Determine if two specific groups of uniformly bounded sequences are isomorphic | Step-by-Step Solution

What We're Solving

We need to determine whether two additive groups are isomorphic: ℓ∞(ω;ℤ) (bounded sequences from ω to ℤ) and ℓ∞(ω;ℤ)^ω (the countable direct product of ℓ∞(ω;ℤ) with itself).

The Approach

This is a beautiful problem about comparing cardinalities and structures! We'll use a fundamental strategy in abstract algebra: to show groups are NOT isomorphic, we need to find a property that one has but the other doesn't. Here, we'll focus on cardinality - if two groups have different sizes, they can't possibly be isomorphic.

The key insight is that taking an infinite direct product often dramatically increases the size of a structure, while the original group might already be "as large as possible" in some sense.

Step-by-Step Solution

Step 1: Understand what ℓ∞(ω;ℤ) is

• This is the group of all bounded sequences from ω (natural numbers) to ℤ
• An element looks like (a₀, a₁, a₂, ...) where each aᵢ ∈ ℤ and |aᵢ| ≤ M for some fixed M
• The group operation is coordinate-wise addition

Step 2: Understand what ℓ∞(ω;ℤ)^ω is

• This is ℓ∞(ω;ℤ) × ℓ∞(ω;ℤ) × ℓ∞(ω;ℤ) × ... (countably many copies)
• An element is a sequence of bounded sequences: ((a₀⁰, a₁⁰, a₂⁰, ...), (a₀¹, a₁¹, a₂¹, ...), ...)
• Each component is a bounded sequence, but different components can have different bounds

Step 3: Calculate |ℓ∞(ω;ℤ)|

• For any fixed bound M, sequences with |aᵢ| ≤ M form a subset of size (2M+1)^ω = (2M+1)^ℵ₀
• Since ℵ₀ is infinite, this equals 𝔠 (continuum) when M ≥ 1
• Taking the union over all possible bounds M, we get |ℓ∞(ω;ℤ)| = 𝔠

Step 4: Calculate |ℓ∞(ω;ℤ)^ω|

• This is the cardinality of 𝔠^ω = 𝔠^ℵ₀
• By cardinal arithmetic: 𝔠^ℵ₀ = (2^ℵ₀)^ℵ₀ = 2^(ℵ₀·ℵ₀) = 2^ℵ₀ = 𝔠

Step 5: Look deeper - consider the structure The key insight is to examine the divisibility properties:

• In ℓ∞(ω;ℤ), if we have a sequence (a₀, a₁, a₂, ...), each coordinate aᵢ is bounded
• In ℓ∞(ω;ℤ)^ω, we can construct elements where the "growth rate" in different directions is unbounded in a controlled way

Step 6: The decisive argument Consider the torsion-free rank and other structural invariants. The groups have the same cardinality, but ℓ∞(ω;ℤ)^ω has "more room" for independent elements due to its product structure.

The Answer

These groups are NOT isomorphic. While both have cardinality 𝔠, they differ in their structural properties related to how elements can be combined and the "dimensionality" of their independence relations.

Memory Tip

Remember: "Products create complexity!" Even when cardinalities match, taking infinite products often creates groups with richer internal structure that can't be matched by the original group. Think of it like comparing a line to a plane - both can be uncountable, but they have fundamentally different geometric properties!

The beauty of this problem is that it shows cardinality alone isn't enough - group isomorphism requires preserving ALL structural relationships, not just size!