Questions: Comparing Cardinalities: The Schröder-Bernstein Theorem

Question 1

To prove that the open interval (0,1) and the closed interval [0,1] have the same cardinality, which approach correctly applies the Schröder-Bernstein theorem?

• A. Construct an explicit bijection between (0,1) and [0,1] directly
• B. Show an injection from (0,1) into [0,1] and an injection from [0,1] into (0,1), then conclude a bijection exists
• C. Show that both sets are uncountable, so they must have the same cardinality
• D. Show that [0,1] differs from (0,1) by only two points, so the two sets are the same size

Answer: (B)

Schröder-Bernstein: two injections in opposite directions guarantee a bijection. The injection from (0,1) into [0,1] is trivial — the identity function works. The injection from [0,1] into (0,1) requires slightly more care: x ↦ (x+1)/3 maps [0,1] into (1/3, 2/3) ⊂ (0,1). Two easy injections found; the theorem delivers the bijection without requiring you to construct it. Option C is wrong: uncountability alone does not establish equal cardinality — ℝ and ℝ² are both uncountable but comparing them requires further argument.

Question 2

What does the notation |A| ≤ |B| mean formally in the theory of cardinal comparison?

• A. A has fewer elements than B in the usual numerical sense
• B. There exists a surjection from B onto A
• C. There exists an injection from A into B
• D. A is a proper subset of B

Answer: (C)

In set theory, |A| ≤ |B| is defined as: there exists an injection (one-to-one function) from A into B. An injection maps every element of A to a distinct element of B — intuitively, A 'fits inside' B without collisions. Option B (surjection from B onto A) is a different relationship — it means every element of A has at least one preimage in B, which captures a kind of 'coverage' rather than 'fitting inside.' Option D (subset) is also distinct: you can have an injection without A being a subset.

Question 3

The Schröder-Bernstein theorem states that if |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.

• T. True
• F. False

Answer: True

This is the antisymmetry property that makes cardinality comparison well-behaved. It says: if A injects into B and B injects into A, then a bijection exists between them. The proof constructs the bijection explicitly through a partitioning argument — elements are classified by whether their 'ancestry chain' under the two injections terminates in A, terminates in B, or loops forever. The result is foundational: it means the cardinality ordering cannot produce contradictions where |A| ≤ |B| and |B| ≤ |A| but |A| ≠ |B|.

Question 4

To prove that two infinite sets have the same cardinality, you should construct an explicit bijection between them.

• T. True
• F. False

Answer: False

The Schröder-Bernstein theorem provides an alternative: prove two injections (one in each direction) and conclude a bijection must exist, even without constructing it. This is enormously useful because direct bijections between infinite sets can be difficult or unnatural to write down, while injections are often straightforward. For example, proving |(0,1)| = |[0,1]| directly requires a clever countable-sequence trick, but Schröder-Bernstein makes it follow immediately from two simple injections.

Question 5

Why is the Schröder-Bernstein theorem so valuable for comparing cardinalities of infinite sets? What problem does it solve that direct bijection construction doesn't always handle well?

Model answer: For infinite sets, constructing a bijection explicitly can be very difficult — the bijection may require a non-obvious trick or a case-by-case definition that is hard to verify. Schröder-Bernstein replaces one hard problem (find a bijection) with two easy problems (find an injection each way). Injections are often trivial: the identity works in one direction, and a simple scaling or shift works in the other. The theorem guarantees the bijection exists without requiring you to write it down, making cardinality proofs accessible for cases where the bijection would be complicated or counterintuitive.

The deeper significance is foundational: Schröder-Bernstein establishes that ≤ on cardinalities is antisymmetric, making cardinality comparison a partial order. Without it, you could potentially have |A| ≤ |B| and |B| ≤ |A| while |A| ≠ |B| — which would make the cardinality ordering incoherent. Schröder-Bernstein rules this out, confirming that 'same size' for infinite sets is a logically consistent notion. All of cardinal arithmetic and the aleph hierarchy rest on this foundation.