\cap B=\phi$.

Let $z=w+\frac{r}{2}$, then $x,y,z$ satisfy the desired properties.

EOP

Chapter 3: Numerical Sequences and Series

Numerical Sequences

Notations

Rudin use $\{p_n\}$ to denote a sequence $p_1,p_2$.

To avoid confusion with sets, we use $(p_n)_{n=1}^\infty$ or $(p_n)$

Definition 3.1

Let $(X,d)$ be a metric space. Let $(p_n)$ be a sequence in $X$.

Let $p\in X$. We say $(p_x)$ converges to $p$ if $\forall \varepsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(p_n,p)<\varepsilon$. ($p_n\in B_\varepsilon (p)$)

Notation $\lim_{n\to \infty} p_n=p$, $p_n\to p$

We say $(p_n)$ converges if $\exists p\in X$ such that $p_n\to p$.

i.e. $\exists p\in X$ such that $\forall\varepsilon>0,\exists N\in\mathbb{N}$ such that $\forall n\geq N,d(p_n,p)<\varepsilon$

We say $(p_n)$ diverges if $(p_n)$ doesn't converge.

i.e. $\forall p\in X$, $p_n\cancel{\to} p$

i.e. $\forall p\in X$ such that $\exists \varepsilon>0,\forall N\in\mathbb{N}$ such that $\exists n\geq N,d(p_n,p)\geq\varepsilon$

Definition 3.2

We say a sequence $(p_n)$ is bounded if $\exists x\in X$, $\forall r>0$ such that $\forall n\in \mathbb{N},p_n\in B_r(x)$

Example:

$X=\mathbb{C}$, $s_n=\frac{1}{n}$

Then $s_n\to 0$ i.e. $\forall \varepsilon>0 \exists N\in \mathbb{N}$ such that $\forall n\geq N$, $|s_n-0|<\varepsilon$.

Proof:

Let $\varepsilon >0$ (arbitrary)

Let $N\in \mathbb{N}$ be greater than $\frac{1}{\varepsilon}$ (by Archimedean property) e.g. $N=\frac{1}{\varepsilon}+1$ (we choose $N$)

Let $n\geq N$ (arbitrary)

Then $|s_n-q|=\frac{1}{n}\leq \frac{1}{N}\leq \varepsilon$

EOP