Porous sets
There are several ways to define a small set, like Meager and Measure-zero. Porous sets are an alternate way to measure how small a set is.
Definition
A set is porous if every interval I in \mathbb{R} contains an interval I_2 completly missing the set, such that |I_2| = \alpha |I| where 0 < \alpha < 1.
Formalizing this definition requires a few steps.
Let E be a set in \mathbb{R}, and x \in \mathbb{R}. The right porosity of E at x is denoted p^+(E,x) and is given by
p^+(E,x) = \limsup_{\varepsilon \to 0^+} \frac{k}{k+h}
where (x + h, x + h + k) \cap E = \varnothing, h + k < \varepsilon and h,k > 0, and \sup refers to the lowest greater bound. Similary, the left porosity of E, denoted p^-(E,x) is defined by (x - h - k, x - h) \cap E = \varnothing
Then the porosity of E at x is defined by
p(E,x) = \max\{p^-(E,x),p^+(E,x)\}
It is clear the value of p is inbetween [0,1]. We say that E is porous at \bf x if p(E,x) > 0, and that E is porous if p(E,x) > 0 for all x \in E.
Are the following sets porous at x?
- Is the set E = [0,1] porous at x = \frac{1}{2}?
Let’s start by trying to find the left porosity of E at x. It’s clear that when \varepsilon < \frac{1}{2}, no values of h,k satisfy (x + h, x + h + k) \cap E = \varnothing, h + k < \varepsilon. Therefore, the set does not have a defined left porosity. Similary the set does not have a defined right porosity, so the set is not porous at x = \frac{1}{2}.
- Is the set E = C porous at x = \frac{1}{3}?
In this case C refers to the Cantor ternary set. At x = \frac{1}{3}, it is clear that p^+(C,\frac{1}{3}) = 1, because h can get arbitrarily small while completly avoiding C. Because p^- can’t possibly be greater than 1, p(C,\frac{1}{3}) = 1, and the Cantor ternary set is porous at x = \frac{1}{3}.
Beta-porous
Let A \subseteq \mathbb{R}, \beta \in (0,1). Then A is \beta-porous iff for every (a,b) \ in \mathbb{R}, \exists(a^\prime,b^\prime) \subseteq (a,b) such that (b^\prime - a^\prime) = \beta(b-a)
f-porous
A set A is f-porous if for some f, there exists an \varepsilon_n such that for all intervals |J| \subset |I|, there exists an interval |K| \subset |J| such that $$
A set A is f-porous \forall f iff A is discrete
If a set is discrete, then we can
By way of contradition, assume A is not discrete, and f-porous for every f. Then
\exists x \in A \mid \forall \varepsilon > 0,\quad \mathcal{B}(x,\varepsilon) - \{x\} \cap A \not = \varnothing
Without loss of generality, let
\lvert x - x_{n-1} \rvert < \lvert x - {x_n}\rvert
and
|x_{n-1} - x_n| < \frac{x_n - x}{2}
For each n \in \mathbb{N}, let f_n be such that
f_n(|x_n - x|) > |x_n - x_{n+1}|
We choose a random interval x \in I \subset \mathbb{R}, and let \varepsilon_{I,n} > 0 be this intervals associated epsilon for any f_n.
Let J \subset I such that |J| < \varepsilon_{I,n} and n \in \mathbb{N}.
Need to show that |x_n - x_{n-1}| < |x_n - x|/2. Can’t prove.
Suppose A is uncountable. Uncountable sets have the property that there exists some point x that is a limit point from both sides, symbollically.
\exists x \in A \mid \forall \varepsilon > 0
(x-\varepsilon,x) \cap A \not = \varnothing \land (x, x + \varepsilon) \cap A \not = \varnothing
We can now construct a continually decreasing sequence. Let \{x_n\} \to x. Without loss of generality, let
|x - x_n{+1}| < |x_{n+1} - x_n|
Repeating for y. approaching from the oppsite side.
Let \ell n = |x_n - y_n|, c_n = |y_n - y_{n+1}| and m_n = |x_n - x_{n+1}|.
Let I \subseteq \mathbb{R}, x \in I. For each n \in \mathbb{N}, let
f_n = \mathbb{R}^+ \to \mathbb{R}^+ \mid f_n(\ell n) > \max\{c_n,m_n\}
And let \varepsilon_{I_n} be the associated value of epsilon. Let J \subset I and n \in \mathbb{N} such that |J| < \varepsilon_{I,n}.
Let J be such that x_n,y_n \in J. Since f_n is a decreasing function,
f_n(|J|) > f_n(\ell n) > \max\{c_n,m_n\}
Because either c_n or m_n is the biggest hole in (y_n,x_n), there is clearly no hole bigger than \max\{c_n,m_n\}. Thus a contradiction, and if a set is f-porous for every f, then A is not uncountable.