The real number system, $\displaystyle \mathbb{R}\times \mathbb{R}$ contains order axioms that show relations and properties of the system that add completeness to the ordered field algebraic laws.

The properties are as follows.

Trichotomy Property:

Given $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$ only one of the following statements holds true $\displaystyle a< b$, $\displaystyle b< a$, or $\displaystyle a=b$.

Transitive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle b< c$ then this implies $\displaystyle a< c$.

Additive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c\ \epsilon\ \mathbb{R}$ then this implies $\displaystyle a+b< b+c$.

Multiplicative Properties:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c>0$ then this implies $\displaystyle ac< bc$ and  $\displaystyle a< b$ and $\displaystyle c< 0$ then this implies $\displaystyle bc< ac$.

Therefore looking at the options the Trichotomy Property identifies the property in this particular question.

The real number system, $\displaystyle \mathbb{R}\times \mathbb{R}$ contains order axioms that show relations and properties of the system that add completeness to the ordered field algebraic laws.

The properties are as follows.

Trichotomy Property:

Given $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$ only one of the following statements holds true $\displaystyle a< b$, $\displaystyle b< a$, or $\displaystyle a=b$.

Transitive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle b< c$ then this implies $\displaystyle a< c$.

Additive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c\ \epsilon\ \mathbb{R}$ then this implies $\displaystyle a+b< b+c$.

Multiplicative Properties:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c>0$ then this implies $\displaystyle ac< bc$ and  $\displaystyle a< b$ and $\displaystyle c< 0$ then this implies $\displaystyle bc< ac$.

Therefore looking at the options the Transitive Property identifies the property in this particular question.

The real number system, $\displaystyle \mathbb{R}\times \mathbb{R}$ contains order axioms that show relations and properties of the system that add completeness to the ordered field algebraic laws.

The properties are as follows.

Trichotomy Property:

Given $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$ only one of the following statements holds true $\displaystyle a< b$, $\displaystyle b< a$, or $\displaystyle a=b$.

Transitive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle b< c$ then this implies $\displaystyle a< c$.

Additive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c\ \epsilon\ \mathbb{R}$ then this implies $\displaystyle a+b< b+c$.

Multiplicative Properties:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c>0$ then this implies $\displaystyle ac< bc$ and  $\displaystyle a< b$ and $\displaystyle c< 0$ then this implies $\displaystyle bc< ac$.

Therefore looking at the options the Additive Property identifies the property in this particular question.

The real number system, $\displaystyle \mathbb{R}\times \mathbb{R}$ contains order axioms that show relations and properties of the system that add completeness to the ordered field algebraic laws.

The properties are as follows.

Trichotomy Property:

Given $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$ only one of the following statements holds true $\displaystyle a< b$, $\displaystyle b< a$, or $\displaystyle a=b$.

Transitive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle b< c$ then this implies $\displaystyle a< c$.

Additive Property:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c\ \epsilon\ \mathbb{R}$ then this implies $\displaystyle a+b< b+c$.

Multiplicative Properties:

For $\displaystyle a$, $\displaystyle b$, and $\displaystyle c\ \epsilon\ \mathbb{R}$ where $\displaystyle a< b$ and $\displaystyle c>0$ then this implies $\displaystyle ac< bc$ and  $\displaystyle a< b$ and $\displaystyle c< 0$ then this implies $\displaystyle bc< ac$.

Therefore looking at the options the Multiplicative Properties identifies the property in this particular question.

According to the Well-Ordered Principal this statement is true. The following proof illuminate its truth.

Suppose $\displaystyle A\subseteq \mathbb{N}$ is nonempty. From there, it is known that $\displaystyle -A$ is bounded above, by $\displaystyle -1$.

Therefore, by the Completeness Axiom the supremum of $\displaystyle -A$ exists.

Furthermore, if $\displaystyle A\subset \mathbb{Z}$ has a supremum, then $\displaystyle \text{sup}A\ \epsilon\ A$, thus in this particular case $\displaystyle \text{sup}(-A)\ \epsilon\ -A$.

Thus by the Reflection Principal,

$\displaystyle \text{inf}A=-\text{sup}(-A)$ 

exists and 

$\displaystyle \text{inf}A\ \epsilon\ -(-A)=A$.

Therefore proving the statement in question true.

This statement:

A sequence of sets $\displaystyle \begin{Bmatrix} I_n \end{Bmatrix}_{n\ \epsilon\ \mathbb{N}}$ is __________ if and only if $\displaystyle I_1\supseteq I_2\supseteq ...$

is the definition of nested.

This means that the sequence $\displaystyle I_n$ for all $\displaystyle n$ elements, for which $\displaystyle n$ belongs to the natural numbers, is considered a nested set if and only if the subsequent sets are subsets of it. 

Other theorems in intro analysis build off this understanding.

Determine this statement is false by showing a contradiction when actual values are used.

Let 

$\displaystyle a=1, b=2, c=-3, d=-2$

First make sure the inequalities hold true.

$\displaystyle \\a< b \\1< 2$ 

and

$\displaystyle \\c< d< 0 \\-3< -2< 0$

Now find the products.

$\displaystyle \\ac< bd \\(1)(-3)< (2)(-2) \\-3\nless -4$

Therefore, the statement is false.

Using the definition for Riemann sums to define the upper and lower integrals of a function answers the question.

According the the Riemann sum where $\displaystyle U(f,P)$ represents the upper integral and $\displaystyle L(f,P)$ the following are defined:

1. The upper integral of $\displaystyle f$ on $\displaystyle [a,b]$ is

$\displaystyle (U)\int_a^bf(x)dx:=\text{inf}\begin{Bmatrix} U(f,p)) \end{Bmatrix}$ where $\displaystyle P$ is a partition of $\displaystyle [a,b]$.

2. The lower integral of $\displaystyle f$ on $\displaystyle [a,b]$ is

$\displaystyle (L)\int_a^bf(x)dx:=\text{sup}\begin{Bmatrix} L(f,p)) \end{Bmatrix}$where $\displaystyle P$ is a partition of $\displaystyle [a,b]$.

3. If 1 and 2 are the same then the integral is said to be

$\displaystyle \int_a^bf(x):=(U)\int_a^bf(x)dx=(L)\int_a^bf(x)dx$

if and only if $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$, $\displaystyle a< b$,  and $\displaystyle f:[a,b]\rightarrow \mathbb{R}$ be bounded.

Therefore the necessary condition for the proving the upper and lower integrals of a bounded function exists is if and only if $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$, $\displaystyle a< b$,  and $\displaystyle f:[a,b]\rightarrow \mathbb{R}$ be bounded.

By definition 

If $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$ and $\displaystyle a< b$.

A partition over the interval $\displaystyle [a,b]$ is a set of points $\displaystyle P=\begin{Bmatrix} x_0,x_1,...,x_n \end{Bmatrix}$ such that

$\displaystyle a=x_0< x_1< ...< x_n=b$.

Therefore, the term that describes this statement is partition.

By definition 

If $\displaystyle a$, $\displaystyle b\ \epsilon\ \mathbb{R}$ and $\displaystyle a< b$.

A partition over the interval $\displaystyle [a,b]$ is a set of points $\displaystyle P=\begin{Bmatrix} x_0,x_1,...,x_n \end{Bmatrix}$ such that

$\displaystyle a=x_0< x_1< ...< x_n=b$.

Furthermore, 

The norm of the partition 

$\displaystyle P=\begin{Bmatrix} x_0,x_1,...,x_n \end{Bmatrix}$ is

$\displaystyle ||P||=\max_{1\leq j\leq n}|x_j-x_{j-1}|$

Therefore, the term that describes this statement is norm.