User:Arbowen1/Some Elementary Transfinite Arithmetic

SOME ELEMENTARY TRANSFINITE ARITHMETIC

Author: ArB Owen

ABSTRACT: This article features some familiar elementary results of transfinite arithmetic proved in an unfamiliar way, along with:

i. A definition of ``infinite" set which allows us to prove that the cardinal of $ \mathbb{N}$ is the ``least" transfinite cardinal.

ii. Defining cardinal multiplication in terms of $ ``\bigcup" $ instead of $``\times" $. Which provides:

iii. A ``non-diagonal" proof that $\aleph_o^2 = \aleph_o$.

This article is not intended to be a comprehensive introduction to cardinal arithmetic, just some interesting results that can be obtained without ordinal arithmetic. To begin the article,we observe that transfinite(infinite) numbers behave very

differently than real numbers. For example, as will be elaborated upon later, sums and products, of two transfinite numbers are always just the larger of the two. A definition of ``larger" will be presented later. In this discussion, the usual ``college Algebra" conception of set theory along with the axiom of choice, will be assumed. First, a definition of ``infinite set"

DEFINITION 1: The set E is infinite if and only if there is a 1-1 function from E onto a proper subset of E.

Example 1: The function f(n)=n+1 is a 1-1 function from the set of natural numbers $\mathbb{N}$ onto the set $\mathbb{N}-\{1\}$. Thus, $\mathbb{N}$ is an infinite set.

THEOREM 0: $\emptyset\neq A \subseteq \mathbb{N} $ is infinite iff A is not bounded. That is, for r $\in \mathbb{R} $ there is n $\in A $ such that n $ > $ r.

PROOF: If A is bounded there is an r $\in \mathbb{R} $ such that r $ \geq $ n for all n $\in $ A, then clearly A is finite.

On the other hand, if A is not bounded, then for n $\in A $ there is n* $\in A $ such that n* $ > $ n, then n$_o$, n$_1$ are in A along with n$_2$, n$_3$ ...n$_j$ for all j$\in \mathbb{N}$.

f:$\mathbb{N}\rightarrow$A by; f(n) = m$_n $ for n $\in \mathbb{N}$ is a 1-1 function from $\mathbb{N}$ into A. Since A is not bounded, there is an m$_n \in$ A for n$\in\mathbb{N}$.

Since A$\subseteq \mathbb{N}$, f(A)$\subseteq$ A$-\{m_o\}$. Thus we have a 1-1 function from A onto a proper subset of A which proves that A is infinite. $\blacksquare$

It appears that definition 1 will create difficulties in comparing the ``size" of infinite sets. The ``size" of sets is addressed by the following :

DEFINITION 2: The ``size" of a set E is indicated by its Cardinality or Cardinal of E denoted by C(E).

For $ \mathbb{N}$, C($\mathbb{N}$)=$\aleph_{o}$, where $\aleph$ is the first letter of the Hebrew alphabet. C($\mathbb{R})=\aleph_{1}$.

The cardinality of a finite set is the number of elements in the set. Any set which is finite, or has cardinality $\aleph_o$ is said to be countable. Sets with larger cardinals are said to be uncountable. A well-known result is that $\mathbb{R}$ is uncountable.

The cardinal numbers separate infinite sets into equivalence classes along with the usual properties of an equivalence relation.

DEFINITION 3: Two sets A,B are equivalent if and only if there is a 1-1 function from either set onto the other. In this case we write C(A)=C(B) and A$\thickapprox$ B.

Example 2: We see from the definition of infinite set that every infinite set is equivalent to a proper subset of itself. This property of infinite sets makes it difficult to compare relative sizes of cardinal numbers. However, the Trichotomy Law applies to cardinal numbers. Given sets A,B either: i) C(A)$ <$ C(B), ii) C(A) = C(B), or iii) C(B)$<$C(A).

DEFINITION 4: Given two sets A,B we say that C(A) is less than C(B) i.e,C(A)$<$C(B) if and only if there is a 1-1 function from A into B and there is  no 1-1 function from A onto B. That is, every 1-1 function from A to B is strictly into B. Thus the cardinal of any finite set is less than the cardinal of any infinite set.

In some situations equivalence can be established using the:

Schroeder-Bernstein Theorem which states that two infinite sets A,B are equivalent if A is equivalent to a proper subset of B, and B is equivalent to a proper subset of A. An example to illustrate this theorem is the following:

f(n) = 4n is a 1-1 function from $\mathbb{N}$ into the set of even naturals $\mathbb{E}$ and g(2n) = 2n+1 is a 1-1 function from $\mathbb{E}$ into $\mathbb{N}$. Thus, by the Schroeder-Bernstein theorem $\mathbb{N}$ and $\mathbb{E}$ are equivalent, i.e. have the same cardinal,$\aleph_o$.

THEOREM 1 : Every infinite set contains a countably infinite subset i.e. $\aleph_o$ is the smallest transfinite (infinite) cardinal.

PROOF: Given an infinite set E, there is an x$_1$ such that E is equivalent to a subset of E-\{x$_1$\}. Also, there is an x$_2$ such that E$\thickapprox$E-\{x$_1$\}$\thickapprox$E-\{x$_1$,x$_2$\}. Let I$_n$=\{x$_1$,x$_2$...x$_n$\}

LEMMA: For each n $\in\mathbb{N}$ E$\thickapprox$E-I$_n$

PROOF OF LEMMA: The lemma is true for I$_1$=\{x$_1$\}. Assume true for I$_k$. Since E$\thickapprox$E-I$_k$, it follows that E-I$_k$, being infinite, is equivalent to a proper subset of itself. Thus there is an x$_{k+1}\in$ E-I$_k\thickapprox$ E. Let I$_{k+1}$=I$_k$$\bigcup$\{x$_{k+1}\}$. We see from the result above that removing one element doesn't change equivalence for infinite sets, so it follows that E-I$_{k+1}\thickapprox$ E.

Thus, by Math Induction, for n $\in\mathbb{N}$ there is an I$_n$ such that E-I$_n\thickapprox$ E which establishes the Lemma. $\square$

Proceeding with the proof of the theorem, let I$_\infty$=$\bigcup_{n\in\mathbb{N}}$I$_n$. Clearly, I$_\infty$ is a countably infinite subset of E, hence proves  theorem 1. $\blacksquare$

DEFINITION 5: Given a set A, the collection of all subsets of A will be called the power set of A, \underline{$\mathbb{P}$(A)}. With this definition we can now prove that one specific infinite cardinal is less than another.

THEOREM 2: For any set A$\neq \emptyset$, C(A)$<$C($\mathbb{P}$(A)).

PROOF: The function g(x)=\{x\} is a 1-1 function from A into $\mathbb{P}$(A). Now suppose that f:A$\rightarrow\mathbb{P}$(A) and f is 1-1.We consider two cases:

i) For all x$\in$A, x$\in$f(x). In this case, there is no x such that f(x)=$\emptyset$.

ii) E=\{x:x$\notin$f(x)\}$\neq\emptyset$. In this case, E is a subset of A, so is there an x such that f(x)=E?

If for some x$\in$A, f(x)=E; then by definition of E, x$\notin$ E. Since x$\notin$E, x$\notin$f(x). Thus by definition of E, x$\in$E. Since the assumption that f(x)= E for some x$\in$A leads to the paradox that x is both in E and not in E, we conclude that for all x, f(x)$\neq$E. It follows then that no 1-1 function from A to $\mathbb{P}$(A) is onto $\mathbb{P}$(A). Hence C(A)$<$C($\mathbb{P}$(A)). $\blacksquare$[1]

RUSSELL'S PARADOX: If we allow U to be the `` set of all sets", then Theorem 1 creates a paradox. i.e. Since if U is ``the set of all sets", then U=$\mathbb{P}$(U) and C(U) $< \mathbb{P}(U)$ can't  be true.Thus, the concept of ``set of all sets" is avoided.

Now to discuss some cardinal arithmetic. For sets A,B:

DEFINITION 6: The sum of the cardinals C(A)+C(B) is the cardinal of the union of two disjoint sets one equivalent to A and the other equivalent to B.

NOTE: Given a set A an equivalent disjoint set can be constructed by replacing the elements of A by ordered pairs of the elements.

EXAMPLE 3: If we define f(n)=2n, and g(n)=2n+1, we see that the set of even natural numbers is equivalent to $\mathbb{N}$ as is also the set of odds. Thus we have:

$\aleph_{o}$+$\aleph_{o}$=$\aleph_{o}$.

Using advanced techniques it can be proved that in general, the sum of finitely many cardinals is just the largest of them.Also, as will be proved later,

(1) if $\{A_i\}_{i\in I}$ is disjoint and A$_i$ such that C(A$_i$)=$\aleph_o$=C(I).Then $\sum$ C(A$_i)_{i\in I}$ =$\aleph_o$.

Cardinal addition satisfies the usual properties of finite addition: commutativity, associativity, etc. C($\emptyset$)=0 is the additive identity, however, there is no subtraction because the difference of two cardinals could be anything from the larger cardinal to 0. Now we discuss multiplication of cardinals. One definition of the product of C(A)$*$C(B) is:

DEFINITION 7: C(A)$*$C(B)=C(A$\times$B) for any pair of sets A,B.

NOTE: C(A)$*$C($\emptyset$)=C(A$\times\emptyset$)=C($\emptyset$)=0. And C(A)$*$C(\{1\})=C(A$\times\{1\}$)=C(A).

Obviously, any singleton set will serve as the identity. With this definition, multiplication satisfies  the usual properties  except for inverses. Next, multiplication as repeated addition:

DEFINITION 8: alternative definition of cardinal multiplication: Given two sets A,B; C(A)$*$C(B)=C(A$\times$B)=C($\bigcup(\{a\}\times$B)$_{a\in A}$)

=C($\bigcup(\{b\}\times$A)$_{b\in B}$). The equality follows from f:$\bigcup(\{a\}\times$B)$_{a\in A}\rightarrow \bigcup(\{b\}\times$A)$_{b\in B}$

EXAMPLE 4: Let A=\{1,2,3\}, B=\{a,b\}. C(A)$*$C(B)

=C($\bigcup$\{\{(1,a),(1,b)\},\{(2,a),(2,b)\},\{(3,a),(3,b)\}\})=C( \{(1,a),(1,b),(2,a),(2,b),\{(3,a),(3,b)\}) =6.

Also,

C(B)$*$ C(A)=C($\bigcup$\{\{(a,1),(a,2),(a,3)\},\{(b,1),(b,2),(b,3)\}\})=C(\{(a,1),(a,2),(a,3),(b,1),(b,2),(b,3)\})=6

Next, an application of the alternate definition.

THEOREM 3: $\aleph_{o}*\aleph_{o}=\aleph_{o}$

PROOF: For each prime p let $\mathbb{P}_p=\{p^n:n\in\mathbb{N}\}$. Since powers of primes are different for different exponents, it follows that for a fixed prime p, C$(\mathbb{P}_p)=\aleph_{o}$.

Also, C($\{\mathbb{P}_p:p$ is  prime$\})=\aleph_{o}$.

It follows then, that $\bigcup\{\mathbb{P}_p:p$ is prime$\}$ is a countable union of disjoint countable sets and is contained in $\mathbb{N}$,Thus, by the alternative definition of cardinal multiplication, we conclude that $\aleph_{o}*\aleph_{o}=\aleph_{o}. \blacksquare $

COROLLARY: The union of countably many countable sets is itself countable,as was mentioned earlier in (1).

NOTE: For infinite sets A,B both C(A)+C(B) and C(A)$*$C(B) equal the larger cardinal. Due to this property, we need exponentiation to progress to a larger infinite cardinal by means of a binary operation.

DEFINITION 9:Given two sets A,B the power $C(B)^{C(A)}$ is C(\{functions f: f:A$\rightarrow$B\}). In particular, for  a set A, 2$^{C(A)}$=C$(\{$functions f : f:A$\rightarrow\{0,1\}$).

THEOREM 4: For any set A, C($\mathbb{P}$(A))=2$^{C(A)}$.

PROOF: For a set A and B$\subseteq$A, let f$_B$(x)=1 if x$\in$B and 0 if not in B. If E,D are two distinct sets, it follows that $ f_E \neq f_D$. Conversely, if $ f_E \neq f_D$, for E,D subsets of A,   it follows that E$\neq$D. Hence there is a 1-1 correspondence between $\mathbb{P}$(A) and $\{$functions f : f:A$\rightarrow\{0,1\}.$ Thus C($\mathbb{P}$(A))=C$(\{$functions f : f:A$\rightarrow\{0,1\}$)=2$^{C(A)}$. $\blacksquare$

DEFINITION 10: For any non-empty set A, C(A)!=C(\{functions f: f:A$\rightarrow$ A, where f is 1-1 and onto\}). This is included as simply another link between cardinals and sets of functions.

NOTE: The GENERALIZED CONTINUUM HYPOTHESIS [2] assumes that for any set A there is no cardinal between C(A) and 2$^{C(A)}$.Thus, the transfinite cardinals for a set A are : C(A), 2$^{C(A)}$, 2$^{2^{C(A)}}$, etc.

As initially stated, some elementary results have been established. Some with familiar, and some with ``unfamiliar" proofs.

1. SET THEORY, Thomas Jech;1978. pp3,7.

2. TOPOLOGY, James Dugundji, 1966 p.49.

3. http://planetmath.org/CardinalArithmetic.html

4.http://thetwomeatmeal.wordpress.com/2010/12/08/cardinal-arithmetic/ [Category;Set Theory]