The set of positive even numbers E = {2, 4, 6, 8, ...} is a proper subset of ℕ = {1, 2, 3, ...}. What does this tell us about the cardinality of E compared to ℕ?
AE has strictly smaller cardinality than ℕ — a proper subset is always smaller
BE has the same cardinality as ℕ — the bijection n ↦ 2n maps ℕ onto E one-to-one
CE is uncountable since its elements grow without bound
DThe cardinality of E is undefined because it is a proper subset of an infinite set
For infinite sets, cardinality is defined by bijections, not by the subset relation. The function f(n) = 2n maps ℕ → E bijectively: every natural number maps to a distinct even number, and every even number is hit. So |E| = |ℕ| = ℵ₀. For infinite sets, a proper subset can have the same cardinality as the whole — in fact this property (having a proper subset of equal cardinality) is Dedekind's definition of what it means to be infinite.
Question 2 Multiple Choice
Which of the following sets has cardinality strictly greater than ℵ₀?
AThe set of all integers ℤ
BThe set of all rational numbers ℚ
CThe set of all finite binary strings
DThe set of all real numbers in the interval [0, 1]
ℤ, ℚ, and all finite strings over any finite alphabet are countably infinite — each can be listed in a sequence via a bijection with ℕ. The real numbers in [0, 1] are uncountably infinite; Cantor's diagonalization argument shows no such listing is possible. Their cardinality is 2^ℵ₀, which is strictly greater than ℵ₀.
Question 3 True / False
The rational numbers ℚ are countably infinite even though between any two rationals there are infinitely many more rationals.
TTrue
FFalse
Answer: True
Density and cardinality are different properties. 'Dense' means no gaps in the ordering — between any two rationals lie infinitely many more. 'Countably infinite' means a bijection with ℕ exists. Cantor's diagonal enumeration constructs this bijection by arranging all fractions in an infinite grid and tracing a diagonal path, visiting every rational exactly once. ℚ is dense in ℝ but still countable; density is a topological property, not a cardinality property.
Question 4 True / False
If A is a proper subset of B (A ⊊ B), then A has strictly fewer elements than B.
TTrue
FFalse
Answer: False
This is true for finite sets but fails completely for infinite ones. The even integers are a proper subset of ℤ, yet both have cardinality ℵ₀. The natural numbers are a proper subset of the rationals, yet |ℕ| = |ℚ| = ℵ₀. Having a proper subset of the same cardinality is not just possible for infinite sets — it is their defining characteristic (Dedekind's definition of infinity). Intuitions built on finite sets do not transfer to infinite ones.
Question 5 Short Answer
The rational numbers seem far denser than the integers, yet both are countably infinite. Explain what 'countably infinite' means and why density doesn't determine countability.
Think about your answer, then reveal below.
Model answer: A set is countably infinite if there exists a bijection with ℕ — equivalently, all its elements can be arranged in a sequence s₁, s₂, s₃, ... with no repetitions or omissions. Density describes how elements are spaced within their natural ordering (ℚ is dense — no gaps between rationals; ℤ is not). But density says nothing about whether a bijection with ℕ can be constructed. Cantor showed such a bijection exists for ℚ by arranging all fractions in a grid and tracing a diagonal path. What distinguishes countable from uncountable sets is whether any listing is possible at all.
The moral is that infinite cardinality is determined by bijections, not by intuitive notions of size, density, or subset. ℚ being dense in ℝ says something about the order topology — rationals cluster near every real number. That's a topological statement, not a cardinality statement. The two properties are independent: ℚ is dense and countable; the Cantor set is uncountable but nowhere dense. Cantor's diagonalization (the next topic) shows why ℝ escapes every attempted listing.