NoteNextra-origin/content/Math4121/Math4121_L16.md

Math4121 Lecture 16

Continue on Patches for Riemann Integrals

Harnack's Mistake

Theorem 3.6 Heine-Borel Theorem

If \{U_\alpha\}_{\alpha \in A} is a collection of open sets which cover [a, b], then there exists a finite subcover, i.e. \{U_{\alpha_1}, U_{\alpha_2}, \cdots, U_{\alpha_n}\} such that [a, b] \subseteq \bigcup_{i=1}^n U_{\alpha_i}.

Using the fact that [a, b] is compact.

Proof:

Define S = \{x \in [a, b] : \exists U_\alpha \text{ s.t. } [a, x] \subseteq \bigcup_{i=1}^n U_{\alpha_i}\}.

If we can show that b\in S, then we are done.

Clearly S\neq \emptyset since a\in S.

Let \beta=\sup S\geq a. If we can show that \beta \geq b, then we are done. (taking x=b to take finite subcover)

Suppose toward contradiction that \beta < b.

Then \exists U_\alpha and (c,d)\subseteq U_\alpha such that a<c<\beta<d<b.

Since c<\beta, \exists U_{\alpha_1},\ldots,U_{\alpha_n} s.t. [a,x]\subseteq \bigcup_{i=1}^n U_{\alpha_i}.

So [a,d]\subseteq U_\alpha\cup \bigcup_{i=1}^n U_{\alpha_i}.

So d\in S, this contradicts the definition of \beta as the supremum of S.

So \beta \geq b.

QED

Reviewing sections for Math 4111

Definition: Cardinality

Two sets A and B have the same cardinality if there exists a bijection f:A\to B.

We can imaging the cardinality as the number of elements in the set. However, this is not rigorous, for example, the set of rational and irrational numbers are both infinite, but the set of rational numbers are countable while the set of irrational numbers are uncountable.

• the cardinality of a finite set is the number of elements in the set.
• the cardinality of \mathbb{N} is \aleph_0, such a set is called "countable".

Definition: Countable

A set is countable if there exists an injection f:\mathbb{N}\to A.

Example:

1. The set of integers \mathbb{Z} is countable.

   We can construct a bijection f:\mathbb{N}\to \mathbb{Z} by mapping $f(n)=\begin{cases} \frac{n}{2} & \text{if } n \text{ is even} \ -\frac{n+1}{2} & \text{if } n \text{ is odd} \end{cases}$

2. The set of rational numbers \mathbb{Q} is countable.

   We can construct a bijection f:\mathbb{N}\to \mathbb{Q} using the

3. The algebraic numbers \mathbb{A} (roots of polynomials with integer coefficients) are countable.

   Proof using the union of countable sets is countable.

Definition: Uncountable

A set is uncountable if it is not countable.

Theorem: \mathbb{R} is uncountable

Easy proof using Cantor's diagonal argument.

A new one

Proof:

Suppose \mathbb{R}=\{r_1,r_2,\ldots\}. Let a_0<b_0 be the first two entries a_0,b_0\in \{r_1,r_2,\ldots\} in our list which lie in (0,1).

Let a_1<b_1 be the first two entries such that a_1,b_1\in (a_0,b_0).

Continue this process, we can construct [0,1]\supsetneq [a_0,b_0]\supsetneq [a_1,b_1]\supsetneq \cdots sets.

By the nested interval theorem of real numbers, \bigcap_{n=1}^\infty [a_n,b_n]\neq \emptyset.

Setting r=\sup a_n (by the least upper bound property of real numbers), r\in [a_n,b_n] for all n\in\mathbb{N}. Since r\in \mathbb{R}, r=r_m for some m\in\mathbb{N}. Then find a_n,b_n which come after r_m in the list.

This contradicts the assumption that a_n,b_n as the first element in the list.

QED