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.
Nowhere dense sets
A set A is nowhere dense (nwd) if for every interval I, there exists a subinterval J \subset I such that
J \cap A = \varnothing
Thus, a set is nowhere dense if the set is full of ‘holes’. Lets look at an example of a nowhere dense set.
Lets see if the set \mathbb{Z} is nowhere dense. Take any interval I = (a,b).
If I \cap \mathbb{Z}, then we are done. If I contains any point x \in \mathbb{Z}, we can construct subinterval J such that x \not \in J.
There are a finite amount of integers in any interval I, so this process can be repeated until J \cap \mathbb{Z} = \varnothing.
And an example of a not-nwd set
The set \mathbb{Q} is dense in \mathbb{R}. This means that for any interval I = (a,b), there exists a q \in \mathbb{Q} such that q \in I. This means that every subinterval J will intersect \mathbb{Q}, and \mathbb{Q} is not a nowhere dense set.
f-porous sets
We define a set to be f-porous as follows
Let f \colon \mathbb{R}^+ \to \mathbb{R}^+ such that f(x) < \frac{x}{2}.
A set A is f-porous if and only if \forall I, \exists \varepsilon_I > 0 such that for any
J \subseteq I where |J| < \varepsilon_I, \exists K \subseteq J such that |K| = f(|J|) and
K \cap A =\varnothing.
Lets take for example the set of all integers \mathbb{Z} and test if the function is f-porous for
f(x) = \frac{x}{3}.
For any interval I = (a,b), there should exists an \varepsilon_I such that |J| < \varepsilon_I. We want to choose \varepsilon_I such that |K| = f(|J|). I’ll choose
\varepsilon_I = 1.
We want to show that for any J \subset I where J < 1, there exists an interval K \subseteq J such that |K| = f(|J|) and K \cap A = \varnothing.
If J does not intersect \mathbb{Z}, we are done. J intersects the set at most once, as |J| < 1 and the elements in \mathbb{Z} are spaced apart by a length of 1. The intersection point between J and \mathbb{Z} will be denoted x.
We want to find a K \subseteq J of length |K| = f(|J|) = \frac{|J|}{3} for every J. We can explicitly construct K by dividing J into three equal subintervals J_1, J_2, J_3 each with length \frac{|J|}{3}. Because x is the singular intersection point, let K = J_n for whichever J_n \cap \mathbb{Z} = \varnothing.
Thus, we have proved \mathbb{Z} is f-porous for f(x) = \frac{x}{3}.
A set is f-porous if and only if the set is nowhere dense.
If a set A is f-porous for some f, then clearly the set is nowhere dense, as for any interval I, we have some interval K which misses the set.
The harder direction is proving that for some nowhere dense A, that A is f-porous. We will need to construct a function f for any arbitrary nowhere dense set.
Take any interval I = (a,b). Lets divide the interval into three equal subintervals
I_1 = (a,\frac{b-a}{3})\quad I_2 = (\frac{b-a}{3},\frac{2(b-a)}{3}) \quad I_3 = (\frac{2(b-a)}{3},b)
If a set A is f-porous for every f, then A is countable.