A student argues that the set of integers ℤ must have strictly greater cardinality than the natural numbers ℕ, because ℤ contains all of ℕ plus all the negative integers. What is wrong with this reasoning?
Aℤ is actually a finite set, so the argument about size doesn't apply
BThe student is correct — ℤ has strictly greater cardinality than ℕ
CA bijection can be constructed between ℕ and ℤ, so they have the same cardinality despite ℤ appearing larger
DCardinality comparisons are only meaningful for finite sets — infinite sets cannot be compared
The student's intuition that 'more elements → larger cardinality' works for finite sets but fails for infinite ones. Cardinality is defined by bijection existence, not by subset relations. The function f(n) = n/2 for even n and -(n+1)/2 for odd n maps ℕ → ℤ bijectively: 0↔0, 1↔−1, 2↔1, 3↔−2, ... Every integer appears exactly once. Because a bijection exists, |ℕ| = |ℤ| — they are both countably infinite, even though ℕ ⊂ ℤ. This is the signature property of infinite sets: they can be put in bijection with proper subsets of themselves.
Question 2 Multiple Choice
Which of the following sets is uncountable — having strictly greater cardinality than the natural numbers?
AThe set of all integers ℤ
BThe set of all rational numbers ℚ
CThe set of all real numbers ℝ
DThe set of all pairs of natural numbers ℕ × ℕ
ℤ, ℚ, and ℕ × ℕ are all countably infinite — explicit bijections to ℕ can be constructed for each (sign-interleave for ℤ, Cantor's zigzag for ℕ × ℕ and then ℚ). The real numbers ℝ are provably uncountable: Cantor's diagonal argument shows that any attempted listing of real numbers in [0,1] is incomplete — a real number can always be constructed that differs from the nth listed number in the nth decimal place. |ℝ| > |ℕ|, representing a genuinely larger infinite cardinality.
Question 3 True / False
Two infinite sets can have the same cardinality even when one appears to contain 'more elements' — cardinality equality is determined by bijection existence, not by subset relations or apparent density.
TTrue
FFalse
Answer: True
True. This is the counterintuitive core of set-theoretic cardinality. ℕ ⊂ ℤ ⊂ ℚ, and the rationals are dense (between any two rationals lies another), yet |ℕ| = |ℤ| = |ℚ|. All three are countably infinite. The bijection criterion replaces intuitive 'size' with a precise matching condition. For infinite sets, being a proper subset is consistent with having equal cardinality — a fact that troubled mathematicians when Cantor first proposed it.
Question 4 True / False
Because the rational numbers are dense — between any two rationals there is generally another rational — the set of rational numbers is expected to be uncountable, having strictly greater cardinality than the natural numbers.
TTrue
FFalse
Answer: False
False. Density and cardinality are independent properties. The rational numbers are dense in ℝ (no gaps between rationals) but are countably infinite — the same cardinality as ℕ. Cantor's zigzag argument constructs an explicit enumeration of all positive rationals by traversing the grid ℕ × ℕ diagonally, then extending to all rationals. Density tells you about the order structure (no isolated points, no gaps in the ordering); cardinality tells you about the size of the set in the bijection sense. These two notions are logically independent.
Question 5 Short Answer
What does it mean for two sets to have the same cardinality, and why does the Cantor-Bernstein-Schroeder theorem make establishing cardinality equality practically useful?
Think about your answer, then reveal below.
Model answer: Two sets A and B have the same cardinality (|A| = |B|) if and only if there exists a bijection between them — a function that is both injective (one-to-one) and surjective (onto). The Cantor-Bernstein-Schroeder theorem makes this practical by allowing a two-step shortcut: rather than constructing an explicit bijection (which can be technically difficult), you can establish |A| = |B| by finding any injection from A into B and any separate injection from B into A. The theorem guarantees these two injections can be combined into a bijection, even without constructing it explicitly. This is analogous to the squeeze theorem: bounding |A| ≤ |B| ≤ |A| pins down equality.
Explicit bijections between infinite sets are often hard to construct directly. For example, proving |(0,1)| = |ℝ| is straightforward via CBS: embed (0,1) into ℝ via inclusion, and embed ℝ into (0,1) via arctan rescaled. Without CBS, you would need to write down an explicit bijection from all of ℝ to the open interval — a more demanding construction. The theorem's non-constructive nature (it proves the bijection exists without exhibiting it) is a powerful feature of modern set theory.