Skip to main content
MathOverflow

Return to Question

Post Closed as "Not suitable for this site" by Noah Schweber, bof, Martin Sleziak, Max Horn, Gro-Tsen
Became Hot Network Question
`\DeclareMathOperator`
Source Link
LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

Let$\DeclareMathOperator\Card{Card}$Let $A$ be an arbitraryinfinite set, $A$ is not finite, and let $B=\{X\subset A | Card(X)<Card(A)\}$$B=\{X\subset A \mid \Card(X)<\Card(A)\}$. Can it be proven that $Card(A)=Card(B)$$\Card(A)=\Card(B)$?

Let $A$ be an arbitrary set, $A$ is not finite, and let $B=\{X\subset A | Card(X)<Card(A)\}$. Can it be proven that $Card(A)=Card(B)$?

$\DeclareMathOperator\Card{Card}$Let $A$ be an infinite set, and let $B=\{X\subset A \mid \Card(X)<\Card(A)\}$. Can it be proven that $\Card(A)=\Card(B)$?

edited tags
Link
Noah Schweber
Noah Schweber
  • 20.5k
  • 10
  • 110
  • 331
added 19 characters in body
Source Link
Vuk Jovovic
Vuk Jovovic
  • 211
  • 1
  • 5

Let $A$ be an arbitrary set, $A$ is not finite, and let $B=\{X\subset A | Card(X)<Card(A)\}$. Can it be proven that $Card(A)=Card(B)$?

Let $A$ be an arbitrary set, and let $B=\{X\subset A | Card(X)<Card(A)\}$. Can it be proven that $Card(A)=Card(B)$?

Let $A$ be an arbitrary set, $A$ is not finite, and let $B=\{X\subset A | Card(X)<Card(A)\}$. Can it be proven that $Card(A)=Card(B)$?

Source Link
Vuk Jovovic
Vuk Jovovic
  • 211
  • 1
  • 5