# On the nature of math — abstraction

Abstraction is such a central part of modern mathematics that one forgets that it was not until Frechet’s 1906 thesis that sets of points with no a priori underlying structure (not assumed points in or functions on $\Re^{n}$) are considered and given a structure a posteriori (Frechet first defined abstract metric spaces). And after its success in analysis, abstraction took over significant parts of algebra, geometry, topology and logic.

Cheers,

Nalin Pithwa.

# Some v basic stuff from Rudin’s Analysis

Section 2.18:

Definition: Let X be metric space. All points and sets mentioned before are understood to be elements and subsets of X.

(a) A neighbourhood of a point p is a set $N_{r}(p)$ consisting of all points q such that $d(p,q). The number r is called the radius of $N_{r}(p)$.

(b) A point p is a limit point of the set E if every neighbourhood of p contains a point $q \neq p$ such that $q \in E$.

(c) If $p \in E$ and p is not a limit point of E, then p is called an isolated point of E.

(d) E is closed if every limit point of E is a point of E.

(e) A point p is an interior point if there is a neighbourhood N of p such that $N \subset E$.

(f) E is open if every point of E is an interior point of E.

(g) The complement of E (denoted by $E^{c}$) is the set of all points $p \in X$ such that $p \notin E$.

(h) E is perfect if E is closed and if every point of E is a limit point of E.

(i) E is bounded if there is a real number M and a point $q \in X$ such that $d(p,q) for all points p in E.

(j) E is dense in X if every point of X is a limit point of E, or a point of E (or both).

Let us note that in $R^{1}$ neighbourhoods are segments, whereas in $R^{2}$ neighbourhoods are interiors of circles.

2.19 Theorem: Every neighbourhood is an open set.

2.20 Theorem If p is a limit point of a set E, then every neighbourhood of p contains infinitely many points of E.

2.20 Corollary A finite point set has no limit points

2.21 Examples Let us consider the following subsets of $R^{2}$.

2.21 a: The set of all complex z such that $|z|<1$ This is not closed, it is open, it is not perfect set but it is bounded

2.21b: The set of all integers z such that $|z| \leq 1$ This is closed set, it is not open set, it is a perfect set and it is also bounded.

2.21c: A finite set: Such a set is vacuously closed, it is not open, it is not perfect, it is bounded.

2.21d: The set of all integers. This is a closed set in $R^{2}$, this set is not open in $R^{2}$, this is not a perfect set in $R^{2}$, and it is not bounded

2.21e: The set consisting of the numbers $\frac{1}{n}$ where $n \in \mathcal{N}$. Let us note that this set E has a limit point, namely, $z=0$ but that no point of E is a limit point of E, we wish to stress the difference between having a limit point and containing one. What can we say about the basic topological properties of this set ? It is not closed in $R^{2}$, it is not open in $R^{2}$, it is not perfect in $R^{2}$ but it is bounded in $R^{2}$.

2.21f: The set of all complex numbers that is $R^{2}$: this is closed in $R^{2}$, this is also open in $R^{2}$, this is also perfect set in $R^{2}$, but this set is not bounded.

2.21g: the segment $(a,b)$. This set is not closed in $R^{2}$, it is not open if we regard it as an open set in $R^{2}$, whereas it is open in $R^{1}$, and it is also bounded in $R^{2}$.

NOTE: It is very enriching to compare all the above definitions with the following basic starting point of topology: A set S is said to be topologized if we can answer the following question: Given any real point p and any subset X we can answer the question: is p a limit point of X ? There are two extreme cases here: each point is a limit point….in this case there are simply too many limit points, this is a worthless topology; on the other extreme, there is the case in which no point is a limit point. This is called discrete topology. The very fact that this is endowed with a name means this is not as worthless as the other case !! Now this in view, compare all above definitions with this. Remember Rudin’s emphasis is analysis whereas here it is topology which is emphasised.

2.22 Theorem Let $\{ E_{\alpha}\}$ be an arbitrary (finite or infinite) collection of sets $E_{\alpha}$. Then $(\bigcup_{\alpha}E_{\alpha})^{c} = \bigcap_{\alpha}(E_{\alpha}^{c})$

2.23 A set E is open if and only if its complement is closed.

2.23 Corollary: A set F is closed if and only if its complement is open.

2.24 Theorem:

2.24a: For any collection $\{ G_{\alpha}\}$ of open sets, $\bigcup_{\alpha}G_{\alpha}$ is open.

2.24b: For any collection $\{ F_{\alpha}\}$ of closed sets, $\bigcap_{\alpha}F_{\alpha}$ is closed.

2.24c: For any finite collection $G_{1}, G_{2}, \ldots, G_{n}$ of open sets $\bigcap_{i=1}^{n}G_{i}$ is open.

2.24d: For any finite collection $F_{1}, F_{2}, \ldots, F_{n}$ of closed sets, $\bigcup_{i=1}^{n}F_{i}$ is closed.

2.25 Example: In parts (c) and (d) of the preceding theorem, the finitiness of the collections is essential. For, let $G_{n}$ be the segment $(-\frac{1}{n}, \frac{1}{n})$ where $n=1,2,3,\ldots$. Then, $G_{n}$ is an open subset of $R^{1}$. Put $G=\bigcap_{i=1}^{\infty}G_{n}$. Then G consists of a single point (namely, x=0) and is, therefore not an open subset of $R^{1}$.

Thus, the intersection of an infinite collection of open sets need not be open. Similarly, the union of an infinite collection of closed sets need not be closed.

2.26 Definition: If X is a metric space, if $E \subset X$ and if $E^{'}$ denotes the set of all limit points of E in X, then the closure of E is the set $\overline{E}=E \bigcup E^{'}$

(Remark: The following few theorems, in my theorem, give an approach to solving problems in analysis: )

(Remark: I have tried to fill the missing gaps in Prof Rudin’s terse, distilled proofs!)

2.27 Theorem: If X is a metric space and $E \subset X$, then prove that

2.27a: $\overline{E}$ is closed

2.27b: $E=\overline{E}$ if and only if E is closed.

2.27c: $\overline{E} \subset F$ for every closed set $F \subset X$ such that $E \subset F$.

By (a) and (c), $\overline{E}$ is the smallest closed subset of X that contains E.

Proof:

2.27a: If $p \in X$ and $p \notin \overline{E}$, then p is neither a point of E nor a limit point of E. Hence, p has a neighbourhood which does not intersect E. The complement of $\overline{E}$ is therefore open. Hence, $\overline{E}$ is closed.

2.27b: If $E = \overline{E}$, from part a, we conclude that E is closed. Conversely, if E is closed, then $E^{'} \subset E$ and by definition of closed and 2.26 theorem above, we get that $\overline{E} =E$.

2.27c: Given that F is a closed set, and that $F \supset E$, then $F \supset F^{'}$ hence, $F \supset E^{'}$. Thus, $F \supset E$.

2.28 Theorem : Let E be a nonempty set of real numbers which is bounded above. Let $y = \sup {E}$. Then prove that $y \in \overline{E}$. Hence, $y \in E$ if E is closed.

****************************************************

Remark: The author Prof Rudin wants us to pause now and compare above with the examples in Sec 1.9. I reproduce Sec 1.8 and Sec 1.9 below for easy reference and understanding:

Sec 1.8: Definition: Suppose S is an ordered set, with $E \subset S$, and E is bounded above. Suppose there exists an $\alpha \in S$ with the following properties:

(i) $\alpha$ is an upper bound of E.

(ii) If $\gamma < \alpha$ then $\gamma$ is not an upper bound of E. Then $\alpha$ is called the least upper bound of E (that there is at most one such $\alpha$ is clear from (ii)) or the supremum of E, and we write:

$\alpha = \sup{E}$.

The greatest lower bound or infimum of a set E which is bounded below is defined in the same manner. The statement $\alpha = \inf{E}$ means that $\alpha$ is a lower bound of E and that no $\beta$ with $\beta > \alpha$ is a lower bound.

Sec 1.9 Examples:

Sec 1.9a example: Let $llatex p^{2}=2$. Let A be the set of all positive rationals p such that $p^{2}<2$ and let B consist of all positive rationals p such that $p^{2}=2$. A and B are subsets of the ordered set Q. The set A is bounded above. In fact, the upper bounds of A are exactly the members of B. Since B contains no smallest member, A has no least upper bound in Q.

Similarly, B is bounded below. The set of all lower bounds of B consists of A and of all $r \in Q$ with $r \leq 0$. Since A has no largest member, B has no greatest lower bound in Q.

Sec 1.9b example: If $\alpha = \sup {E}$ exists, then $\alpha$ may or may not be a member of E. For instance, let $E_{1}$ be the set of all $r \in Q$ with $r <0$. Let $E_{2}$ be the set of all $r \in Q$ with $r \leq 0$. Then

$\sup {E_{1}}= \sup{E_{2}}=0$

and $0 \notin E_{1}$ and $0 \in E_{2}$.

Sec 1.9c example: Let E consist of all numbers $\frac{1}{n}$ where $n=1,2,3,\ldots$. Then $\sup {E}=1$, which is in E and inf E = 0, which is not in E.

****************************************************

Now, we come back to our current discussion:

2.28 Theorem: Let E be a nonempty set of real numbers which is bounded above. Let $y = \sup {E}$. Then prove that $y \in E$ if E is closed.

Proof: If $y \in E$, then $y \in \overline{E}$. Assume $y \notin E$. For every $h >0$, there exists then a point $x \in E$ such that $y-h, for otherwise $y-h$ would be an upper bound of E. Thus, y is a limit point of E. Hence, $y \in \overline{E}$

2.29 Remark Suppose $E \subset Y \subset X$ where X is a metric space. To say that E is an open subset of X means that to each point $p \in E$, there is associated a positive number r such that the conditions $d(p,q), $q \in X$ imply that $q \in E$. But we have already observed (Sec 2.16) that Y is also a metric space, so that our definitions may equally well be made within Y. To be quite explicit, let us say that E is open relative to Y if to each $p \in E$ there is associated an $r>0$ such that $q \in E$ whenever $d(p,q) and $q \in Y$. Example 2.21g showed that a set may be open relative to Y without being an open subset of X. However, there is a simple relation between these concepts, which we now state.

2.30 Theorem Suppose $Y \supset X$. A subset E of Y is open relative to Y if and only if $E =Y \bigcap G$ for some open subset G of X.

Proof:

Suppose E is open relative to Y. To each $p \in E$, there is a positive number $r_{p}$ such that the conditions $d(p,q) and $q \in Y$ imply that $q \in E$. Let $V_{q}$ be the set of all $q \in X$ such that $d(p,q) and define

$G = \bigcup_{p \in E}V_{p}$

Thus G is an open subset of X, by theorems 2.19 and 2.24.

Since $p \in V_{p}$ for all $p \in E$, it is clear that $E \subset G \bigcap Y$.

By our choice of $V_{p}$, we have $V_{p} \bigcap Y \subset E$ for every $p \in E$, so that $G \bigcap Y \subset E$. Thus, $E = G \bigcap Y$, and one half of the theorem is proved.

Conversely, if G is open in X and $E = G \bigcap Y$, every $p \in E$ has a neighbourhood $V_{p} \subset G$. Then $V_{p} \bigcap Y \subset E$, so that E is open relative to Y.

Cheers,

Nalin Pithwa

# Is Math really abstract? I N Herstein answers…

Reference: Chapter 1: Abstract Algebra Third Edition, I. N. Herstein, Prentice Hall International Edition:

For many readers/students of pure mathematics, such a book will be their first contact with abstract mathematics. The subject to be discussed is usually called “abstract algebra,” but the difficulties that the reader may encounter are not so much due to the “algebra” part as they are to the “abstract” part.

On seeing some area of abstract mathematics for the first time,be it in analysis, topology, or what not, there seems to be a common reaction for the novice. This can best be described by a feeling of being adrift, of not having something solid to hang on to. This is not too surprising, for while many of the ideas are fundamentally quite simple, they are subtle and seem to elude one’s grasp the first time around. One way to mitigate this feeling of limbo, or asking oneself “What is the point of all this?” is to take the concept at hand and see what it says in particular concrete cases. In other words, the best road to good understanding of the notions introduced is to look at examples. This is true in all of mathematics.

Can one, with a few strokes, quickly describe the essence, purpose, and background for abstract algebra, for example?

We start with some collection of objects S and endow this collection with an algebraic structure by assuming that we can combine, in one or several ways (usually two) elements of this set S to obtain, once more, elements of this set S. These ways of combining elements of S we call operations on S. Then we try to condition or regulate the nature of S by imposing rules on how these operations behave on S. These rules are usually called axioms defining the particular structure on S. These axioms are for us to define, but the choice made comes, historically in mathematics from noticing that there are many concrete mathematical systems that satisfy these rules or axioms. In algebra, we study algebraic objects or structures called groups, rings, fields.

Of course, one could try many sets of axioms to define new structures. What would we require of such a structure? Certainly we would want that the axioms be consistent, that is, that we should not be led to some nonsensical contradiction computing within the framework of the allowable things the axioms permit us to do. But that is not enough. We can easily set up such algebraic structures by imposing a set of rules on a set S that lead to a pathological or weird system. Furthermore, there may be very few examples of something obeying the rules we have laid down.

Time has shown that certain structures defined by “axioms” play an important role in mathematics (and other areas as well) and that certain others are of no interest. The ones we mentioned earlier, namely, groups, rings, fields, and vector spaces have stood the test of time.

A word about the use of “axioms.” In everyday language, “an axiom means a self-evident truth”. But we are not using every day language; we are dealing with mathematics. An axiom is not a universal truth — but one of several rules spelling out a given mathematical structure. The axiom is true in the system we are studying because we forced it to be true by “force” or “our choice” or “hypothesis”. It is a licence, in that particular structure to do certain things.

We return to something we said earlier about the reaction that many students have on their first encounter with this kind of algebra, namely, a lack of feeling that the material is something they can get their teeth into. Do not be discouraged if the initial exposure leaves you in a bit of a fog.Stick with it, try to understand what a given concept says and most importantly, look at particular, concrete examples of the concept under discussion.

Follow the same approach in linear algebra, analysis and topology.

Cheers, cheers, cheers,

Nalin Pithwa

# Constructing numbers from sets

Continued from previous blog: A fast review of set theory; same reference: A Second Course in Analysis by M Ram Murty, Hindustan Book Agency.

Mathematicians and philosophers of the nineteenth century pondered deeply into the nature of a number. The question of “what is a number?” is not a simple one. But since mathematicians decided to give foundations of mathematics using the axiomatic method and sets as the basic building blocks, we are led to define numbers using sets. We follow Richard Dedekind (1831-1916) and Giuseppe Peano (1858-1932) in the following construction. It was as late as 1888 and 1889 when this construction was described in two papers written independently by Dedekind and Peano.

We construct a sequence of sets to represent the natural numbers. As noted earlier, zero is represented by the empty set. We have already described the construction of the natural numbers using the empty set. For each natural number n, the successor of n is denoted by $n+1$ (and sometimes as $n^{'}$) and defined as

$n \bigcup \{ n \}$

Thus, each natural number is a set with n elements, namely,

$\{ 0,1,2, \ldots, n-1\}$

We designate the set of natural numbers by the symbol $\mathcal{N}$. (It is a matter of personal convenience whether to include zero as a natural number or not. In this discussion, zero is a natural number. In other settings, it may not be. There is no universal convention regarding this and the student is expected to understand depending on the context. Some authors use the term “whole numbers” to indicate that zero is included in the discussion.)

The arithmetic operations on $\mathcal{N}$ are now defined recursively. Addition is defined as a function from $\mathcal{N} \times \mathcal{N}$ to $\mathcal{N}$:

$+ (m,n) = m+n$

where $m+n$ is defined recursively by $0+n=n$ and $m'+n = (m+n)^{'}$. A similar definition is given for multiplication x by defining $0 \times n = 0$ and

$m^{'} \times n = (m \times n) + n$

We also define $m \times n$ as simply mn which is the familiar symbology.

An equivalence relation on a set S is a subset R of $S \times S$ satisfying:

1. (reflexive axiom) $(a,a) \in R$ $\forall {x} \in S$.
2. (symmetry axiom) $(a,b) \in R \Longleftrightarrow (b,a) \in R$.
3. (transitive axiom) $(a,b) \in R$ and $(b,c) \in R$ implies $(a,c) \in R$.

The notion of an equivalence relation is an abstaction of our concept of equality, or at least what we implicitly expect of the notion of equality. It is more suggestive to write the equivalence relationn, not as a subset of $S \times S$ as indicated above, but rather more symbolically as $\sim$ that our axioms become:

1. (reflexive) $a \sim a$, $\forall {a} \in$.
2. (symmetry) $a \sim b$ if and only if $b \sim a$.
3. (transitive) $a \sim b$ and $b \sim c$ implies $a \sim c$.

Equivalence relations play a fundamental role in all of mathematics. They allow us to understand aspects of sets by grouping them using certain properties.

To construct negative integers, we define an equivalence relation on $\mathcal{N} \times \mathcal{N}$. We write

$(m,n)\sim (j,k) \Longleftrightarrow m+k=j+n$.

Intuitively, we think of $(m,n)$ as $m-n$ so that it becomes evident that our definition is now in terms of concepts that have been defined earlier. This is very similar to how the ancients worked with negative numberss that appeared in an equation. They usually moved them to the other side so that the equation became an equation of non-negative numbers. However, with our set theoretic definition, we have reached a more fundamental and higher level of abstraction. Thus, with our equivalence relation above on the natural numbers, we define the set of integers as the set of equivalence classes of such ordered pairs. It is now easy to see that the following lemma holds:

Lemma 1.1 If $(j,k)$ is an ordered pair of non-negative integers, then exactly one of the following statements holds:

(a) $(j,k)$ is equivalent to $(m,0)$ for a unique non-negative integer m;

(b) $(j,k)$ is equivalent to $(0,m)$ for a unique non-negative integer m;

(c) $(j,k)$ is equivalent to $(0,0)$.

Sometimes, we denote by $|(j,k)|$ the equivalence class of $(j,k)$. With this lemma in place, we now denote by m the set of pairs of non-negative integers equivalent to $(m,0)$ ; by -m the set of pairs equivalent to $(0,m)$ and by 0 the set of pairs equivalent to $(0,0)$. We denote these equivalence classes by $\mathcal{Z}$.This gives us set theoretic construction of the set of integers.

We can define the operations of addition and multiplication by setting:

$|(j_{1}, k_{1})|+|(j_{2}, k_{2})|=|(j_{1}+j_{2}, k_{1}+k_{2})|$

$|(j_{1}, k_{1})| \times |(j_{2}, k_{2})| = |(j_{1}j_{2}+k_{1}k_{2}, j_{1}k_{2}+j_{2}k_{1})|$

This latter definition is best understood if we recall that the symbol $(j,k)$ represents j-k so that the left hand side of the above equation is

$(j_{1}-k_{1})(j_{2}-k_{2}) = j_{1}j_{2}+k_{1}k_{2}-(j_{1}k_{2}+j_{2}k_{1})$

One needs to check that these definitions are “well-defined” in the sense that they are independent of the representatives chosen for the equivalence class. This can be done as exercises.

In this way, we have now extended the notion of addition and multiplication from the set of natural numbers to the set of integers. Subtraction of integers can be defined by

$|(j_{1}, k_{1})|-|(j_{2}, k_{2})| = |(j_{1}, k_{1})| +(-1)|(j_{2}, k_{2})|$

where -1 represents the equivalence class $(0,1)$. All of these definitions correspond to our usual notion of addition, subtraction and multiplication. Their virtue lies in their pure set-theoretic formulation.

We can also order the set of integers in the usual way. Thus,

$j_{1}+k_{2} < k_{1}+j_{2} \Longrightarrow |(j_{1}, k_{1})|< |(j_{2}, k_{2})|$

and

$j_{1}+k_{2} \leq k_{1}+j_{2} \Longleftrightarrow |(j_{1}, k_{1})| \leq |(j_{2}, k_{2})|$.

This corresponds to our usual notion of “less than” and “less than or equal to”.

Finally, we can define the absolute value on the set of integers by setting

$|k|=k$, if $0

$|k|=0$, if $k=0$

$|k|=-k$, if $k<0$

We can now construct the rational numbers $\mathcal{Q}$ from the set of integers. We do this by defining an equivalence relation on the set $\mathcal{Z} \times \mathcal{Z}^{+}$ by stating that two pairs $(j_{1}, k_{1})$ and $(j_{2}, k_{2})$ are equivalent if and only if $j_{1}k_{2}=j_{2}k_{1}$. Intuitively, we think of $(j_{1}, k_{1})$ as representing the “fraction” $\frac{j_{1}}{k_{1}}$ and examining what we would mean by $\frac{j_{1}}{k_{1}} = \frac{j_{2}}{k_{2}}$ by reducing it to notions already defined. The set of rational numbers $\mathcal{Q}$ is then defined as the set of such equivalence classes.

The expected operations of addition and multiplication are now evident:

$|(j_{1}, k_{1})|+|j_{2}, k_{2}| = |(j_{1}k_{2}+j_{2}k_{1}, k_{1}k_{2})|$

$|(j_{1}, k_{1})||(j_{2}, k_{2})| = |(j_{1}j_{2}, k_{1}k_{2})|$

Again, these definitions are easily verified to be well-defined. Finally, we can now define “division”.If $|(j_{1}, k_{1})|, |(j_{2}, k_{2})| \in \mathcal{Q}$ with $j_{2} \neq 0$, we define:

$\frac{|(j_{1}, k_{1})|}{|(j_{2}, k_{2})|} = |(j_{1}k_{2}, j_{2}k_{1})|$

These operations satisfy the familiar laws of associativity, commutativity and distributivity. Subtraction of rational numbers then can be written as :

$|(j_{1}, k_{1})|-|(j_{2}, k_{2})| = |(j_{1}, k_{1})|+(-1,1)|(j_{2}, k_{2})|$

The ordering of rational numbers can also be written as:

$|(j_{1}, k_{1})|< |(j_{2}, k_{2})| \Longleftrightarrow j_{1}k_{2}< j_{2}k_{1}$

$|(j_{1}, k_{1})| \leq |(j_{2}, k_{2})| \Longleftrightarrow j_{1}k_{2} \leq j_{2}k_{1}$.

These definitions agree with out usual notions of ordering of the rational numbers.

Finally, the definition of absolute value can be extended as:

$|[(j,,k)]|= |(j,k)|$ if $[(0,1)] < [(j,k)]$

$|[(j,k)]| = |(0,1)|$ if $|(j,k)|=|(0,1)|$

$|[(j,k)]| = -|(j,k)|< |(0,1)|$.

Again, our familiar properties of the absolute value of rational numbers hold. With this foundational construction in place, we can conveniently represent the equivalence class of $(j,k)$ as simply the fraction j/k and continue to work with these numbers as we were hopefully taught from childhood.

In the next sections/blogs we construct the real numbers from this axiomatic framework.

Exercises:

Hint (generic): keep the meaning of the symbols in mind and meaning of equivalence relations and equivalence classes. Also note that our basic object is a class and a set is a member of a class.

1. let $|(j_{1}, k_{1})|, |(j_{2}, k_{2})|$ be two elements of $\mathcal{Z}$. Show that the addition:

$|(j_{1}, k_{1})|+|(j_{2}, k_{2})| = |(j_{1}+j_{2}, k_{1}+k_{2})|$ is well-defined. That is, prove that for any $(j_{1}^{'}, k_{1}^{'}) \in |(j_{1}, k_{1})|$ and $(j_{2}^{'},k_{2}^{'}) \in |(j_{2}, k_{2})|$, we have that $(j_{1}^{'}+j_{2}^{'}, k_{1}^{'}+k_{2}^{'})$ is equivalent to $(j_{1}+j_{2}, k_{1}+k_{2})$.

2. For $j_{1}, j_{2}, k \in \mathcal{Z}$, prove the distributive law: $(j_{1}+j_{2}).k = j_{1}k+j_{2}k$.

3. Show that the relations < and $\leq$ on $\mathcal{Z}$ have the following properties:

(a) $|(0,j)|< |(0,0)|$ for all $j \in \mathcal{Z}^{+}$

(b) $|(0,j)|< |(k,0)|$ for all $j, k \in \mathcal{Z}^{+}$

(c) $|(0,j)|< |(0,k)|, j ,k \in \mathcal{Z}^{+}$ if and only if $k

(d) $|(0,0)| < |(j,0)|$ for all $j \in \mathcal{Z}^{+}$

(e) $|(j,0)|<|(k,0)|, j, k \in \mathcal{Z}_{\geq 0}$if and only if $j.

(f) $|(0,j)| \leq |(0,0)|$ for all $j \in \mathcal{Z}_{\geq 0}$.

(g) $|(0,j)| \leq |(k,0)|$ for all $j,k \in \mathcal{Z}_{\geq 0}$

(h) $|(0,j)| \leq |(0,k)|$ for $j,k \in \mathcal{Z}_{\geq 0}$ if and only if $k \leq j$

(i) $|(0,0)| \leq |(j,0)|$ for all $j \in \mathcal{Z}_{\geq 0}$

(j) $|(j,0)| \leq |(k,0)|$ where $j, k \in \mathcal{Z}_{\geq 0}$ if and only if $j \leq k$.

Cheers,

Nalin Pithwa.