Definition:Smith-Volterra-Cantor Set

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Abstract

The ``Cantor Set" is a classic example of a perfect nowhere dense set.

It 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.


Objective

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


Note

The only measure considered in this article is countably additive measure.

That is, a nonnegative real-valued set function:

$\mu: \Sigma \to \R$

where $\Sigma$ is a $\sigma$-algebra of sets; the measurable sets.

Further:

$\map \mu \O = 0$

and if:

$\set {A_i}_1^\infty \subset \Sigma$

then:

$\displaystyle \map \mu {\bigcup_1^\infty \set {A_i} } \le \sum_1^\infty \map \mu {A_i}$


Definition

Let $\struct {X, \Sigma, \mu}$ be a complete finite countably additive measure space.

Let $\struct {X, \tau}$ be a topological space on the same underlying set such that $\tau \subseteq \Sigma$.

Let $G$ be a topological base consisting of sets with positive measure contained in $\struct {X, \tau}$.

Let $G$ be such that whenever $g \in G$ and $n \in \N$, there exists a pairwise disjoint set:

$\set {g_i}_1^n \subseteq G$ which satisfies the following properties:
$(1): \quad \displaystyle \bigcup_1^n \set {g_i} \subseteq g$
$(2): \quad \mu g_i \le \mu g / n$ for all $i$ such that $1 \le i \le n$
$(3): \quad \displaystyle \map \mu {g - \bigcup_1^n \set {g_i} } = 0$
$(4): \quad$ Whenever $x \in X, x \in g* \in G$, there exists a natural number $n*$ such that if:
$x \in g \in G$
and:
$\mu g \le 1 / n*$
then $g \subseteq g*$.

Then $G$ forms a Cantor collection.



Example

The subintervals of the open interval $\openint 0 1$ form a Cantor collection.


Exposition (to be refactored and rebuilt in $\mathsf{Pr} \infty \mathsf{fWiki}$ style)

This article will establish the following:


Theorem

Let $G$ be a Cantor collection.

Let $g_0 \in G$ such that $\map \mu {g_0} = b$.

Let $p$ be a natural number.


Then 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 $\map \mu P \le b / p$.


Proof

Let $n = p + 1$ for $G, g_0, p, b$ as 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 $\le b / n$.

The proportion of sets removed is $1 / n$ and the measure of the set removed is $\le b/n$.


Step 2

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

We now have $\paren {n - 1} n^2$ 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:

$\dfrac {n - 1} {n^2 \paren {n - 1} } = \dfrac 1 {n^2}$

And the measure of the union of the removed sets is $\le \paren {n - 1} b / n^3 \le b / n^2$.

These two steps suggest an induction argument for:


Lemma

Given the sets and conditions of the Theorem hypothesis.

Let $m$ be a natural number.

Then we can remove from $g_0$ a pair-wise disjoint collection of sets whose total measure is $\le b / n^m$.


Proof

The preceding 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 $\dfrac 1 {n^{k - 1} }$.

The measure of the removed sets is:

$M_{\left({k - 1}\right)} = \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right) N m_0 \le \dfrac b {n^{\left({k - 1}\right)} }$


Step $k$

After step $\left({k - 1}\right)$ there are $\left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N$ sets of type $\left({k - 1}\right)$.

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

$n^k \left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N$ sets of type $k$, each with measure $\dfrac {m_0} {n^k}$.

Removing one type $k$ set from each type $\left({k - 1}\right)$ set results in the removal of:

$\left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N$ type $k$ sets.

Thus the proportion removed is $\dfrac 1 {n^k}$.


Let $M_{\left({k - 1}\right)}$ be the measure of the union of the sets removed at step $\left({k - 1}\right)$.

Let $M_k$ be the measure removed at step $k$.

Since:

\(\displaystyle M_k\) \(=\) \(\displaystyle \left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N \dfrac {m_0} {n^k}\)
\(\displaystyle \) \(\le\) \(\displaystyle N \dfrac {m_0} {n^k}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 n \dfrac 1 {n^{\left({k - 1}\right)} } N m_0\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {M_{\left({k - 1}\right)} } n\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac b {n^k}\)



By the Principle of Mathematical Induction, if $m$ is a natural number, there exists a pairwise disjoint subcollection of $G$ with total measure $\le \dfrac b {n^m}$ that can be removed from $g_0$.

Thus the Lemma is proved.

$\Box$


Let $P$ denote the union of all the removed sets.

Let $N_p = g_0 - P$.

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 $\left\{ {g_{i_1} }\right\}_{i_1 = 1}^n$ such that:

$(1.1): \quad \displaystyle \bigcup_1^n \left\{ {g_i}\right\} \subseteq g$
$(1.2): \quad \displaystyle \mu \left({g_0 - \bigcup_{i_1 \mathop = 1}^n \left\{ {g_{i_1} }\right\}}\right) = 0$
$(1.3): \quad \mu \left({g_{i_1} }\right)_{i_1 = 1}^n \le \dfrac b n$

Remove from the collection the set $g_n$.

The measure of the set removed is $\le \dfrac b n$.

The measure of the union of the remaining $n - 1$ sets $\left\{ {g_{i_1} }\right\}_{ i_1 = 1}^{n - 1}$ is $\leq \left({n - 1}\right) \dfrac b n$ 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 $\left\{ {g_{i_1 i_2} }\right\}_{i_2 = 1}^{n^2}$ such that:

$\displaystyle \bigcup_{i_2 \mathop = 1}^{n^2} \left\{ {g_{i_1 i_2} }\right\} \subseteq g_{i_1}$

and which satisfies:

$\displaystyle \mu \left({g_{i_1} - \bigcup_{i_2 \mathop = 1}^{n^2} \left\{ {g_{i_1 i_2} }\right\} }\right) = 0$

The sets of the collection:

$\left\{ {g_{i_1 i_2} }\right\}_{i_1 = 1;}^{n-1;} {}_{i_2 = 1}^{n^2}$

each have measure $\le \left({\dfrac b n}\right) / n^2 = \dfrac b {n^3}$ and they number $n^2 \left({n - 1}\right)$.



Remove from this collection the subcollection:

$\left\{ {g_{i_1 n^2} }\right\}_{i_1 = 1}^{n - 1}$

The measure of the union of the removed sets is:

$\le \left({n - 1}\right) \dfrac b {n^3} \le \dfrac b {n^2}$

The remaining $\left({n - 1}\right) \left({n^2 - 1}\right)$ 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:

$\left\{ {g_{i_1 i_2 i_3} }\right\}_{i_3 = 1}^{n^3}$

such that:

$\displaystyle \mu \left({g_{i_1 i_2} - \bigcup \left\{ {g_{i_1 i_2 i_3} }\right\}_{i_3 = 1}^{n^3} }\right) = 0$

There are $\left({n - 1}\right) \left({n^2 - 1}\right)n^3$ sets in this collection each having measure $\le \dfrac b {n^6}$.

Now remove the subcollection:

$\left\{ {g_{i_1 i_2 n^3} }\right\}_{i_1 = 1}^{n - 1}; {}_{i_2 = 1}^{n^2 - 1}$

The measure of the union of the sets thus removed is $\le \left({n - 1}\right) \left({n^2 - 1}\right) \dfrac b {n^6} \le \dfrac b {n^3}$.

The remaining $\left({n - 1}\right) \left({n^2 - 1}\right) \left({n^3 - 1}\right)$ 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.

Let $m$ be a natural number.

Then we can remove from $g_0$ a pair-wise disjoint collection of sets whose total measure is $\le \dfrac b {n^m}$.


Proof

The preceding work has established the first part of the inductive argument.


Step $\left({k - 1}\right)$

There are $\displaystyle \prod_{j \mathop = 1}^{k \mathop - 1} \left({n^j - 1}\right)$ sets remaining after removing $\displaystyle \prod_{j \mathop = 1}^{k \mathop - 2} \left({n^j - 1}\right)$ sets each having measure $\le \dfrac b {\displaystyle \prod_{j \mathop = 1}^{k \mathop - 1} \left({n^j - 1}\right)}$.

The proportion of sets removed is $\le \dfrac 1 {n^{\left({k - 1}\right)} }$.

The measure of the union of the removed sets is:

$\displaystyle \le \dfrac{\displaystyle \left({\prod_{j \mathop = 1}^{k \mathop - 2} \left({n^j - 1}\right) }\right) b} {\displaystyle \prod_{j \mathop = 1}^{k \mathop - 1} n^j} \le \dfrac b {n^{\left({k - 1}\right)} }$


Step $k$

Let step $\left({k - 1}\right)$ be been completed.

There are $\displaystyle \prod_{j \mathop = 1}^{k \mathop - 1} \left({n^j - 1}\right)$ sets of type $\left({k - 1}\right)$ each having measure $\le \dfrac b {\displaystyle \prod_{j \mathop = 1}^{k \mathop - 1} \left({n^j - 1}\right)}$.

Since $G$ is a Cantor Collection each of these sets contains a pair-wise disjoint collection each having measure $\le \dfrac b {\displaystyle \prod_{j \mathop = 1}^k \left({n^j}\right)}$.

Next,remove the subcollection:

$\left\{ {g_{i_1 i_2 \dotsm {i_{\left({k - 1}\right)} } n^k} }\right\}_{i_1 = 1; i_2 = 1; i_3 = 1; \dotsc; i_{\left({k - 1}\right)} = 1}^{n - 1; n^2 - 1; n^3 - 1; \dotsc; n^{\left({k - 1}\right)} - 1}$

The measure of the set removed is:

$\displaystyle \le \prod_{j \mathop = 1}^{k \mathop - 1} \left({n^j - 1}\right) \dfrac b {\displaystyle \prod_{j \mathop = 1}^k n^j} \le \dfrac b {n^k}$

Thus by the Principle of Mathematical Induction, we can remove from $g_0$ a collection of sets with total measure $\le \dfrac b {n^m}$ 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:

$\displaystyle \le \sum_{k \mathop = 1}^\infty \frac b {n^k} = \dfrac b {n - 1} = \dfrac b p$

Let $P$ denote the union of all the removed sets and let $N_p = g_0 - P$.

Since $\mu P \le \dfrac b p$, 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 this page 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 $x \in X$, we have $x \in g \bigcap N_q$.

From the construction of $N_p$, $\left\{ {x}\right\} \cap g$ 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 $\mu g^* \le \dfrac 1 {n^*}$ then $g^* \subseteq g$.

The sets of type $k$ have measure $\le \dfrac b {n^k}$.

Let $k$ be chosen so that $\dfrac b {n^k} \le \dfrac 1 {n^*}$.

Then it follows that there is a set of type $k$ for which $g_k \subseteq g^* \subseteq g$.

Moreover, since $g_k$ contains a subset with measure $\le \dfrac b {n^{k+1} }$ 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 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.

$\blacksquare$


Source of Name

This entry was named for Henry John Stephen Smith, Vito Volterra and Georg Cantor.


Sources