alexandræ · discrete sets, sets with discrete closure, and discontinuities

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alexandræ · discrete sets, sets with discrete closure, and discontinuities

discrete sets, sets with discrete closure, and discontinuities

june 19th, 2023

    i noticed discontinuities in functions are much more well-behaved when they form discrete sets, especially sets with discrete closure. a set is called discrete if all of its elements are isolated : S is discrete if, for all x ∈ S, there exists ε > 0 such that B(x, ε) ∩ S = {x}. if a set S has a discrete closure set, that means that its intersection with any bounded set is finite. here are two instances i came across where these sets made my life easier to study discontinuity-related questions.

    when it comes to so-called "counting functions", discontinuities arise when there are things that are being counted some way or another, like their norm or something. that's mostly a number theory thing but it also arised in my research on lattice groups. usually, we want counting functions to have their coset be ℕ (with the min(ℕ) = 0 convention ~~ehehe~~). however, this set doesn't include +∞, so i searched for a minimal way to ensure such a counting function to exist. and sure enough, i found one : if you have a set S⊂Ω such that sup(Ω) = +∞, if f:Ω→ℕ has its set of discontinuities be S, then cl(S) must not have a cluster point... in other words, its closure set must be discrete !
    proof. i used a proof by negation for this (the constructively valid version of proofs by contradiction, since it simply uses the definition of negation as ¬p:≡[p⇒⊥]) ; we'll assume that there exists a cluster point ℓ in cl(S). then, there exists (a_i)_i∈ℕ∈S^ℕ that converges to ℓ, while never reaching it. since the sequence is real-valued and converges, it is a cauchy sequence, and never reaches ℓ so we can build a strictly increasing (extractive) function φ : ℕ → ℕ such that (|a_φ(i)-ℓ|)_i∈ℕ is strctly discreasing. without loss of generality, we'll assume (a_φ(i))_i∈ℕ to be strictly monotonous, let's say increasing since a similar reasoning applies to when it's decreasing. then, since im(f)=ℕ, that f is increasing, and that S is the set of its discontinuities, then for all ε>0 and x∈S, we have f(x−ε)+1≤f(x) or f(x)+1≤f(x+ε), or both tbh. therefore, we can inductively conclude that f(a_φ(n))≥f(a_φ(0))+n→+∞, so (f(a_φ(n)))_i∈ℕ diverges to +∞, which implies f is undefined over [ℓ,+∞). by contradiction/negation, S can't have any cluster point, which concludes the proof. here's a loose visualisation of the proof.

    another odd place where such sets become life-saving is when it comes to find minimal conditions for the integrability of functions of alternating sign : let ε>0, large enough A>0 and ω>0, and f : Ω⊇(A,+∞) → ℝ a function such that |f(x)| ≤ 1/x^ε for all x>A, then it's integrable whenever it changes sign at least once but a finite number of times over all ω-long interval ⊆ (A,+∞)... which implies sgn(f)'s set of discontinuities has discrete closure. although it's not a direct equivalence, but that's a property that will come in handy when we get to the proof itself !
    proof. by chasles's rule, if ∫_A ∞ f(x) dx does exist, it should be equal to ∫⌈A/ω⌉·ωAf(x) dx + ∑∞k=⌈A/ω⌉ ∫(k+1)·ωk·ωf(x) dx. we can assume without loss of generality that A∈ωℕ* (as it has supremum +∞ so there's necessarily one that's large enough), this way we can just write it ∑∞k=A/ω ∫(k+1)·ωk·ωf(x) dx. since x↦1/x^ε is decreasing, (1/((n+1)ω))^ε ≤ ∫(n+1)·ωn·ωf(x) dx ≤ (1/(nω))^ε, so by the sandwich theorem, (∗) ∫(n+1)·ωn·ωf(x) dx→0 when n→+∞. here comes the discrete closure trick : because I, the set of discontinuities of sgn∘f, has discrete closure by assumption, there exists a successor function suc(I,·):{A}∪I→I such that for all x∈{A}∪I, we have suc(I,x)=min{y∈{A}∪I | x<y}. therefore, we can define a₀:=A=min(I∪{A}) and a_k+1:=suc(I,a_k) for all k∈ℕ this way the integral we search for can be written as ∑∞k=0 ∫a_k+1a_kf(x) dx, of which the absolute values of their summands are bounded by above by their respective n in (∗), which all tend to 0, and by non-negativity of the absolute value and sandwich theorem again, the summands also do converge to zero, yielding an alternating sum with terms that tend to 0, which means it converges, and thus does the integral of f over (A, +∞), which concludes the proof !

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alexandræ

(ɔ) 2023 – 2024, intellectual property is a scam