1 - Ordered Fields, Euclidean Spaces, Cardinality.

Field

A set $F$ with two operators $+, \cdot$ is a Field if:

$(F, +)$ is an Abelian group, whose identity we denote as $0$ .
$(F - {0}, \cdot)$ is an Abelian group, whose identity we denote as $1$ .
There is a distributive law of $\cdot$ over $+$ . That is, $x (y + z) = x y + x z$ for all $x, y, z \in F$ .

About Fields

There are a lot of things to say about fields. We won't say a lot here. We'll rather take our familiarity with real numbers for granted, and capture some important properties.

Ordered Field

An ordered set $F$ that is also a field is an ordered field if:

$x < y$ implies $z + y < z + x$ for any $z \in F$ .
if $x > 0$ and $y > 0$ then $x y > 0$ .

Informal

Essentially, an ordered field is one such that order plays nice with multiplication and addition in a familiar way.
The following statements are true in every ordered field. We will not prove these.

If $x > 0$ then $- x < 0$ , and vice versa.
If $x > 0$ and $y < z$ then $x y < x z$ .
If $x < 0$ and $y < z$ then $x y > x z$ .
If $x \neq 0$ then $x^{2} > 0$ . In particular, $1 > 0$ .
If $0 < x < y$ then $0 < 1 / y < 1 / x$ .

Properties of Reals:

Intuition for the Archimedean property

Pick any real $x$ , and any positive real $y$ , it's clear that at at some point, the natural number multiplies of $y$ , $N y = {n y : n \in N}$ get larger than $x$ . say $n_{0}$ is the smallest number for which $n_{0} y > x$ <ns0:svg xmlns:ns0="http://www.w3.org/2000/svg" x="0" y="0" width="661.45458984375" height="301.4545593261719" style=" width:661.45458984375px; height:301.4545593261719px; background: transparent; fill: none; "> <ns0:svg class="role-diagram-draw-area"><ns0:g class="shapes-region" style="stroke: black; fill: none;"><ns0:g class="arrow-line"><ns0:path class="connection real" stroke-dasharray="" d=" M74,95 L594.82,95" style="stroke: rgb(0, 0, 0); stroke-width: 1; fill: none; fill-opacity: 1;" /><ns0:g stroke="#11111b" transform="matrix(-1,1.2246467991473532e-16,-1.2246467991473532e-16,-1,596.8181762695312,95.00000000000001)" style="stroke: rgb(0, 0, 0); stroke-width: 1;"><ns0:path d=" M10.93,-3.29 Q4.96,-0.45 0,0 Q4.96,0.45 10.93,3.29" /></ns0:g><ns0:g 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Otherwise, as seen in the bold white diagram above, the set $N y$ accumulates in between $y$ and $x$ , and $x$ acts as an upper bound for $N y$ , which means that $N y$ has a supremum in $R$ , which raises an issue:

Archimedean property

For any $x \in R$ and $y > 0 \in R$ , the set $N y > x = {n y > x : n \in N}$ is non empty.

$* p r o o f * :$ For the sake of contradiction, suppose $N y > x$ is empty, which means that for any element $u$ of the set $N y = {n y : n \in N}$ , $u \leq x$ . Hence, for example $x + 1$ is an upperbound for $N y$ . Therefore, there exists $α = sup N y \in R$ . But since $y > 0$ , we have $α - y < α$ . Which means that $α - y$ is not an upper bound of $N y$ . so we have some $m \in N$ such that $m y \geq α - y$ , allow $(m + 1) y > α - y$ . Then, $(m + 2) y > α$ . Which is a contradiction because $(m + 2) y$ is an element of $N y$ , which is greater than the supremum $α$ of $N y$ .

$◻$

Any POSITIVE real number is sandwiched between two consecutive integers. <ns0:svg xmlns:ns0="http://www.w3.org/2000/svg" x="0" y="0" width="661.45458984375" height="185.9091033935547" style=" width:661.45458984375px; height:185.9091033935547px; background: transparent; fill: none; "> <ns0:svg class="role-diagram-draw-area"><ns0:g class="shapes-region" style="stroke: black; fill: none;"><ns0:g class="arrow-line"><ns0:path class="connection real" stroke-dasharray="" d=" M75,95 L593.82,95" style="stroke: rgb(0, 0, 0); stroke-width: 2; fill: none; fill-opacity: 1;" /><ns0:g stroke="#11111b" transform="matrix(-1,1.2246467991473532e-16,-1.2246467991473532e-16,-1,596.8181762695312,95.00000000000001)" style="stroke: rgb(0, 0, 0); stroke-width: 2;"><ns0:path d=" M14.21,-4.28 Q6.45,-0.59 0,0 Q6.45,0.59 14.21,4.28" /></ns0:g><ns0:g stroke="#11111b" transform="matrix(1,0,0,1,72,95)" style="stroke: rgb(0, 0, 0); stroke-width: 2;"><ns0:path d=" M14.21,-4.28 Q6.45,-0.59 0,0 Q6.45,0.59 14.21,4.28" /></ns0:g></ns0:g><ns0:g class="arrow-line"><ns0:path class="connection real" stroke-dasharray="" d=" M315,80.82 L315,115" style="stroke: rgb(0, 0, 0); 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The idea is to choose $y = 1$ and apply the Archimedean property, which will tell us that the set $N > x = {n : n > x}$ is non empty, and then we allow $n_{0}$ to be the least element of $N > x$ by the well ordering property of $N$ .

Any real number is sandwiched between two consecutive integers

let $x \in R$ there exists $m \in Z$ such that $m - 1 \leq x < m$ .

$* p r o o f * :$ Let $x > 0$ , then by Archimedean property, choosing $y = 1$ , the set $N > x = {n : n > x}$ is non empty. Therefore, the well ordering property furnishes $n_{0}$ to be the least element of $N > x$ . which means that $n_{0} - 1 \notin N > x$ . Hence, $n_{0} - 1 \leq x < n_{0}$ .
if $x < 0$ then, we have $n_{0} - 1 \leq - x < n_{0}$ . Equivalently, $- n_{0} + 1 \geq x > - n_{0}$ setting $z = - n_{0} \in Z$ We have $z + 1 \geq x > z$ . In such a case, if $z + 1 > x$ , we are done, as $z + 1 > x \geq z$ . if $z + 1 = x$ , $z + 2 > x \geq z + 1$ , and we are done.
Finally if $x = 0$ , $0 \leq x < 1$ .

$◻$

Q

is dense in

R

let $x < y \in R$ . There exits $q \in Q$ such that $x < q < y$ .
$* p r o o f * :$ First notice that $(y - x) > 0$ . Hence using Archimedean property, we have $n \in N$ such that $n (y - x) > 1$ . Now using the fact that Any real number is sandwiched between two consecutive integers, we have $m - 1 \leq n x < m$ for some $m \in Z$ . Combining the two inequalities, we have $n y > 1 + n x \geq m$ hence $y > \frac{m}{n}$ . On the other hand, $x < \frac{m}{n}$ . Therefore $q = \frac{m}{n}$ is the desired $q$ for which $x < q < y$ .
$◻$

Existence of nth roots in

R

For each real $x > 0$ and natural number $n$ there is exactly one real $y > 0$ such that $y^{n} = x$

$* p r o o f * :$ WLOG assume that $0 < y_{1} < y_{2}$ then clearly, $y_1{#n} < y_2^n$ so at most one of $y_{1}^{n}$ and $y_{2}^{n}$ is equal to $x$ .
Now, consider the set $T = {t \in R : t^{n} < x}$ . First, notice that $\frac{x}{1 + x} \in T$ , due to the following inequality holding, when $0 < x < 1$ and $x > 1$ as well:

(\frac{x}{1 + x})^{n} < x ⟺ x^{n - 1} < (1 + x)^{n} $ $ . S o $ T $ i s n o n e m p t y . M o r e o v e r, $ 1 + x $ i s a n u p p e r b o u n d f o r $ T $ a s $ (1 + x)^{n} > x $ f o r a n y $ x > 0 $ . T h e r e f o r e, $ T $ h a s a s u p r e m u m, s e t $ y = sup T $ . * * * T h e b i g c l a i m h e r e i s t h a t $ y^{n} = x $ . * * * L e t $ y^{n} < x $ . s e t a l l o w $ 0 < h < 1 $, a n d $ h < \frac{x - y^{n}}{n (y + 1)^{n - 1}} $ . T h e i d e n t i t y $ b^{n} - a^{n} = (b - a) \sum_{i = 0}^{n} a^{i} b^{n - 1 - i} $ . T a k i n g i t f r o m t h e t o p, f o r $ b > a > 0 $, w e k n o w t h a t $ b^{n - 1} $ i s g o i n g t o b e l a r g e r t h a n a n y m u l t i p l i c a t i v e c o m b i n a t i o n $ a^{i} b^{n - 1 - i} $ . H e n c e w e o b t a i n t h e i n e q u a l i t y $ $ b^{n} - a^{n} < n b^{n - 1} (b - a)

now, setting $b = y + h$ , and $a = y$ , we first notice that

(y + h)^{n} - y^{n} < h n (y + h)^{n - 1} < h n (y + 1)^{n - 1}

the rightmost inequality owing to $0 < h < 1$ . Now using $h < \frac{x - y^{n}}{n (y + 1)^{n - 1}}$ , taking the only the ends of the inequality, we have

(y + h)^{n} - y^{n} < x - y^{n} ⟺ (y + h)^{n} < x

So clearly, $(y + h) \in T$ , but $y + h > y$ , contradicting the fact that $y$ is an upper bound for $T$ .

On the other hand, if $y > x^{n}$ , put $k = \frac{y^{n} - x}{n y^{n - 1}}$ . Here, $0 < k < y$ if $t \geq y - k$ , we have

y^{n} - t^{n} \leq y^{n} - (y - k)^{n} < k n y^{n - 1} = y^{n} - x

Hence, $t^{n} > x$ therefore $t \notin T$ , $y - k$ is an upper bound for $T$ . but $y - k < y$ hence contradicting the fact that $y$ is the least upper bound of $T$ .

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Extended reals:

We have two extra symbols $- \infty, + \infty$ , where each $x \in R$ has $- \infty < x < + \infty$ . We have the following definitions:

$x + \infty = + \infty$ , $x - \infty = - \infty$ , $0 = \frac{x}{+ \infty} = \frac{x}{- \infty}$
if $x > 0$ , $x \cdot + \infty = + \infty$ , $x \cdot - \infty = - \infty$ . if $x < 0$ the signs get switched.
Due the lack of definitions for additive and multiplicative inverses of $+ \infty, - \infty$ , the extended reals don't form a field, as defined above.

Complex field

the field of complex numbers $C$ is just $R^{2}$ with vector addition, and a special multiplication we are all familiar with. To show field structure is left for the reader.
if $z = a + i b \in C$ the conjugate of $z$ , $\bar{z} = a - i b$ which is a reflection about the real axis.
We know for example that the conjugate distributes over $+$ and $\cdot$ , and that the double conjugate is the identity transform.
The norm of a complex number is given by the positive square root of $z \bar{z}$ which is a positive real number.
The norm distributes over $\cdot$ and has a triangle inequality over $+$ , namely $| z + w | \leq | z | + | w |$ .

Complex Cauchy Schwarz inequality:

Cauchy Schwarz inequality

If $a_{1}, \dots, a_{n}$ and $b_{1}, \dots, b_{n}$ are complex numbers, then

\left|\sum_{j=1}{#n} a_j b_j\right|^2 \leq \sum_{j=1}^n\left|a_j\right|^2 \sum_{j=1}^n\left|b_j\right|^2

$* p r o o f * :$ Put $A = Σ {| a_{j} |}^{2}, B = Σ {| b_{j} |}^{2}, C = Σ a_{j} b_{j}$ (in all sums in this proof, $j$ runs over the values $1, \dots, n$ ). If $B = 0$ , then $b_{1} = \dots = b_{n} = 0$ , and the conclusion is trivial. Assume therefore that $B > 0$ . By Theorem 1.31 we have

\begin{aligned} \sum {| B a_{j} - C b_{j} |}^{2} & = \sum (B a_{j} - C b_{j}) (B {\bar{a}}_{j} - {\overset{―}{C b}}_{j}) \\ = B^{2} \sum {| a_{j} |}^{2} - B \bar{C} \sum a_{j} {\bar{b}}_{j} - B C \sum {\bar{a}}_{j} b_{j} + | C |^{2} \sum {| b_{j} |}^{2} \\ = B^{2} A - B | C |^{2} \\ = B (A B - | C |^{2}) \end{aligned}

Since each term in the first sum is nonnegative, we see that

B (A B - | C |^{2}) \geq 0

Since $B > 0$ , it follows that $A B - | C |^{2} \geq 0$ . This is the desired inequality.

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EUCLIDEAN SPACES

Euclidean spaces

For each positive integer $k$ , let $R^{k}$ be the set of all ordered $k$ -tuples

x = (x_{1}, x_{2}, \dots, x_{k})

where $x_{i}, \dots, x_{k}$ are real numbers, called the coordinates of $x$ . The elements of $R^{k}$ are called points, or vectors, especially when $k > 1$ . We shall denote vectors by boldfaced letters. If $y = (y_{1}, \dots, y_{k})$ and if $α$ is a real number, put

$\begin{aligned} x + y & = (x_{1} + y_{1}, \dots, x_{k} + y_{k}) \\ α x & = (α x_{1}, \dots, α x_{k}) \end{aligned}$
so that $x + y \in R^{k}$ and $α x \in R^{k}$ . This defines addition of vectors, as well as multiplication of a vector by a real number (a scalar). These two operations satisfy the commutative, associative, and distributive laws (the proof is trivial, in view of the analogous laws for the real numbers) and make $R^{k}$ into a vector space over the real field. The zero element of $R^{k}$ (sometimes called the origin or the null vector) is the point 0 , all of whose coordinates are 0.
We also define the so-called "inner product" (or scalar product) of $x$ and y by
$x \cdot y = \sum_{i = 1}^{k} x_{i} y_{i}$
and the norm of $x$ by
$| x | = (x \cdot x)^{1 / 2} = {(\sum_{1}^{k} x_{i}^{2})}^{1 / 2}$

The structure now defined (the vector space $R^{k}$ with the above inner product and norm) is called euclidean $k$ -space.

Properties of Euclidean norm

Suppose $x, y, z \in R^{k}$ , and $α$ is real. Then
(a) $| x | \geq 0$
(b) $| x | = 0$ if and only if $x = 0$ ;
(c) $| α x | = | α | | x |;$
(d) $| x \cdot y | \leq | x | | y |$ ;
(e) $| x + y | \leq | x | + | y |$ ;
(f) $| x - z | \leq | x - y | + | y - z |$ .

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