User:Arbowen1

\textbf{\centerline{NON-DIMENSIONAL CANTOR SET}}

\textbf{AUTHOR : ArB Owen}

Abstract The ``Cantor Set" is a classic example of a perfect nowhere dense set, and is referred to extensively in discussions of both topology and real analysis. the simplest such set is constructed by removing subintervals from the unit interval on the real line. The process has been extended to produce Cantor sets from other intervals. However,these constructions usually rely on the linear order of the real number system. Successively removing from a fixed measurable set a constant proportion of the remaining set produces  a Cantor set of measure zero. for example, The ``no-middle-third", or Ternary set is constructed by repeatedly removing from the unit interval one-third of the set remaining after each reduction. If however, we remove (in an appropriate manner) a diminishing proportion from a fixed set we produce a Cantor set of positive measure. Such sets are referred to in the literature as ``Smith-Volterra" Cantor sets, or ``Fat" Cantor sets.[1]

Objective: The purpose of this article is to construct a Cantor set of positive \emph {countably additive} measure without relying on linear order, dimension, or connectedness.

Note: The only measure considered in this article is\emph{ countably additive} measure. That is, a nonnegative real-valued set function $\mu : \Sigma \rightarrow R $, where $ \Sigma $ is a $ \sigma$-algebra of sets; the \emph{measurable} sets. .Further, $ \mu \emptyset = 0 $. And if $ \{ A_ i \}_1^\infty \subset \Sigma $, then $\mu ( \cup \{ A_ i \}_1^\infty)\leq \sum \mu{( A_i)}_1^\infty.[2] $

Definition: A \emph{Cantor collection} is a topological base G consisting of sets with positive measure contained in a topological space (X,T) which is also a complete finite \emph{countably additive} measure space $(X,\Sigma,\mu)$ with T$ \subseteq \Sigma $ such that whenever $g\in G$ and n is a natural number, there is a pairwise disjoint collection ${{\{g_{i}}\}_{1}^{n}}\subseteq G$ which satisfies the following properties:

1). $ \bigcup {{\{g_{i}}\}_{1}^{n}}\subseteq$ g.

2). $\mu g_{i} \leq \mu g/ n $ for all i such that $1\leq i\leq$ n.

3). $\mu (g-\bigcup{{\{g_{i}}\}_{1}^{n}})= 0$. And:

4). Whenever x $ \in $ X and x $\in$  g* $ \in $ G, there is a natural number n* such that if  x $  \in $ g $ \in$ G and $\mu$g $\leq$  1/n*  then g $ \subseteq$  g*.\

EXAMPLE: The subintervals of (0,1) form a Cantor collection.

This article will establish the following:

Theorem: Whenever G is a Cantor collection and $ g_{0} \in $ G  with $ \mu g_{0} $ = b;

then if p is a natural number there are two nonempty disjoint sets $N_{p}$ and P such that:

$ g_{0} = N_{p} \bigcup P$, where $N_{p}$ is nowhere dense in the relative topology on $g_{0}$ and $\mu P \leqq b/q $.

Proof: Let n = p + 1 for G, $g_{0}$, p, and b given in the hypothesis.

We construct our Cantor set by removing a diminishing proportion of sets available.

The construction will be accomplished using two techniques. First, a ``conversational" approach, then one with more mathematical symbolism, or ``rigor".

Step 1 : By definition of the Cantor collection we express $g_0$ as the union ( with possible exception of a set of measure zero) of n pairwise disjoint sets from G. These are sets of type 1. Remove one of them leaving n-1 sets remaining each with measure $ \leq b/n$. The proportion of sets removed is 1/n and the measure of the set removed is $\leq b/n$.

Step 2 Express each of the n-1 sets of type 1 as the union of $ n^2$ sets from G.

We now have $ (n-1)n^2$ sets. These are sets of type 2. Remove from each of the n-1 sets of type 1 one of the type 2 sets.

The proportion of sets removed is $ (n-1)/[n^2(n-1)] = 1/n^2 .$ And the measure of the union of the removed sets is $ \leq (n-1)b/n^3 \leq b/n^2$.

These two steps suggest an induction argument for:

Lemma: Given the sets and conditions of the Theorem hypothesis, if m is a natural number we can remove from $g_0$ a pair-wise disjoint collection of sets whose total measure is $\leqq b/n^m$.

Proof : The preceding three steps have established the first part of the inductive argument

Step (k-1): Assume that there are N sets of type (k-1) each with measure m$_{0}$. Remove (from the type (k-1) sets) the proportion 1/$n^{(k-1)}$.

The measure of the removed sets is:

M$_{(k-1)}$ = $(1/n^{(k-1)}) $ Nm$_{0} \leq$ b/ $n^{(k-1)}$.

Step k: After step (k-1) there are $(1-(1/n^{(k-1)}))$N\thinspace{  sets of type (k-1)}.

Expressing each of them as the union of n$^k$ subsets of G yields:

n$^{k}((1-(1/n^{(k-1)}))$N sets of type k, each with measure = m$_{0}$/n$^{k}$.

Removing one type k set from each type (k-1) set results in the removal of :

$(1-(1/n^{(k-1)}))$N type k sets. Thus the proportion removed =1/n$^k$.

M$_{(k-1)}$ = the measure of the union of the sets removed at step (k-1), and let M$_k$ = measure removed at step k. Since M$_{k}$ = $(1-(1/n^{(k-1)}))$ Nm$_{0}$/n$^{k} \leq $ Nm$_{0}$/n$^{k}$  = (1/n)(1/n$^{(k-1)})$ Nm$_{0}$ =  M$_{(k-1)}$/n $\leq$  b/n$^{k}$.

By Math Induction, if m is a natural number, there is a pairwise disjoint subcollection of G with total measure $\leq $ b/ n$^m$

that can be removed from g$_0$. Thus the Lemma is proved.$ \Box$

Let P denote the union of all the removed sets and N$_p$ = g$_0 -$P. The proof that N$_p$ and P satisfy the theorem will be presented following the rigorous construction of them.

Rigorous construction

Step 1: Since G is a Cantor collection there is a subcollection $ \{{g_{i_1}}\}_{{i_1} =1}^n$ such that:

1.1) $ \bigcup {{\{g_{i}}\}_{1}^{n}} \subseteq $g.

1.2) $\mu( g_{0} -\bigcup \{{g_{i_1}}\}_{{i_1} =1}^n) = 0$. 1.3) $\mu{(g_{i_1})}_{{i_1} = 1}^n  \leq $b/n. Remove from the collection the set $g_{n} $. The measure of the set removed is $\leq $ b/n.

The measure of the union of the remaining n-1 sets $ \{{g_{i_1}}\}_{{i_1}=1}^{n-1}$ is $ \leq $ (n-1)(b/n) and will be called sets of Type 1.

Step 2: Again, since G is a Cantor collection,for each of the sets of type 1;

$g_{i_1}$ with ${i_1}$ fixed, there is a collection $ \{{g_}\}_{{i_2}=1}^{n^2}$ such that: $\bigcup \{{g_}\}_{{i_2}=1}^{n^2}\subseteq  g_{i_1}$ and which satisfies:

$ \mu({g_{i_1}} - \bigcup \{{g_}\}_{{i_2}=1}^{n^2}) = 0.$ The sets of the collection:

${\{{g_}\}_{{i_1}=1;}^{n-1;}}  { }_^{n^2}$ each have measure $\leq (b/n)/n^2 = b/n^3 $ and they number $n^2(n-1).$

Remove from this collection the subcollection: $\{{g_{i_1 n^2}}\}_{{i_1}=1}^{n-1} $. The measure of the union of the removed sets is: $\leq (n-1)b/n^3 \leq b/n^2$. The remaining $(n-1)(n^2 -1)$ sets will be called sets of Type 2.

Step 3: For each set $g_{{i_1}{i_2}}$ of type 2 with the sequence ${i_1}{i_2}$ fixed, there is a subcollection:

$\{g_\}_{{i_3}=1}^{n^3}$ such that $\mu(g_{{i_1}{i_2}}-\bigcup\{g_\}_{{i_3}=1}^{n^3})=0.$ There are $(n-1)(n^2-1)n^3$sets in this collection each having measure $\leq b/n^6$. Now remove the subcollection: ${}\{g_\}_{{i_1}=1}^{n-1} $    $ {}_ {{i_2}=1}^{{n^2}-1}$. The measure of the union of the sets thus removed is: $\leq (n-1)({n^2}-1)b/{n^6} \leq b/{n^3}$. The remaining $(n-1)({n^2}-1)({n^3}-1)$ sets are sets of Type 3.

Now that the preliminaries have been established, we proceed to prove by induction the following :

Lemma: Given the sets and conditions of the Theorem hypothesis, if m is a natural number we can remove from $g_0$ a pair-wise disjoint collection of sets whose total measure is $\leq b/n^m$. Proof : The preceding work has established the first part of the inductive argument. Step (k-1): There are $ \prod {({n^j}-1)}_{j=1}^{k-1}$ sets remaining after removing $ \prod {({n^j}-1)}_{j=1}^{k-2}$ sets each having measure $\leq b/[{\prod]_{j=1}^{(k-1)}}$. The proportion of sets removed is $ \leq 1/n^{(k-1)} $.The measure of the union of the removed sets is : $\leq[\prod {({n^j}-1)}_{j=1}^{k-2}]b/ [{\prod]_{j=1}^{(k-1)}}\leq b/n^{(k-1)}.$ Step k: Now that step k-1 has been completed. There would be $ \prod {({n^j}-1)}_{j=1}^{k-1}$ sets of type k-1 each having measure $\leq b/(\prod{{n^j})}{}_{j=1}^{(k-1)}$. Since G is a Cantor Collection each of these sets contains a pair-wise disjoint collection each having measure $ \leq b/({\prod{{n^j})}_{j=1}^{k}}$. Next,remove the subcollection: $$\huge\textbf{$\{g_\}_{{i_1}=1;}^{n-1;}{}_{{i_2}=1;}^{{n^2}-1;} {}_{{i_3}=1...}^{{n^3}-1...}{}_^{n^{(k-1)}-1}$}.$$ The measure of the set removed is: {\Large $$\leq \prod {({n^j}-1)}_{j=1}^{k-1}( b/(\prod{({n^j)}_{j=1}^k}))\leq b/{n^{k}}$$}\\Thus by math induction, we can remove from $g_0$ a collection of sets with total measure $\leq b/{n^m}$ for each natural number m.$\Box$ If the foregoing procedure is completed for each natural m, the measure of the union of the removed sets is $\leq{\sum({b/n^k})_{k=1}^\infty} =b/(n-1) =b/p.$ Let P denote the union of all the removed sets and let $N_p$ = $g_0-P$. Since $\mu{P} \leq b/p$, we have established part of the theorem. Clearly, $ \mu N_p > 0.$\\ At this point we observe that both the ``conversational" and the`` rigorous techniques "produce the same two sets N$_p$ and P. The remainder of the article will be devoted to proving that these two sets satisfy the previously stated theorem.

Next we prove that $ N_{p}$ is nowhere dense in the relative topology on $g_{0}.$ Proof: Suppose that $ g \in G $ and for some x $ \in X,x \in g \bigcap N_{q} $. Due to the construction of $N_{p}$, $\{x\} \cap $ g intersects a set $g_{k}$ of type k in $N_{p}$ for every natural number k. Since G is a Cantor collection, by property 4) there is a natural number $ n^*$ such that if x $ \in g^* \in G $, and $ \mu g^* \leq 1/n^* $ ; then $ g^* \subseteq  g $. Since the sets of type k have measure $ \leq b/n^k, $ if we choose k so that $ b/n^k   \leq 1/n^*$; then it follows that there is a set of type k, $g_k \subseteq g^* \subseteq  g$. Moreover, since $ g_k $ contains a subset with measure $ \leq b/n^{(k+1)} $  that will be removed to form the set P, it is clear that $ N_p $ has empty interior,  thus $ N_p $ is nowhere dense in the relative topology on  $g_0 $. $ N_p =  (g_0 -P) $ and P open in the relative topology on $g_0 $ imply that $N_p $ is closed in the relative topology on $ g_0 $, $N_p $ closed with no isolated points means that it is also a \emph{Perfect} set. In conclusion we see that by removing diminishing proportions of the remaining sets we have constructed a nowhere dense set of positive measure, a``Fat" Cantor set.Therefore our objective has been achieved.$ \blacksquare$

[1] Principles of Real Analysis, Aliprantis and Burkinshaw, 1981,199. [2] [Real Analysis, Royden;1963,pg. 191.