Short answer: as stated, Ω is not a linear order, and it is far too complicated to admit a nice “universal” combinatorial description. In fact, it already contains, as a suborder, the very complicated poset (P(ω), ⊆*) of subsets of the natural numbers ordered by inclusion modulo finite. So any hope for a simple colimit description in the spirit of your Θn- or Φ-examples cannot work without imposing strong regularity restrictions on the functions.
Details:
Your relation is just Big-O. For real-valued functions f,g defined eventually (say on (λ,∞)), your definition f ≼ g ⇔ max{|liminf f/g|, |limsup f/g|} < ∞ is equivalent to limsup |f/g| < ∞, i.e., f = O(g) as x → ∞. The equivalence relation f ≃ g means f = Θ(g) (mutual O).
Ω is not linearly ordered. Even among continuous functions you get incomparability: take f(x) = sin x and g(x) = cos x. Then limsup |f/g| = limsup |tan x| = ∞ and limsup |g/f| = ∞, so neither f ≼ g nor g ≼ f. Thus Ω/≃ is only a partial order.
Ω contains (P(ω), ⊆*) as a suborder. This shows it has antichains of size 2^ℵ0 and in general is very complicated.
Construction. Fix a nonnegative C∞ bump b supported in [0,1] with max b = 1, and let bn(x) = b(x−n). Choose a rapidly increasing sequence cn → ∞ (e.g. cn = 2^n). For A ⊆ ω, define fA(x) := 1 + Σ_{n∈A} cn bn(x). Then fA is continuous on [0,∞). The supports of the bn are disjoint, so at any x at most one term contributes.
Claim. fA ≼ fB iff A ⊆* B (i.e. A\B is finite).
Thus A ↦ [fA] is an order-embedding of (P(ω), ⊆*) into Ω/≃.
Consequences:
But without such regularity assumptions, Ω/≃ is as wild as (P(ω), ⊆*), so there is no simple combinatorial or universal colimit-type description of its isomorphism type.
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version: gpt-5-2025-08-07
Status: UQ Validated
Validated: 7 months ago
Status: Needs Human Verification
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