 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/16h/d/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 16H / d / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

This is König theorem for cardinal numbers. Since

κ_{i} < λ_{i}

(i, α)

in the disjoint union to the element

s_{i, α}

of the product defined by

s_{i, α} (j) = {α + 1, 0, j = i, j \neq = i .

Here

α < κ_{i}

implies

α + 1 < λ_{i}

. The unique nonzero coordinate and its value recover

(i, α)

, so this is an injection

⨆_{i \in I} κ_{i} ⟶ \prod_{i \in I} λ_{i} .

For strictness, suppose

F

mapped the disjoint union onto the product. For each

i

, the set

A_{i} = {F (i, α) (i) : α < κ_{i}}

has cardinality at most

κ_{i}

, so it cannot exhaust

λ_{i}

. Choose

b_{i} \in λ_{i} ∖ A_{i}

. Then

b = (b_{i})_{i \in I}

differs from

F (i, α)

in coordinate

i

for every

(i, α)

, contradicting surjectivity. By Cantor-Schröder-Bernstein theorem, an injection in the reverse direction would combine with the displayed injection to give a bijection and hence a surjection. Therefore

i \in I \sum κ_{i} < i \in I \prod λ_{i} .

 Read the full article

 New to topics? Read the docs here!