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Lecture 13

Review

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

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.
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.