# 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](https://notenextra.trance-0.com/Math4111/Math4111_L6#theorem-212).

#### Definition: Uncountable

A set is uncountable if it is not countable.

#### Theorem: $\mathbb{R}$ is uncountable

Easy proof using [Cantor's diagonal argument](https://notenextra.trance-0.com/Math4111/Math4111_L6#theorem-214).

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