Mathematical Morsels : IV

The words set and space are often used in loose context to one another. A set is merely an amorphous collection of elements, without coherence or form. When some kind of algebraic or geometric structure is imposed on a set, so that its elements are organized into a systematic whole, then it becomes a space.

Solutions to System of Sets:

Reference: Introductory Real Analysis, Kolmogorov and Fomin. Dover Publications. Translated and edited by Richard A. Silverman:

Problem 1:

Let X be an uncountable set, and let \mathscr{R} be the ring consisting of all finite subsets of X and their complements. Is \mathscr{R} a \sigma-ring?

Solution 1:

Recall the definition: A ring of sets is called \sigma-ring if it contains the union S= \bigcup_{n=1}^{\infty}A_{n} whenever it contains the sets A_{1}, A_{2}, \ldots, A_{n}, \ldots.

There are countably many finite subsets of X (given uncountable set) and their complements (also especially) so that the union of all these is the whole set X. Hence, X is a \sigma-ring as it is closed under countably many set unions. QED.

Problem 2:

Are open intervals Borel sets?

Solution 2:

Recall the following:

(1) Borel sets or B-sets are the subsets of the real line belonging to the minimal B-algebra  generated by the set of all closed intervals [\alpha, \beta].

Problem 3:But a B-algebra is also a ring and is closed under differences/complements. Open intervals are complements of closed intervals. And, they belong hence to B-algebra. Note we still have to answer the question whether open sets are Borel sets and so we have to check if they are contained in a minimal B-algebra. As per the following theorem: Given any non-empty system of sets \mathscr{S}, there is a unique irreducible (minimal) B-algebra \mathscr{B}(\mathscr{P}) generated by the system \mathscr{S} or the Borel closure of \mathscr{P}. So, yes indeed open sets are Borel sets. QED.

Problem 3A:

Let y=f(x) be a function defined on a set M and taking values in a set N. Let \mathscr{M} be a system of subsets of M, and let f(\mathscr{M}) denote the system of all images f(A) of sets A \in \mathscr{M}. Moreover, let \mathscr{N} be a system of subsets of N, and let f^{-1}(\mathscr{N}) denote the system of all preimages f^{-1}(B) of sets B \in \mathscr{N}. Prove that:

i) If \mathscr{N} is a ring, so is f^{-1}(\mathscr{N}).

ii) If \mathscr{N} is an algebra, so is f^{-1}(\mathscr{N})

iii) If \mathscr{N} is a B-algebra, so is f^{-1}(\mathscr{N})

iv) \mathscr{R}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{R}(\mathscr{N})) where \mathscr{R} stands for a ring.

v) \mathscr{B}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{B}(\mathscr{N})) where \mathscr{B} is irreducible or minimum Borel algebra generated by a non-empty system of sets \mathscr{S}.

Problem 3B:

Prove the following:

i) If \mathscr{M} is a ring, so is f(\mathscr{M}).

ii) If \mathscr{M} is an algebra, so is f(\mathscr{M}).

iii) If \mathscr{M} is Borel-algebra, so is f(\mathscr{M}).

iv) \mathscr{R}(f(\mathscr{M}))=f(\mathscr{R}(\mathscr{M}))

v) \mathscr{B}(f(M))=f(\mathscr{B}(\mathscr{M})) where \mathscr{B} is irreducible minimum Borel algebra generated by a non-empty system of sets \mathscr{S}.

Solution 3A:

Part 1: Given mapping f: M \rightarrow N, that is, y=f(x), Let Y_{1}, Y_{2}, Y_{3}, \ldots be subsets of N and let X_{1}, X_{2}, X_{3}, \ldots be subsets of M. Then, as it is given that \mathscr{N} is a ring then :

Y_{1} \bigcap Y_{2} \in \mathscr{N}

Y_{1}\bigcup Y_{2} \in \mathscr{N}

Y_{1}-Y_{2} \in \mathscr{N}

Y_{1} \triangle Y_{2} = (Y_{1}-Y_{2}) \bigcup (Y_{2}-Y_{1}) \in \mathscr{N}

The above implies:

f^{-1}(Y_{1}\bigcap Y_{2}) \in f^{-1}(\mathscr{N}), that is, f^{-1}(Y_{1}) \bigcap f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N}), that is, X_{1} \bigcap X_{2} \in \mathscr{M}.

Also, Y_{1} \bigcup Y_{2} \in \mathscr{N} so that f^{-1}(Y_{1} \bigcup Y_{2}) \in f^{-1}(\mathscr{N}), that is, f^{-1}(Y_{1}) \bigcup f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N}), that is, we conclude X_{1} \bigcup X_{2} \in \mathscr{M}.

Also, Y_{1}-Y_{2} \in \mathscr{N} as \mathscr{N} is a ring; so that f^{-1}(Y_{1}-Y_{2}) \in f^{-1}(\mathscr{N}) so that for some X_{0}, which is f^{-1}(Y_{1}-Y_{0}) \longrightarrow f^{-1}(Y_{1}) - f^{-1}(Y_{0}) \in \mathscr{M}

So, we have shown that the system of sets \mathscr{M} is closed under set union, set intersection, set difference, and hence, also set symmetric difference. Hence, it is is a ring. QED.

Part ii:

We have already shown that f^{-1}(\mathscr{N}) is a ring, now \mathscr{N} is also an algebra meaning a ring of sets with a unit element. Let it contain a unit E \in \mathscr{N} such that if Y_{0} is any set belonging to \mathscr{N}, then Y_{0} \bigcap E = Y_{0}. so that f^{-1}(Y_{0} \bigcap E) = f^{-1}(Y_{0}), hence, f^{-1}(Y_{0}) \bigcap f^{-1}(E) = f^{-1}(Y_{0}), which in turn implies that f^{-1}(E) is a unit of f^{-1}(\mathscr{N}). Hence, f^{-1}(\mathscr{N}) is a ring of sets with a unit element, or in other words, an algebra of sets.

Part iii:

To prove that: if \mathscr{N} is a Borel algebra, so is f^{-1}(\mathscr{N}). From parts i and ii above, f^{-1}(\mathscr{N}) is a ring with a unit or in other words, f^{-1}(\mathscr{N}) is an algebra. Moreover, if it contains \bigcup_{k=1}^{\infty}Y_{k} whenever it contains the sets Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}, \ldots all of which are members of \mathscr{N}.

But for an arbitrary (finite or infinite) collection of sets:

f^{-1}(\bigcup_{\alpha}A_{\alpha}) = \bigcup_{\alpha}f^{-1}(A_{\alpha})

Now, \bigcup_{k=1}^{\infty}Y_{k} \in \mathscr{N}

hence f^{-1}(\bigcup_{k=1}^{\infty}Y_{k}) \in f^{-1}(\mathscr{N})

Consequently, \bigcup_{k=1}^{\infty}f^{-1}(Y_{k}) \in f^{-1}(\mathscr{N}), which in turn implies that \mathscr{N} is a Borel algebra also. QED.

Part iv and part v:

Using parts i and iii above, these can be very easily proved.

Solutions 3b:

i) Prove: If \mathscr{M} is a ring, so is f(\mathscr{M}). Note that f: M \rightarrow N, y=f(x). Proof (i): As \mathscr{M} is a system of subsets of M. and it is a ring, so if A_{1} \in \mathscr{M}, and A_{2} \in \mathscr{M}, it implies that A_{1} \bigcup A_{2} \in \mathscr{M}; A_{1} \bigcap A_{2} \in \mathscr{M}; (A_{1}-A_{2}) \in \mathscr{M}. We know that f(A_{1}\bigcup A_{2}) = f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M}) and f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M}). And, quite obviously, f(A_{1}-A_{2}) =  f(A_{1}) - f(A_{2}) \in f(\mathscr{M}) so that it is true that if \mathscr{M} is a ring, then f(\mathscr{M}) is also a ring.

ii) Check if it is true or false: if \mathscr{M} is an algebra, so is f(\mathscr{M}). Argument: As \mathscr{M} is an algebra, it is already a ring. But additionally, \mathscr{M} contains an element E such that A_{1} \bigcap E= A_{1} (whenever A_{1} \in \mathscr{M} but f(A_{1} \bigcap E) \subset f(A_{1}) \bigcap f(E) \in f(\mathscr{M}) but it need not be true that f(E) is a unit element of f(\mathscr{M}). So, we can say that in case \mathscr{M} is an algebra, then f(\mathscr{M}) may or may not be an algebra. QED.

iii) Check if it is true or false: If \mathscr{M} is a Borel algebra, then so also f(\mathscr{M}) is also a Borel algebra. Argument: As in previous case, f(\mathscr{M}) may not have a unit element. So, f(\mathscr{M}) need not be a Borel algebra. QED.

iv) \mathscr{R}(f(\mathscr{M})) = f(\mathscr{R}(\mathscr{M})). Check if this is true or false. Argument: Let f(A_{1}) \in f(\mathscr{M}), f(A_{2}) \in \mathscr{M}. By definition of a ring, f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M}), f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M}), f(A_{1}) - f(A_{2}) \in f(\mathscr{M}), hence we get the following: f(A_{1} \bigcup A_{2}) \subset f(\mathscr{M}); f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M}) and f(A_{1}-A_{2}) = f(A_{1}) - f(A_{2}) \in f(\mathscr{M}). This in turn means that A_{1} \bigcup A_{2} \in \mathscr{M}, A_{1} \bigcap A_{2} \in \mathscr{M} and A_{1} - A_{2} \in \mathscr{M}. Hence, LHS \subset RHS. Similarly, we can show that RHS \subset LHS. QED.

v) \mathscr{B}(f(\mathscr(M))) = f(\mathscr{B}(\mathscr{M})). Check if this is true or false. (We will discuss this later; meanwhile, if you wish, please try as homework and let me know).

Cheers,

Nalin Pithwa.

Comments are most welcome.

Introductory Real Analysis: Exercise 3 solutions: related to set theory

Problem 1:

Exhibit both a partial ordering and a simple ordering on the set of complex numbers.

Solution I:

Recall the definition of a partial ordering: A binary relation R on a set M is said to be a partial ordering (and the set M itself is said to be partially ordered) if:

(i) R is reflexive (aRa for every a \in M) (ii) R is transitive (aRb and bRc together imply aRc) (iii) aRb and bRa together imply a=b (anti symmetric).

Recall the definition of a simple ordering or totally ordered or (just) ordered set: a set M is said to be ordered if it is a partially ordered set and if, given any two distinct elements a, b \in M, either a < b or b < a.

Let us consider the following relation R on the set of complex numbers, C:

consider any two elements z_{1}, z_{2} \in C such that z_{1}=x_{1}+i y_{1} and z_{2} = x_{2}=iy_{2}. We define a relation z_{1}Rz_{2} if and only if x_{1} \leq x_{2} and y_{1} \leq y_{2}. Clearly, this is both a partial ordering on C as well as total ordering on C.

Now consider the following relation K on the set of complex numbers:

We define the relation K as: z_{1}Kz_{2} if and only if \frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}} where x_{1}, y_{1}, x_{2}, y_{2} \in \Re. Clearly, if any one of y_{1} or y_{2} is zero, then the complex numbers z_{1} and z_{2} are non-comparable. So this could be a partial ordering but not a total ordering on the set of complex numbers.

Problem 2:

What is the minimal element of the set of all subsets of a given set X partially ordered by set inclusion. What is the maximal element?

Solution 2:

Recall the definition: An element “a” of a partially ordered set is said to be maximal if aRb implies b=a and minimal if bRa implies b=a.

Note that in the above definition, element “a” is our given/chosen element to be compared with other elements of the set. So, clearly, if the partial ordering is set inclusion, the minimal element is the null set \phi and the maximal element is the given set X itself.

Problem 3:

A partially ordered set M is said to be a directed set if, given any two elements a, b \in M, there is an element c \in M such that a \leq c and b \leq c. Are the following partially ordered sets directed sets also ?

PS: in the above we use the notation aRb or equivalently, a \leq b to mean one and the same thing.

Question 3a:  Any set M can be trivially partially ordered by setting a \leq b if and only if a=b.

Answer 3a: Clearly, this is a trivial directed set.

Question 3b: Let M be the set of all continuous functions f, g, \ldots defined in a closed interval [\alpha, \beta]. Then, we get a partial ordering by setting f \leq g if and only if f(t) \leq g(t) for every t \in [\alpha, \beta]. Is this also a directed set ?

Answer 3b: We need to choose two arbitrary elements say f_{1} and f_{2} of the given set of continuous functions on [\alpha, \beta]. Our desired “element” could be f_{3}= |f_{1}(t)|+|f_{2}(t)| again defined on [\alpha, \beta]. Then, the given set becomes a directed set.

Note that the function y=f(x) = |x| is continuous at zero also (but not differentiable at zero).

Question 3c:

Set inclusion.

Answer 3: Clearly, given any two arbitrary subsets, both are contained always in the given big set M or X. So, this is again a directed set.

Question 3d: The set of all integers greater than 1 is partially ordered if a \leq b means that “b is divisble by a.”

Answer 3: So let us consider two arbitrary positive integers, both greater than 1, call them a and b. We want to know if there exist a positive integer, also greater than 1, call it c such that a \leq c and b \leq c? That is, “c is divisible by a” and “c is divisible by b”. Clearly, c can be least common multiple of a and b. So, yes, this is also a directed set.

Problem 4:

By the greatest lower bound of two elements a and b of a partially ordered set M, we mean an element c \in M such that c \leq a and c \leq b and there is no element d \in M such that c < d \leq a, d \leq b. Similarly, by the least upper bound of a and b, we mean an element c \in M such that a \leq c, b \leq c, and there is no element d \in M such that a \leq d < c and b \leq d. By a lattice is meant a partially ordered set any two elements of which have both a greatest lower bound and a least upper bound. Prove that the set of all subsets of a given set M, partially ordered by set inclusion, is a lattice. What is the set theoretic meaning of the greatest lower bound and least upper bound of two elements of this set ?

Solution 4:

It can be easily checked that the null set is the greatest lower bound and the set M itself if the least upper bound. (PS: these are the minimal and maximal elements of this set M respectively.)

Problem 5:

Prove that an order-preserving mapping of one ordered set onto another is automatically an isomorphism.

Solution 5:

By definition.

Problem 6:

Prove that ordered sums and products of ordered sets are associative, that is, prove that if M_{1}, M_{2} and M_{3} are ordered sets, then

(M_{1}+M_{2})+M_{3}=M_{1}+(M_{2}+M_{3})

(M_{1}.M_{2}).M_{3} = M_{1}.(M_{2}.M_{3})

PS: comment: this allows us to drop the parentheses in writing ordered sums and products.

where the operations are as defined below:

Let M_{1} and M_{2} be two ordered sets of type \theta_{1} and \theta_{2} respectively. Then, we can introduce an ordering in the union M_{1} \bigcup M_{2} of the two sets by assuming that: (i) a and b have the same ordering as in M_{1} if a, b \in M_{1} (ii) a and b have the same orderiing as in M_{2} if a, b \in M_{2} (iii) a \leq b if a \in M_{1} and b \in M_{2}.

In this case/example, the ordered sum or union of three ordered sets M_{1}, M_{2}, M_{3} would mean that any two elements a, b would have the same ordering if a, b \in M_{1}, or a, b \in M_{2} or a,b \in M_{3}. Also, comparing with ordered sum of two ordered sets, we can say that a \leq b if and only if a \in M_{1}, b \in M_{2} or a \in M_{1}, b \in M_{3}, or a \in M_{2}, b \in M_{3}. In such a case, the ordered sum is also associative.

The operation of ordered product of two ordered sets is defined as follows: M_{1}. M_{2} is the set of all pairs (a,b) where a \in M_{1} and b \in M_{2} ordered in such a way that (i) (a_{1}, b_{1}) < (a_{2}, b_{2}) if b_{1} < b_{2} for arbitrary a_{1}, a_{2}, and (ii) (a_{1},b) < (a_{2}, b) if a_{1} < a_{2}.

So, if we have three ordered sets M_{1}, M_{2}, M_{3}, we can define their ordered product M_{1}.M_{2}.M_{3} as consisting of all ordered triples such that (i) (a_{1}, b_{1}, c_{1}) < (a_{2}, b_{2}, c_{2}) < (a_{3}, b_{3}, c_{3}) if and only if b_{1}<b_{2}<b_{3} and c_{1} < c_{2} < c_{3} for arbitrary a_{1}, a_{2}, a_{3}, and (ii) (a_{1}, b, c) < (a_{2}, b, c) < (a_{3}, b, c) if a_{1}<a_{2}<a_{3}; (a, b_{1}, c) < (a, b_{2}, c) < (a, b_{3}, c) if b_{1}<b_{2}<b_{3}; (a,b, c_{1})< (a,b,c_{2}) < (a,b, c_{3}) of c_{1} < c_{2} < c_{3}. In such a case, the product is well-defined and is also associative. (any comments ??)

Problem 7:

Construct well-ordered sets with ordinals

\omega + n, \omega + \omega, \omega + \omega + n, \omega + \omega + \omega, \ldots

Show that the sets are all countable.

Solution 7:

\omega + n: \{ 1,2,3,\ldots, a_{1}, a_{2}, \ldots, a_{n}\}. This is countable because there can be bijection from the set Z^{+} to the given set: 1 \rightarrow a_{1}, 2 \rightarrow a_{2}, \ldots, n \rightarrow a_{n}, n+1 \rightarrow 1, n+2 \rightarrow 2, \ldots

\omega + \omega: \{1,2,3, \ldots, 1,2,3, \ldots \}, this is also countable as it can be put in one-to-one correspondence with Z^{+}.

\omega + \omega + n; \{ 1,2, 3, \ldots, 1, 2, 3, \ldots, a_{1}, a_{2}, a_{3}, \ldots, a_{n}\}. This is also countable in the same manner as the first example.

\omega + \omega + \omega: \{ 1,2, 3, \ldots, 1,2, 3, \ldots, 1,2,3, \ldots\}: this can be put in one-to-one correspondence with Z^{+}

Problem 8:

Construct well-ordered sets with ordinals

\omega.n, \omega^{2}, \omega^{2}.n, \omega^{3}, \ldots.

Show that the sets are all countable.

Solution 8:

First let us recall the definition: An ordered set M is said to be well-ordered if every non empty subset A of M has a smallest (or, first) element, that is, an element \mu such that \mu < a for every a \in A.

Ans 1: \omega.n is set : M= \{(1,a_{1}), (2,a_{2}), (3, a_{3}), \ldots, (n, a_{n}), (n+1, a_{1}), (n+2, a_{2}), \ldots  \}. This set is ordered clearly by the definition of ordered product of two ordered sets; and this is well-ordered if we define the first element to be (1, a_{1}).

Ans 2: \omega^{2} = \omega. \omega: this set can be constructed as M= \{ (1,1), (1,2),(1,3), \ldots, (2,1), (2,2), (2,3), \ldots, \} and we can define the first element as (1,1) so that it becomes well-ordered under the usual definition of product of two ordered sets.

Ans 3: \omega^{2}.n = \omega. \omega. n: this set can be constructed as follows as an ordered product of three ordered sets: M = \{ (1,1,a_{1}), (1,1,a_{2}), (1,1,a_{3}), \ldots, (1,1,a_{n}), (1,2,a_{1}), (1,2,a_{2}), (1,2,a_{3}), \ldots, (1, 2, a_{n}), \ldots\}. We see that this is also well-ordered if we define the first element to be (1,1,a_{1}) and “order or list or count” them as shown. (any comments ? )

Ans 4: \omega^{3}= \omega.\omega.\omega: this can be well-ordered as above with clearly the first element to be (1,1,1). (any comments?)

PS: the explicit listing shows that all the above sets are clearly countable.

More later,

Cheers,

Nalin Pithwa

Exercises based on system of sets

Reference: Introductory Real Analysis, Kolomogorov and Fomin, Dover Publications, Translated and edited by Richard A. Silverman:

Problem 1:

Let X be an uncountable set, and \mathscr{R} be the ring consisting of all finite subsets of X and their complements. Is \mathscr{R} a \sigma -ring also?

Problem 2:

Are open intervals Borel sets ?

Problem 3:

Let y=f(x) be a function defined on a set M and taking values in a set N. Let \mathscr{M} be a system of subsets of M, and let f(\mathscr{M}) denote the system of all images f(A) of sets A \in \mathscr{M}. Moreover, let \mathscr{N} be a system of subsets of N, and let f^{-1}(\mathscr{N}) denote the system of all preimages of f^{-1}(B) of sets B \in \mathscr{N}. Prove that

(i) If \mathscr{N} is a ring, so is f^{-1}(\mathscr{N})

(ii) If \mathscr{N} is an algebra, so is f^{-1}()\mathscr{N}

(iii) If \mathscr{N} is a borel algebra, then so is f^{-1}(\mathscr{N})

(iv) \mathscr{R}(f^{-1}(\mathscr{N}))=f^{-1}(\mathscr{R}(\mathscr{N}))

(v) \mathscr{R}(f^{-1}(\mathscr{N}))=f^{-1}(\mathscr{R}(\mathscr{N})).

Which of these assertions remain true if \mathscr{N} os replaced by \mathscr{M} and f^{-1} by f?

Regards,

Nalin Pithwa

Introductory Real Analysis: System of Sets

Reference: Introductory Real Analysis by Kolomogorov and Fomin, Dover Pub. 

Section 4:System of Sets: 

Section 4.1 Rings of sets:

By a system of sets, we mean any set whose elements are themselves sets. Unless the contrary is explicitly stated, the elements of a given system of sets will be assumed to be certain subsets of some fixed set X. System of sets will be denoted by capital script letters like \mathscr{R}, \mathscr{S}, etc. Our chief interest will be systems of sets which have certain closure properties under some set operations.

DEFINITION 1:

A non-empty system of sets \mathscr{R} is called a ring of sets if A \triangle B \in \mathscr{R} and A \bigcap B \in \mathscr{R} whenever A \in \mathscr{R} and B \in \mathscr{R}.

Since

A \bigcup B = (A \triangle B) \triangle (A \bigcap B),

A - B = A \triangle (A \bigcap B)

we also have A \bigcup B \in mathscr{R} and A - B \in \mathscr{R} whenever A \in \mathscr{R} and B \in \mathscr{R}. Thus, a ring of sets is a system of sets closed under the operations of taking unions, intersections, differences, and symmetric differences. Clearly, a ring of sets is also closed under the operations of taking finite unions and intersections: \bigcup_{k=1}^{n}A_{k} and \bigcap_{k=1}^{n}A_{k}.

A ring of sets must contain the empty set \phi since A-A=\phi.

A set E is called the unit of a system of sets \mathscr{S} if E \in \mathscr{S} and A \bigcap E =A for every A \in \mathscr{S}. Clearly, E is unique (why?). Thus, the unit of \mathscr{S} is just the maximal set of \mathscr{S}, that is, the set containing all other sets of \mathscr{S}. A ring of sets with a unit is called an algebra of sets.

Example 1: Given a set A, the system \mathscr{M}(A) of all subsets of A is an algebra of sets, with unit E=A.

Example 2: The system \{ \phi, A\} consisting of the empty set \phi and any nonempty set A is an algebra of sets, with E=A.

Example 3: The system of all finite subsets of a given set A is a ring of sets. This ring is an algebra if and only if A itself is finite.

Example 4: The system of all bounded subsets of the real line is a ring of sets, which does not contain a unit.

Theorem 1:

The intersection \mathscr{R} = \bigcap_{\alpha} \mathscr{R}_{\alpha} of any set of rings is itself a ring.

Proof 1: This follows from definition 1. QED.

Theorem 2:

Given any nonempty system of sets \mathscr{S}, there is a unique ring \mathscr{P} containing \mathscr{S} and contained in every ring containing \mathscr{S}.

Proof 2:

If \mathscr{P} exists, then clearly \mathscr{P} is unique (why?). To prove the existence of \mathscr{P}, consider the union X = \bigcup_{A \in \mathscr{S}}A of all sets A belonging to \mathscr{S} and the ring \mathscr{M}(X) of all subsets of X. Let \Sigma be the set of all rings of sets contained in \mathscr{M}(X) and containing \mathscr{P}. Then, the intersection \mathscr{P}=\bigcap_{\mathscr{R} \in \Sigma}\mathscr{R} of all these rings clearly has the desired properties. In fact, \mathscr{P} obviously contains \mathscr{S}. Moreover, if \mathscr{R}^{*} is any ring containing \mathscr{S}, then the intersection \mathscr{R}=\mathscr{R}^{*} \bigcap \mathscr{M}(X) is a ring in \Sigma and hence, \mathscr{P} \subset \mathscr{R} \subset \mathscr{R}^{*}, as required. The ring \mathscr{P} is called the minimal ring generated by the system \mathscr{S}, and will henceforth be denoted by \mathscr{R}(\mathscr{S}). QED.

Remarks:

The set \mathscr{M}(X) containing \mathscr{R}(\mathscr{S}) has been introduced to avoid talking about the “set of all rings containing \mathscr{S}“. Such concepts as “the set of all sets,” “the set of all rings,” etc. are inherently contradictory and should be avoided. (Recall: Bertrand Russell’s famous set theory paradox).

Section 4.2:

Semirings of sets:

The following notion is more general than that of a ring of sets and plays an important role in a number of problems (especially in measure theory):

Definition 2:

A system of sets \mathscr{S} is called a semiring (of sets) if

i) \mathscr{S} contains the empty set \phi;

ii) A \bigcap B \in \mathscr{S} whenever A \in \mathscr{S} and B \in \mathscr{S}.

iii) If \mathscr{S} contains the sets A and A_{1} \in A, then A can be represented as a finite union A = \bigcup_{k=1}^{n}A_{k} …..(call this (I)) of pairwise disjoint sets of \mathscr{S}, with the given set A_{1}, as its first term.

Remarks:

The representation (I) is called a finite expansion of A, with respect to the sets A_{1}, A_{2}, \ldots, A_{n}.

Example 1:

Every ring of sets \mathscr{R} is a semiring, since if \mathscr{R} contains A and A_{1} \subset A, then A = A_{1}\bigcup A_{2} where A_{2}=A-A_{1} \in \mathscr{R}.

Example 2:

The set \mathscr{S} of all open intervals (a,b), closed intervals [a,b] and half-open intervals [a,b), including the empty interval (a,a) = \phi and the single element sets [a,a] = \{ a\} is a semiring but not a ring.

Lemma 1:

Suppose the sets A_{1}, A_{2}, \ldots, A_{n} where A_{1}, \ldots, A_{n} are pairwise disjoint subsets of A, all belong to a semiring S. Then, there is a finite expansion

A = \bigcup_{k=1}^{s}A_{k}, where s \geq n

with A_{1}, \ldots, A_{n} as its first terms, where A_{k} \in \mathscr{S}, A_{k} \bigcap A_{l}=\phi for all k, l = 1, \ldots, n.

Proof of lemma 1:

The lemma holds for n=1 by the definition of a semiring. Suppose the lemma holds for n=m, and consider m+1 sets A_{1}, \ldots, A_{m}, A_{m+1} satisfying the condition of the lemma. By hypothesis,

A = A_{1} \bigcup \ldots \bigcup A_{m} \bigcup B_{1} \bigcup \ldots \bigcup B_{p} where the sets A_{1}, \ldots, A_{m}, B_{1}, \ldots, B_{p} are pairwise disjoint subsets of A, all belonging to \mathscr{S}. Let B_{q1}=A_{m+1} \bigcap B_{q}

By the definition of a semiring, B_{q} = B_{q1} \bigcup \ldots \bigcup B_{qr_{q}} where the sets B_{qj} (j=1, \ldots, r_{q}) are pairwise disjoint subsets of B_{q}, all belonging to \mathscr{S}. But then it is easy to see that

A = A_{1}\bigcup \ldots A_{m} \bigcup A_{m+1} \bigcup \bigcup_{q=1}^{p}(\bigcup_{j=2}^{r_{q}}B_{qj})

that is, the lemma is true for n = m+1. The proof now holds true by mathematical induction. QED.

Lemma 2:

Given any finite system of sets A_{1}, \ldots, A_{n} belonging to a semiring \mathscr{S}, there is a finite system of pairwise disjoint sets B_{1}, \ldots, B_{t} belonging to \mathscr{P} such that every A_{k} has a finite expansion A_{k}= \bigcup_{s \in M_{k}}B_{s} where k=1,2,\ldots, n with respect to certain of the sets B_{s}. (Note: Here M_{k} denotes some subset of the set \{ 1,2, \ldots, t\} depending on the choice of k).

Proof of Lemma 2:

The lemma is trivial for n=1 since we only need to set t=1 and B_{1}=A_{1}.

Let the lemma be true for n=m and consider a system of sets A_{1}, \ldots, A_{m}, A_{m+1} in \mathscr{P}. Let B_{1}, \ldots, B_{t} be sets of \mathscr{S} satisfying the conditions of the lemma with respect to A_{1}, A_{2}, \ldots, A_{m}, and let B_{s1}=A_{m+1}\bigcap B_{s}.

Then, by Lemma 1, there is an expansion

A_{m+1}=(\bigcup_{s=1}^{t}B_{s1})\bigcup(\bigcup_{p=1}^{q}B_{p}^{'})

where B_{p}^{'} \in \mathscr{S}. while, by definition of a semiring, there is an expansion such that

B_{s}=B_{s1}\bigcup B_{s2} \bigcup \ldots \bigcup B_{sr}, where B_{sj} \in \mathscr{S}.

It is obvious that A_{k}= \bigcup_{s \in M_{k}}(\bigcup_{j=1}^{r_{s}}B_{sj}) where k=1,2, …, m.

for some suitable M_{k}. Moreover, the sets B_{sj}, B_{p}^{'} are pairwise disjoint. Hence, the sets B_{sj}, B_{p}^{'} satisfy the conditions of the lemma with respect to A_{1}, \ldots, A_{m}, A_{m+1}. The proof now follows by mathematical induction. QED.

Section 4.3:

The ring generated by a semiring:

According to Theorem 1, there is a unique minimal ring \mathscr{R}(\mathscr{S}) generated by a system of sets \mathscr{S}. The actual construction of \mathscr{R}(\mathscr{S}) is quite complicated for arbitrary \mathscr{S}. However, the construction is completely straightforward if \mathscr{S} is a semiring, as shown by

Theorem 3:

If \mathscr{S} is a semiring, then \mathscr{R}(\mathscr{S}) coincides with the system \mathscr{Z} of all sets A which have finite expansions

A = \bigcup_{k=1}^{n}A_{k}

with respect to the sets A_{k} \in \mathscr{S}.

Proof of Theorem 3:

First we prove that \mathscr{Z} is a ring. Let A and B be any two sets in \mathscr{Z}. Then, there are expansions

A = \bigcup_{i=1}^{m}A_{i} where A_{i} \in \mathscr{S}

B = \bigcup_{j=1}^{n}B_{j} where B_{i} \in \mathscr{S}

Since \mathscr{S} is a semiring, the sets C_{ij}=A_{i} \bigcap B_{j} also belong to \mathscr{S}. By Lemma 1, there are expansions as follows:

A_{i}=(\bigcup_{j=1}^{n}C_{ij})\bigcup(\bigcup_{k=1}^{r_{i}}D_{ik}), where D_{ik} \in \mathscr{S}

B_{j}=(\bigcup_{r=1}^{m}C_{ij})\bigcup(\bigcup_{l=1}^{s_{j}}E_{jl}), where E_{jl} \in \mathscr{S}

Let us call the above two relations as (II).

It follows from (II) that A \bigcap B and A \triangle B have the expansions

A \bigcap B = \bigcup_{i,j}C_{ij}

A \triangle B = (\bigcup_{i,k}D_{ik})\bigcup (\bigcup_{j,l}E_{jl}).

and hence, belong to \mathscr{Z}. Therefore, \mathscr{Z} is a ring. The fact that \mathscr{Z} is a minimal ring generated by \mathscr{S} is obvious. QED.

Section 4.4

Borel Algebras:

There are many problems (particularly in measure theory) involving unions and intersections not only of a finite number of sets, but also of a countable number of sets. This motivates the following concepts:

Definition 3: 

\sigma - ring and \sigma - algebra:

A ring of sets is called a \sigma -ring if it contains the union S = \bigcup_{n=1}^{\infty}A_{n} whenever it contains the sets A_{1}, A_{2}, \ldots, A_{n}, \ldots.

A \sigma -ring with a unit E is called a \sigma – algebra.

Definition 4:

\delta -ring and \delta– algebra:

A ring of sets is called a \delta -ring if it contains the intersection D = \bigcap_{n=1}^{\infty}A_{n}

whenever it contains the sets A_{1}, A_{2}, \ldots, A_{n}, \ldots.

A \delta -ring with a unit E is called a \delta – algebra.

Theorem 4:

Every \sigma-algebra is a \delta-algebra and conversely.

Proof of theorem 4:

These are immediate consequences of the “dual” formulae:

\bigcup_{n}A_{n}=E - \bigcap_{n}(E-A_{n})

\bigcap_{n}A_{n}= E -\bigcup_{\alpha}(E-A_{n}).

QED.

The term Borel algebra or briefly B-algebra is often used to denote a \sigma -algebra (equivalently, a \delta-algebra). The simplest example of a B-algebra is the set of all subsets of a given set A.

Given any system of sets \mathscr{S}, there always exists at least one B-algebra containing \mathscr{S}. In fact, let

X = \bigcup_{A \in \mathscr{S}}A

Then, the system \mathscr{B} of all subsets of X is clearly a B-algebra containing \mathscr{S}.

If \mathscr{B} is any Borel-algebra containing \mathscr{S} and if E is its unit, then every A \in \mathscr{S} is contained in E and hence,

X = \bigcup_{A \in \mathscr{s}}A \subset E.

A borel-algebra \mathscr{B} is called irreducible (with respect to the system \mathscr{S}) if X=E, that is, an irreducible Borel-algebra is a Borel-algebra containing no points that do not belong to one of the sets A \in \mathscr{S}. In every case, it will be enough to consider only irreducible Borel-algebras.

Theorem 2 has the following analogue for irreducible Borel-algebras:

Theorem 5: 

Given any non empty system of sets \mathscr{S}, there is a unique irreducible (to be precise, irreducible with respect to \mathscr{S}) B-algebra \mathscr{B}(\mathscr{S}) containing \mathscr{S} and contained in every B-algebra containing \mathscr{S}.

Proof of theorem 5:

The proof is virtually identical with that of Theorem 2. The B-algebra \mathscr{B}(\mathscr{S}) is called the minimal B-algebra generated by the system \mathscr{S} or the Borel closure of \mathscr{S}.

Remarks:

An important role is played in analysis by Borel sets or B-sets. These are the subsets of the real line belonging to the minimal B-algebra generated by the set of all closed intervals [a,b].

Exercises to follow,

Regards,

Nalin Pithwa

Introductory real analysis: exercise 3:

Based on previous two blogs. (Reference: Introductory Real Analysis by Kolmogorov and Fomin, Dover Publishers):

Problem 1:Exhibit both a partial ordering and a simple ordering of the set of all complex numbers.

Problem 2:What is the minimal element of the set of all subsets of a given set X, partially ordered by set inclusion. What is the maximal element?

Problem 3: A partially ordered set M is said to be a directed set if, given any two elements a, b \in M, there is an element c \in M such that a < c, b<c. Are the partially ordered sets in the previous blog(s) Section 3.1 all directed sets?

Problem 4: By the greatest lower bound of two elements a and b of a partially ordered set M, we mean an element c \in M such that c \leq a, c \leq b and there is no element d \in M such that a < d \leq a, d \leq b. Similarly, by the least upper bound of a and b, we mean an element c \in M such that a \leq c, b \leq c and there is no element d \in M such that a \leq d <c, b \leq d. By a lattice is meant a partially ordered set any two elements of which have both a greatest lower bound and a least upper bound. Prove that the set of all subsets of a given set X, partially ordered by set inclusion, is a lattice. What is the set theoretic meaning of the greatest lower bound and least upper bound of two elements of this set?

Problem 5: Prove that an order preserving mapping of one ordered set onto another is automatically an isomorphism.

Problem 6: Prove that ordered sums and products of ordered sets are associative, that is, prove that if M_{1}, M_{2}, M_{3} are ordered sets, then

(M_{1}+M_{2})+M_{3}=M_{1}+(M_{2}+M_{3}),

(M_{1}.M_{2}).M_{3}=M_{1}.(M_{2}.M_{3}) where the operations + and . are the same as defined in previous blog(s).

Comment: This allows us to drop the parentheses in writing ordered sums and products.

Problem 7: 

Construct well-ordered sets with ordinals

\omega + n, \omega + \omega, \omega + \omega + n, \omega + \omega + \omega, \ldots.

Show that the sets are all countable.

Problem 8:

Construct well-ordered sets with ordinals

\omega . n, \omega^{2}, \omega^{2}.n, \omega^{3}, \ldots.

Show that the sets are all countable.

Problem 9:

Show that \omega + \omega  = \omega. 2, \omega + \omega + \omega = \omega. 3, \ldots

Problem 10:

Prove that the set W(\alpha) of all ordinals less than a given ordinal \alpha is well-ordered.

Problem 11:

Prove that any non-empty set of ordinals is well-ordered.

Problem 12:

Prove that the set M of all ordinals corresponding to a countable set is itself uncountable.

Problem 13:

Let \aleph_{1} be the power of the set M in the previous problem. Prove that there is no power m such that \aleph_{0} <m< \aleph_{1}.

More later,

Nalin Pithwa.

 

 

 

 

 

 

Introductory Real Analysis: The well-ordering theorem, the axiom of choice,equivalent assertions

Reference: Introductory Real Analysis by Kolmogorov and Fomin, Dover Publications.

Section 3.7 The well-ordering theorem, the axiom of choice, and equivalent assertions:

Theorem 4 of previous blog shows that the powers of two well-ordered sets are always comparable. In 1904, Zermelo succeeded in proving the

Well-ordering theorem: Every set can be well-ordered.

It follows from the well-ordering theorem and Theorem 5 (of previous blog) that the powers of two arbitrary sets are always comparable, a fact already used earlier. Zermelo’s proof, which will not be given here, rests on the following basic:

AXIOM OF CHOICE:

Given any set M, there is a “choice function” f such that f(A) is an element of A for every non-empty subset A \subset M.

We will assume the validity of the axiom of choice without further ado. In fact, without the axiom of choice we would be severely hampered in making set-theoretic constructions. However, it should be noted that from the standpoint of the foundations of set theory, there are still deep and controversial problems associated with the use of axiom of choice.

There are a number of assertions equivalent to the axiom of choice, that is, assertions each of which both implies and is implied by the axiom of choice. One of these is the well-ordering theorem, which obviously implies the axiom of choice. In fact, if an arbitrary set M can be well-ordered, then, by merely choosing the “first” element in each subset A \subset M, we get the function f(A) figuring in the statement of the axiom of choice. On the other hand, the axiom of choice implies the well-ordering theorem, as already noted without proof.

To state further assertions equivalent to the axiom of choice, we need some more terminology:

DEFINITION 3:

Let M be a partially ordered set, and let A be any subset of M such that a and b are comparable for every a, b \in A. Then, A is called a chain (in M). A chain C is said to be maximal if there is no other chain C in M containing C on a proper subset.

DEFINITION 4:

An element a of a partially ordered set M is called an upper bound of a subset M^{'} \subset M, if a^{'} < a for every a^{'} \in M^{'}.

We now have the vocabulary needed to state two other assertions equivalent to the axiom of choice:

Hausdorff’s Maximal Principle: Every chain in a partially ordered set M is contained in a maximal chain in M. 

Zorn’s lemma:  If every chain in a partially ordered set M has an upper bound, then M contains a maximal element. 

For the proof of the equivalence of the axiom of choice, the well-ordering theorem, Hausdorff’s maximal principle and Zorn’s lemma, we refer the reader elsewhere. Of these various equivalent assertions, Zorn’s lemma is perhaps the most useful.

Section 3.8 Transfinite Induction:

Mathematical propositions are very often proved by using the following familiar:

THEOREM 4: MATHEMATICAL INDUCTION:

Given a proposition P(n) formulated for every positive integer n, suppose that :

i) P(1) is true.

ii) The validity of P(k) for all k<n implies the validity of P(n+1). Then, P(n) is true for all n=1,2,\ldots.

PROOF 4:

Suppose P(n) fails to be true for all n=1,2,\ldots and let n, be the smallest integer for which P(n) is false (the existence of such n_{1} follows from the well-ordering of the positive integers). Clearly, n_{1}>1, so that n_{1}-1 is a positive integer. Therefore, P(n) is valid for all k \leq n_{1}-1 but not for n_{1}. Contradiction!!

QED.

Replacing the set of all positive integers by an arbitrary well-ordered set, we get

THEOREM 4 (Transfinite induction):

Given a well-ordered set A, let P(a) be a proposition formulated for every element a \in A. Suppose that

i) P(a) is true for the smallest element of A

ii) The validity of P(a) for all a < a^{*} implies the validity of P(a^{*}). Then, P(a) is true for all a \in A.

PROOF 4:

Suppose P(a) fails to be true for all a \in A. Then, P(a) is false for all a in some non empty subset A^{*} \subset A. By the well-ordering, A^{*} has a smallest element a^{*}. Therefore, P(a) is valid for all a< a^{*} but not for a^{*}. Contradiction !!

QED.

Remark:

Since any set can be well-ordered, by the well-ordering theorem, transfinite induction can in principle be applied to any set M whatsoever. In practice, however, Zorn’s lemma is a more useful tool, requiring only that M be partially ordered.

Section 3.9: Historical Remarks:

Set theory as a branch of mathematics in its own right stems from the pioneer work of Georg Cantor (1845-1918). Originally met with disbelief, Cantor’s ideas subsequently became widespread. By now, the set theoretic point of view has become standard in the most diverse fields of mathematics. Basic concepts, like groups, rings, fields, linear spaces etc are habitually defined as sets of elements of an arbitrary kind obeying appropriate axioms.

Further development of set theory led to a number of logical difficulties, which naturally gave rise to attempts to replace “naive” set theory by a more rigorous axiomatic set theory. It turns out that certain set theoretic questions, which would at first seem to have “yes” or “no” answers, are in fact of a different kind. Thus, it was shown by Godel in 1940 that a negative answer to the question :”Is there an uncountable set of power less than that of the continuum” is consistent with set theory (axiomatized in a way we will not discuss here), but it was recently shown by Stephen Cohen that an affirmative answer to the question is also consistent in the same sense !!

We will continue to exercises based on last two blogs now,

Regards,

Nalin Pithwa.

Ordered Sets and Ordinal Numbers: Introductory Real Analysis: Kolmogorov and Fomin

Section 3: Ordered sets and ordinal numbers:

Section 3.1: Partially ordered sets.

A binary relation R on a set M is said to be a partial ordering (and the set M is said to be partially ordered) if

i) R is reflexive (aRa for every a \in M).

ii) R is transitive (aRb and bRc together imply aRc)

iii) R is antisymmetric in the sense that aRb and bRa together imply a=b.

For example, if M is the set of all real numbers and aRb means that a<b, then R is partial ordering. (please verify!) This suggests writing a \leq b (or equivalently b \geq a) instead of aRb whenever R is a partial ordering, and we will do so from now on. Similarly, we write a < b if a \leq b, but a \neq b; and b > a if b \geq a but b \neq a.

The following examples give some idea of the generality of the concept of a partial ordering:

Example 1:

Any set M can be trivially partially ordered by setting a \leq b iff a=b.

Example 2:

Let M be the set of all continuous functions f, g, \ldots defined in a closed interval [\alpha, \beta]. Then, we get a partial ordering by setting f \leq g iff f(t) \leq g(t) for every t \in [\alpha, \beta].

Example 3:

The set of all subsets M_{1}, M_{2}, \ldots is partially ordered if and only if M_{1} \leq M_{2} means that M_{1} \subset M_{2}.

Example 4:

The set of all integers greater than 1 is partially ordered if “a \leq b” means that “b is divisible by a.”

Remark: An element of a of a partially ordered set is said to be maximal if a \leq b implies b=a and minimal if b \leq a implies b=a. Thus, in Example 4 above, every prime number (greater than 1) is a minimal element. (Of course, 1 is neither prime nor composite.)

Section 3.2: Order-preserving mappings. Isomorphisms:

Let M and M^{'} be any two partially ordered sets, and let f be a one-to-one mapping of M onto M^{'}. Then, f is said to be order-preserving if a \leq b (where a, b \in M) implies f(a) \leq f(b) (in M^{'}). An order preserving mapping f such that f(a) \leq f(b) implies a \leq b is called an isomorphism. In other words, an isomorphism between two partially ordered sets M and M^{'} is a one-to-one mapping of M onto M^{'} such that f(a) \leq f(b) if and only if a \leq b. Two partially ordered sets M and M^{'} are said to be isomorphic (to each other) if there exists and isomorphism between them.

Example:

Let M be the set of positive integers greater than 1 partially ordered as in example 4 earlier. Let M^{'} be the same set partially ordered in the natural way, that is, in such a way that a \leq b if and only if b-a >0. Then, the mapping of M onto M^{'} carrying every integer n into itself is order-preserving, but NOT an isomorphism.

Some remarks:

Isomorphism between partially ordered sets is an equivalence relation as defined in an earlier blog of this series, being obviously reflexive, symmetric, and transitive. Hence, any given family of partially ordered sets can be partitioned into disjoint classes of isomorphic sets. Clearly, two isomorphic partially ordered sets can be regarded as identical in cases where it is the structure of the partial ordering than the specific nature of the elements of the sets that is of interest.

Some other remarks:

For those of you who are aware, compare now the concepts and some examples of homomorphisms and isomorphisms between two algebraic objects called groups with the above concepts.  Two references for the same are: (i) Topics in Algebra by I N Herstein (2) Abstract Algebra by Dummit and Foote.

Some further remarks:

At this stage, it might even be useful to compare the concepts and examples of a cartesian product of two (non-empty) sets A and B; that of a relation R on set A and set B; that of a binary relation on a set A; that, of an equivalence relation on a set A; and also the basic meaning of a function from set A to set B. You will see that there are slight subtle differences.

Sec 3.3 Ordered Sets. Order Types:

Given two elements a and b of a partially ordered set M, it may turn out that neither of the relations a \leq b or b \leq a holds. In this case, a and b are said to be noncomparable. Thus, in general, the relation \leq is defined only for certain pairs of elements, which is why M is said to be partially ordered. However, if M has no non-comparable elements, then M is said to be simply ordered or linearly ordered or ordered. In other words, a set M is ordered if it is partially ordered and if, given any two distinct elements a, b \in M, either a <b, or b<a. Obviously, any subset of an ordered set is itself ordered.

Each of the sets figuring in Examples 1 to 4 of Section 3.1 in this blog, is partially ordered, but not ordered. Simple examples of ordered sets are the set of all positive integers, the set of all rational numbers, the set of all real numbers in the [0,1] and so on (with the usual relations of “greater than” and “less than”.)

Since an ordered set is a special kind of partially ordered set, the concepts of order-preserving mapping and isomorphism apply equally well to ordered sets. Two isomorphic ordered sets are said to have the same order type. Thus “type” is something shared by all isomorphic ordered sets, just as “power” is something shared by all equivalent sets (considered as ‘plain’ sets, without regard for possible orderings).

The simplest example of an ordered set is the set of all positive integers 1,2,3,\ldots arranged in increasing order, with the usual meaning of the symbol <. The order type of this set is denoted by the symbol \omega. Two isomorphic ordered sets obviously have the same power (because an isomorphism is a one-to-one correspondence). Thus, it makes sense to talk about the power corresponding to a given order type. The converse is not true, since a set of a given power can in general be ordered in many different ways. It is only in the finite case that the number of elements in a set uniquely determines its type, designated by the same symbol n as the number of elements in the set. For example, besides the “natural” order type \omega of the set of positive integers, there is another order type corresponding to the sequence: 1,3,5, \ldots, 2,4,6, \ldots where odd and even numbers are separately arranged in increasing order but any odd number precedes any even number. It can be shown that the number of distinct order types of a set of power \aleph_{0} is infinite and in fact, uncountable.

Section 3.4 Ordered sets and products of ordered sets. 

Let M_{1} and M_{2} be two ordered sets of types \theta_{1} and \theta_{2} respectively. Then, we can introduce an ordering in the union M_{1} \bigcup M_{2} of the two sets by assuming that :

(i) a and b have the same ordering as in M_{1} if a,b \in M_{1}.

(ii) a and b have the same ordering as in M_{2} if a,b \in M_{2}

(iii) a <b if a \in M_{1} and b \in M_{2}.

(Verify that this is actually an ordering of M_{1} \bigcup M_{2}). The set M_{1} \bigcup M_{2} ordered in this way is called the ordered sum of M_{1} and M_{2} denoted by M_{1}+M_{2}. Note that the order of terms matters here, that is, in general M_{2}+M_{1} is not isomorphic to M_{1}+M_{2}. More generally, we can define the ordered sum of any finite number of ordered sets by writing (this will be an exercise problem after this blogged section) :

M_{1}+M_{2}+M_{3}=(M_{1}+M_{2})+M_{3}

M_{1}+M_{2}+M_{3}+M_{4}=(M_{1}+M_{2}+M_{3})+M_{4},

and so on.

By the ordered sum of the types \theta_{1}+\theta_{2}, denoted by \theta_{1}+\theta_{2} we mean the order type of the set M_{1}+M_{2}.

Example:

Consider the order types \omega and n. It is easy to see that n + \omega = \omega. In fact, if finitely many terms are written to the left of the sequence 1,2,3,\ldots, k, \ldots, we again get a set of the same type (Quiz: why?). On the other hand, the order type \omega + n, that is the order type of the set (***) :

\{ 1,2,3, \ldots, k, \ldots, a_{1}, a_{2}, \ldots, a_{n}\} is obviously not equal to \omega.

(Note: ***: In the above listing of elements of the set, we have used the usual curly braces notation, but with the emphasis on the order as shown.)

Again, let M_{1} and M_{2} be two ordered sets of types \theta_{1} and \theta_{2} respectively, Suppose we replace each element of M_{2} by a “replica” of the set M_{1}. Then, the resulting set, denoted  by M_{1}.M_{2} is the set of all pairs (a,b) where a \in M_{1} and a \in M_{2} ordered in such a way that:

(i) (a_{1},b_{1})<(a_{2},b_{2}) if b_{1}<b_{2} for arbitrary a_{1},a_{2}

(ii) (a_{1},b)<(a_{2},b) if a_{1}<a_{2}.

Note that the order of factors matters here, that is, in general M_{2}.M_{1} is not isomorphic to M_{1}.M_{2}. The ordered product of any finite number of ordered sets can be defined by writing (again a problem in the exercise later):

M_{1}.M_{2}.M_{3}=(M_{1}.M_{2}).M_{3}

M_{1}.M_{2}.M_{3}.M_{4}=(M_{1}.M_{2}.M_{3}).M_{4}

and so on. By the ordered product of the types \theta_{1} and \theta_{2}, denoted by \theta_{1}.\theta_{2}, we mean the order type of the set M_{1}.M_{2}.

Section 3.5. Well-ordered sets. Ordinal numbers.

A key concept in the theory of ordered sets is given by:

DEFINITION 1: An ordered set M is said to be well-ordered if every non-empty subset A of M has a smallest (or, “first“) element, that is, an element \mu such that \mu <a for every a \in A.

Example 1: Every finite ordered set is obviously well-ordered.

Example 2: Every non-empty subset of a well-ordered set is itself well-ordered.

Example 3: The set M of numbers in the interval [0,1] is ordered but not well-ordered. It is true that M has a smallest element, namely, the number 0 but the subset of M consisting of all positive rational numbers has no smallest element.

DEFINITION 2: The order type of a well-ordered set is called an ordinal number or simply, an ordinal. If the set is infinite, the ordinal is said to be transfinite. (NB: This is a good place to point out that the terms “cardinal number” and “power” of a set are synonymous).

Example 4: The set of positive integers 1,2,\ldots, k, \ldots arranged in increasing order is well-ordered, and hence, its order type \omega is a transfinite ordinal. The order type \omega + n of the set \{1,2,\ldots, k, \ldots, a_{1}, a_{2}, \ldots, a_{n} \} is also an ordinal.

Example 5: The set \{ \ldots, -k, \ldots, -3,-2,-1\} is ordered but not well-ordered. It is true that any non empty subset A of this set has a largest element (that is, an element \mathscr{v} such that a <mathscr{v}) for every a \in A), but in general set A will not have a smallest element. In fact, the given set itself has no smallest element. Hence, the order type of given set is denoted by \omega^{*} is not an ordinal number.

THEOREM 1The ordered sum of a finite number of well-ordered sets M_{1}, M_{2}, \ldots, M_{n} is itself a well-ordered set.

PROOF 1: Let M be an arbitrary subset of the ordered sum M_{1}+M_{2}+\ldots+M_{n} and let M_{k} be the first of the sets M_{1}, M_{2}, \ldots, M_{n}, (namely, the set with the smallest index) containing elements of M. Then, M \bigcap M_{k} is a subset of the well-ordered set M_{k} and as such has a smallest element \mu. Clearly, \mu is the smallest element of M itself. QED.

COROLLARYThe ordered sum of a finite number of ordinal numbers is itself an ordinal number. 

Thus, new ordinal numbers can be constructed from any given set of ordinal numbers. For example, starting from the positive integers, (that is, the finite ordinal numbers) and the ordinal number \omega, we can construct the new ordinal numbers :

\omega + n, \omega + \omega, \omega + \omega + n, \omega + \omega + \omega, and so on.

THEOREM 2The ordered product of two well-ordered sets M_{1}, M_{2} is itself a well-ordered set. 

Proof 2:

Let M be an arbitrary subset of M_{1}\dot M_{2} so that M is a set of ordered pairs (a,b) with a \in M_{1}, b\in M_{2}. The set of all second elements b of pairs in M is a subset of M_{2}, and as such has a smallest element since M_{2} is well-ordered. Let b_{1} denote this smallest element, and consider all pairs of the form (a,b_{1}) contained in M. The set of all first elements a of these pairs is a subset of M_{1}, and as such has a smallest element since M_{1} is well-ordered. Let a_{1} denote this smallest element. Then, the pair (a,b) is clearly the smallest element of M. QED.

COROLLARY 1:

The ordered product of a finite number of well-ordered sets is itself a well-ordered set.

COROLLARY 2:

The ordered product of a finite number of ordinal numbers is itself an ordinal number.

Thus it makes sense to talk about ordinal numbers \omega . n, \omega^{2}, \omega^{2} . n, \omega^{3}, \ldots and so on. It is also possible to define such ordinal numbers as \omega^{\omega}, \omega^{\omega^{\omega}}, \ldots.

Section 3.6 Comparison of ordinal numbers:

If n_{1} and n_{2} are two ordinal numbers, then they either coincide or else one is larger than the other. As we now show, the same is true for transfinite ordinal numbers. We begin by observing that every element a of a well-ordered set M determines an (initial section) P, the set of all x \in M such that x<a, and a remainder Q, the set of all x \in M such that x \geq a. Given any two ordinal numbers \alpha and \beta, Let M and N be well-ordered sets of order type \alpha and \beta. Then, we say that

i) \alpha=\beta if M and N are isomorphic;

ii) \alpha < \beta if M is isomorphic to some section of N

iii) \alpha > \beta if N is isomorphic to some section of M.

(note that this definition makes sense for finite \alpha and \beta).

LEMMA:

Let f be an isomorphism of a well-ordered set A onto some subset B \subset A. Then, f(a) \geq a for all a \in A.

Proof of lemma:

If there are elements a \in A such that f(a)<a, then there is a least such element since A is well-ordered. Let a_{0} be this element, and let b_{0}=f(a_{0}). Then b_{0}<a_{0}, and hence, f(b_{0})<f(a_{0})=b_{0}, since f is an isomorphism. But, then a_{0} is not the smallest element such that f(a)<a. Contradiction! QED.

It follows from the lemma that a well-ordered set A cannot be isomorphic to any of its sections, since if A were isomorphic to the section determined by a, then clearly f(a)<a. In other words, the two relations

\alpha=\beta, \alpha<\beta are incompatible, and so are \alpha=\beta, \alpha>\beta. Moreover, the two relations \alpha<\beta, \alpha>\beta are incompatible, since otherwise we could use the transitivity to deduce \alpha < \alpha, which is impossible the lemma. Therefore, if one of the three relations

\alpha<\beta, \alpha=\beta, \alpha>\beta….II

holds, the other two are automatically excluded. We must still show that one of the relations in II above always holds, thereby proving that any two ordinal numbers are always comparable.

THEOREM 3:

Two given ordinal numbers \alpha and \beta always satisfy one and only of the relations: \alpha=\beta, \alpha < \beta and \beta<\alpha.

Proof 3:

Let W(\alpha) be the set of all ordinals <\alpha. Any two numbers \gamma and \gamma^{'} in W(\alpha) are comparable (Recall the meaning of \gamma < \alpha, \gamma^{'}<\alpha and use the fact that a section of a section of a well-ordered set is itself a section of a well-ordered set) and the corresponding ordering of W(\alpha) makes it a well-ordered set of type \alpha. In fact, if a set A = \{ \ldots, a, \ldots, b, \ldots\} is of type \alpha, then by definition, the ordinals less than \alpha are the types of well-ordered sets isomorphic to sections of A. Hence, the ordinals themselves are in one-one correspondence with the elements of A. In other words, the elements of a set of type \alpha can be numbered by using the ordinals less than \alpha: A = \{ a_{1}, a_{2}, \ldots, a_{n}, \ldots\}.

Now, let \alpha and \beta be two ordinals. Then, W(\alpha) and W(\beta) are well-ordered sets of types \alpha and \beta, respectively. Moreover, let C=A \bigcap B be the intersection of the sets A and B, that is, the set of all ordinals less than both \alpha and \beta. Then, C is well-ordered, of type \gamma, say. We now show that \gamma \leq \alpha. If C=A, then obviously \gamma=\alpha. On the other hand, if C \neq A, then C is a section of A and hence, \gamma < \alpha. In fact, let \xi \in C, \eta \in A - C. Then, \xi and \eta are comparable, that is, either \xi < \eta or \xi > \eta. But, \eta < \xi < \alpha is impossible, since then \eta \in C. Therefore, \xi < \eta and hence, C is a section of A, which implies \gamma < \alpha. Moreover, \gamma is the first element of the set A-C. Thus, \gamma < alpha, as asserted, and similarly, \gamma < \beta. The case \gamma < \alpha, \gamma < \beta is impossible, since then \gamma \in A-C and \gamma \in B-C. But then \gamma \notin C on the one hand and \gamma \in A \bigcap B = C on the other hand. It follows that there are only three possibilities:

\gamma=\alpha and \gamma = \beta and \alpha=\beta

\gamma=\alpha and \gamma< \beta and \alpha < \beta

\gamma < \alpha and \gamma = \beta and \alpha > \beta,

that is, \alpha and \beta are comparable.

QED.

THEOREM 4:

Let A and B be well-ordered sets. Then, either A is equivalent to B or one of the sets is of greater power than the other, that is, the powers of A and B are comparable.

PROOF 4:

There is a definite power corresponding to each ordinal. But we have just seen that ordinals are comparable, and so are the corresponding powers.

[[recall the definition of inequality of powers given in Sec 2.5 in the blog series of Kolmogorov and Fomin). (Reproduced here for the sake of helping the memory: Given any two sets M and N, suppose M and N are equivalent. Then, M and N are said to have the same power. Roughly speaking, “power” is something shared by equivalent sets. If M and N are finite, then M and N have the same number of elements, and the concept of the power of a set reduces to the usual notion of the number of elements in a set. The power of the set \mathcal{Z_{+}} is \aleph_{0} and sets which are equivalent to the set [0,1] are said to have the power of the continuum, c. For the powers of finite sets, that is, for the set of positive integers, we have the notions of “greater than” or “less than”, as well as the notion of equality. These concepts can be extended to infinite sets. If A and B are two sets with powers m(A) and m(B) respectively, either m(A)=m(B), or A is equivalent to a subset of B such that no subset of B is equivalent to A, then by analogy with the final case, m(A) is less than m(B).]]

To be continued soon,

Regards,

Nalin Pithwa