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Lecture 2

Exterior, Interior, Boundary

Let \(A\subset \R^n\) and \(x\in \R^n\)

  1. \(\exists U_x^\varepsilon\subset A\implies x\) is an interior point of \(A\)

    \(\text{Int}A\) is the set of all interior points

  2. \(\exists U_x^\varepsilon\subset \R^n\setminus A\,(U_x^\varepsilon\cap A=\emptyset)\implies x\) is an exterior point of \(A\)

    \(\text{Ext}A\) is the set of all interior points

  3. \(\forall U_x^\varepsilon\cap A\neq \empty,U_x^\varepsilon\cap (\R^n\setminus A)\neq \empty\implies x\) is an boundary point of \(A\)

    \(\partial A\) is the set of all interior points

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Property

\(\partial(\R^{n}\setminus A)=\partial A,\,\text{Int}(\R^n\setminus A)=\text{Ext} A,\,\text{Ext}(\R^n\setminus A)=\text{Int} A\)

\(\text{Int}A\) is open, \(\text{Ext}A\) is open and \(\partial A=\R^n\setminus (\text{Int}A\cup \text{Ext}A)\)

Example

\(A=(a,b]\), then \(\text{Int}A=(a,b),\,\partial A=\{a,b\}\)

Definition

\(\bar A=A\cup \partial A=\text{Int}A\cup \partial A\) is the closure of \(A\)

Proposition

\(A\) is closed\(\iff A=\bar A\iff A\supset \partial A\)

\(A\) is open\(\iff A=\text{Int}A\)

Compactness

Definitions

Given \(A\subset \R^{n}\)

  • A collection ball of open sets is called an open caves of \(A\iff\forall x\in A\) is in some set from ball (ball covers \(A\))

image - \(A\) is compact\(\iff\)every open caves of \(A\) contains a finite sub-collection which also covers \(A\) (Any open caves of \(A\) contains a finite subcaves) - Given collection ball 1 and collection ball 2, they are collections (sets) of subsets and collection ball 1 \(\subset\) collection ball 2 (collection ball 1 is a sub-collection of collection ball 2)

In this case, collection ball 1 is a subcave of collection ball 2 if they both cover \(A\) - \(A\) is bounded\(\iff \exists M>0:A\subset U_0^\varepsilon\)

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Theorem 1

\(A\subset \R^n\) is compact\(\implies A\) is bounded and closed

Proof

  1. \(A\) is compact \(\implies\)\(A\) is bounded

    Consider collection of balls \(O=\{U_0^k\subset \R^n:k=1,2,3,...\}\)

    Since \(A\) is compact, then there exists finite \(k\) s.t. \(O\) covers \(\R^n\) and \(A\), then it's bounded

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  2. \(A\) is not closed\(\implies\)\(A\) is not compact

    Since \(A\) is not closed, then there exists \(x\in\partial A\) s.t. \(x\notin A\)

    \(U= \{ \text{Ext}(U_{x}^{1/k}) \mid k=1,2,3,... \}\), then \(\bigcup_{k=1}^{\infty}\text{Ext}(U_{x}^{1/k}) = \mathbb{R}^{n} \setminus \{x\} \supset A\)

    \(U\) does not contain a finite subcave because any \(U_x^{1/k}\) contains a point from \(A\)

    imageimage

Heine-Bael Theorem

A closed interval \([a,b]\subset \R\) is compact

Proof

Let \(O\) be an open cave of \([a,b]\)

Let \(A=\{x\in [a,b]:[a,x]\text{ is covered by a finite subcollection of }O\}\)

\(A\) is bounded and \(A\neq \empty\) since \(a\in A\) and a point is compact

Then \(\exists\)​\(\alpha\in[a,b]\) which is the least upper bound of \(A\)

Then to prove \(b\in A\), we need to prove

  1. \(\alpha \in A\)

    \(\exists U \in O: \alpha\in U\), then \(\exists x\in A\cap U\), then \([a, x]\) has a finite subcover of \(O\) (adding \(U\))

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    Then \([a, \alpha]\) has a finite subcover of \(O\) since \(\alpha\in U\)

  2. \(\alpha = b\)

    Suppose \(\alpha < b\), then \(\exists y \in U: \alpha < y < b\). Since \(\alpha\in A\), then \([a, \alpha]\) has a finite subcave

    Then we add \(U\), \([a,y]\) has a finite subcave, then \(y\in A\). Contradiction.

Theorem

\(A\subset \R^n,\,B\subset \R^n\) are compact\(\implies A\times B\subset\R^{m+n}\) is compact

Proof

Let \(O\) be a special open cave of \(A \times B\): \(\forall W \in O\,, W = U \times V\) where \(U, V\) are open

\(x \times B\)
Since \(B\) is compact, then \(\exists\)​\(\{V_1, V_2, \ldots\}\) covers \(B\) where \(\{V_1, V_2, \ldots, V_n\}\) is a finite subcover
Consider \(U_x = U_1 \cap U_2 \cap \ldots \cap U_n\), then \(\forall x \in A ,\exists U_{x}\) covers \(x\) since \(A\) is compact
Then \(U_x \times B\) is covered by a finite subcollection of \(O\)

Since \(\{U_x | x \in A\}\) is an open cave of \(A\), then it contains a finite subcave \(U_{x_1}, U_{x_2}, ..., U_{x_m}\)
Then \(U_{x_1}\cup U_{x_2}\cup ... \cup U_{x_m}\supset A\)
Then \(U_{x_j} \times B\) is covered by a finite subcollection of \(\mathcal{O}\)

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Theorem 2

\(A\) is bounded and closed\(\implies\)​\(A\) is compact

Proof

  1. A bounded set is a subset of the direct product of \(n\) closed intervals \(I^{n}=[a_{1},b_{1}]\times ...\times [a_{n},b_{n}]\subset \R^{n}\) (\(n\)-dimensional rectangle)

    We are allowed to do that because image where \(a_1, a_2, \ldots, a_n = -M\) and \(b_1, b_2, \ldots, b_n = +M\)

  2. \(I^n\) is compact(follows from 1 and 2 )

  3. Let \(A\subset B\subset \R^n\), \(B\) is compact, \(A\) is closed\(\implies A\) is compact

    Proposition: A closed subset of a compact set is compact

    Proof

    Let \(O\) is an open caves of \(A\), then \(\mathcal{O}=O\cup \{\R^n\setminus A\}\) is an open caves of \(\R^n\) and \(B\)

    \(\mathcal{O}\) contains finite subcaves of \(B\) consisting of \(\underbrace{U_1,U_2,...,U_N}_{O},\R^n\setminus A\)

    Then \(B\subset U_1\cup ...\cup U_N\cup (\R^n\setminus A)\), then since \(A\subset B\), then \(A\subset U_1\cup ...\cup U_N\)

    Then \(\{U_1,...,U_N\}\) is a finite subcaves of \(A\)

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