Exercise 1558Logic, sets and proofs › Countability and cardinalityL1Added 3 Nov 2026

Cantor–Bernstein theorem

#Cantor-Bernstein#bijection#injections

Statement

State Cantor–Bernstein–Schröder and give an application.

Hint

Injections both ways give a bijection.

Show solutionHide solution

Solution

Theorem. If there is an injection $A\to B$ and an injection $B\to A$ , then there is a bijection $A\to B$ .

Application. $[0,1]$ and $(0,1)$ have the same cardinality: inclusion $(0,1)\hookrightarrow[0,1]$ , and $x\mapsto x/2+1/4$ injects $[0,1]\to(0,1)$ .

Conclusion. Cantor–Bernstein reduces bijection problems to injections.

Cantor–Bernstein theorem

Statement

Solution

More problems in Countability and cardinality