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Why are these sets interesting? My first thought was a line in $\mathbb{R}^2$ should be porous. Then I thought to myself -- this doesn't seem so interesting.

(I'm using the definition here.)

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    $\begingroup$ You would give Readers a better impression of your interest in this topic if the Question's body includes the definition, not merely a link, and if you explained in your own words how you encountered it and/or understood it. $\endgroup$ – hardmath Sep 18 at 14:43
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    $\begingroup$ Countable unions of porous sets (especially upper porous sets) provide tools for strengthening various Baire category co-meager results, while simultaneously strengthening "meager and Lebesgue measure zero" exceptional set results in ${\mathbb R}^n,$ and porous sets (especially lower porous sets) provide a useful tool (in addition to Hausdorff dimension) for measuring the size of measure zero sets in dynamical systems theory (such as Julia sets). See also this survey paper. $\endgroup$ – Dave L. Renfro Sep 18 at 15:40
  • $\begingroup$ @DaveL.Renfro This is basically the answer I was looking for. Feel free to add it as an answer (along with an idea for what "strengthening" means) so that I can accept it! $\endgroup$ – yoshi Sep 18 at 17:17
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Primarily restricted to metric spaces, porosity is a metric-dependent strengthening of the topological property nowhere dense that allows for various hierarchical classifications of nowhere dense sets and meager sets (= first category sets). Porosity notions have been applied in a number of fields, including quasiconformal mapping theory, boundary interpolation theory, harmonic analysis (exceptional sets for trigonometric series), convex geometry (Baire-typical results in various hyperspaces of sets), Julia set theory, nonlinear functional analysis (especially the differentiability of convex and Lipschitz functions on Banach spaces), real analysis, metric number theory (normal numbers and the like), and potential theory.

In every complete metric space that has no isolated points (in particular, in any Banach space) there exist meager sets that are “too large” to be expressed as a countable union of (upper, and hence also lower) porous sets (= $\sigma$-porous sets). Therefore, in these spaces the property of being $\sigma$-porous is strictly stronger than the property of being meager. This allows for the possibility of strengthening properties that are known to be typical (i.e. the exceptions form a meager set). Moreover, in ${\mathbb R}^n$ the property of being $\sigma$-porous is strictly stronger than the combined properties of being meager and Lebesgue measure zero, and an example where this aspect is used is given below.

A well known example of a Baire category co-meager result is the fact that “most” continuous functions are nowhere differentiable (e.g. see this answer for some of the many variations on this theme), meaning that the collection of continuous functions $f:{\mathbb R} \rightarrow {\mathbb R}$ (using the $\sup$ metric) that have at least one point of finite differentiabilty is a meager subset of the space of all such continuous functions. It turns out that this can be strengthened so that instead of meager, the set can be shown to be smaller still -- the set is $\sigma$-porous. This is purely a size issue, by the way, in the sense of set containment, and it does not have anything to do with Borel complexity or something else. Incidentally, for this “super-most continuous functions are nowhere differentiable” result, we can actually express the set as a countable union of lower porous sets, which is a strictly stronger result than if we were using upper porous sets (not just for some spaces, but for the specific case of the space of continuous functions).

For an example of its use to simultaneously strengthen the properties of meager and Lebesgue measure zero in ${\mathbb R},$ recall that if $f:{\mathbb R} \rightarrow {\mathbb R}$ has a finite (ordinary) derivative at a point, then the symmetric derivative also exists at that point and has the same value (this is a relatively common undergraduate real analysis exercise), but the converse fails (consider $f(x) = |x|$ at $x=0).$ A natural question one might ask is whether the converse can fail at every point. No, it can't. To avoid getting caught up in a seemingly endless variety of nuances that have appeared in the literature, let’s restrict ourselves to continuous functions and let’s assume the function has a symmetric derivative at every point. How large can the set of points of non-differentiability be? In 1927, the Russian mathematician Khintchine proved that this set of points has Lebesgue measure zero. (In fact, he proved this for the more general case in which the function is Lebesgue measurable and is such that, at every point, the $\limsup$ of the symmetric difference quotient is finite. I mention this as an example of what I meant by “a seeming endless variety of nuances”.) In 1964, the Indian mathematician Mukhopadhyay proved that the set is also meager. In 1978, Belna/Evans (see Theorem 3 on p. 266 of this paper) strengthened both results by showing that this set is in fact $\sigma$-porous (upper porous variety). Indeed, not only was this a proper strengthening (because there exist sets simultaneously meager and Lebesgue measure zero that are “too large” to be $\sigma$-porous), but their proof was simpler than the combined separate proofs of “meager” and “Lebesgue measure zero”, an aspect that was found in several other similar strengthenings from “meager & Lebesgue measure zero” to “$\sigma$-porous” (suggesting that porous notions are a more natural tool for measuring smallness for these results). Incidentally, going the other direction, it’s known that this set of non-differentiability for a continuous and everywhere symmetrically differentiable function can be large in the sense of having continuum many points in every interval. Indeed, this set can even have Hausdorff dimension $1$ in every interval (i.e. be “everywhere maximally large” in the sense of Hausdorff dimension; for references, see the comments to this question).

Another application of porous sets (in this case, primarily being lower porous sets) is that it provides a useful tool (in addition to Hausdorff dimension) for measuring the size of measure zero sets that arise in dynamical systems theory (such as Julia sets). Incidentally, unlike upper porous sets, which in ${\mathbb R}^n$ can have Hausdorff dimension $n$ (indeed, they can even have Hausdorff dimension $n$ in every nonempty open set), each lower porous set in ${\mathbb R}^n$ has Hausdorff dimension strictly less than $n,$ and it was this aspect that originally led to their consideration in this area. Indeed, the definition of lower porous arose out of a widely applicable and easy to use geometric condition, discovered by several people independently I think, that implies Hausdorff dimension is strictly less than $n.$ However, for each $\epsilon > 0,$ there exist lower porous sets in ${\mathbb R}^n$ whose Hausdorff dimension is greater than $n - \epsilon.$

For many more applications, primarily those in infinite dimensional Banach space settings, see Luděk Zajíček’s 2005 survey paper.

Briefly, the most basic of the very many variations on the notion of porosity are the following. Let $X$ be a metric space and let $E \subseteq X.$ Then $E$ is nowhere dense in $X$ if and only if for each $x \in X$ and for each open ball $B(x,\epsilon)$ centered at $x,$ there exists an open ball (in fact, infinitely many such open balls, unless the space is really pathological) all of whose points belong to the complement of $X$ (e.g. see this answer). For a fixed $x,$ let’s call this condition “$E$ is nowhere dense at $x$”. Now consider this stronger property: For a given set $E$ and point $x,$ there exists $0 < \alpha < 1$ such that for arbitrarily small $\epsilon > 0$ the open ball $B(x,\epsilon)$ contains an open ball $B(y,\alpha \epsilon)$ all of whose points belong to the complement of $X.$ Note that $\alpha$ is allowed to vary with $x,$ but not with $\epsilon.$ This is the notion “$E$ is upper $\alpha$-porous at $x$”, and its relation to “$E$ is nowhere dense at $x$” is somewhat analogous to how “$f$ is $K$-Lipschitz continuous at $x$$(K$ being a pointwise defined Lipschitz constant) compares to “$f$ is continuous at $x$”. Actually, the Lipschitz analogy is more analogous to “$E$ is lower $\alpha$-porous at $x$”, which means that for all sufficiently small $\epsilon > 0$ the open ball $B(x,\epsilon)$ contains an open ball $B(y,\alpha \epsilon)$ all of whose points belong to the complement of $X.$ We say that the set $E$ is “upper porous” (resp., “upper $\alpha$-porous”) if, for each $x \in E$ (sometimes the requirement “for each $x \in X$” is used, which is equivalent to the requirement that the topological closure of $E$ is porous using the “for each $x \in E$” definition), we have $E$ is upper porous at $x$ (resp., $E$ is upper $\alpha$-porous at $x).$ The lower porous versions are defined similarly.

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