Let X, Y be two nonempty sets and let f : X → Y. For a, b ∈ X we write a ∼ b iff f ( a ) = f ( b ).a. Prove that ∼ is an equivalence relation on X.b: Write [ x ] ∫ for the equivalence class of x ∈ X with respect to ∼. Express [ x ] ∫ in terms of the function f : [ x ] ∫ = { x ′ ∈ X : f ( x ′ ) . . . ? ? . . . } (I do not want to see " [ x ] ∫ = { x ′ ∈ X : x ′ ∼ x } " . )

Question

Let X, Y be two nonempty sets and let   f  :  X  →  Y. For a,   b  ∈  X we write   a  ∼  b iff   f  (  a  )  =  f  (  b  ).a. Prove that   ∼ is an equivalence relation on X.b: Write   [  x      ]          ∫       for the equivalence class of   x  ∈  X with respect to   ∼. Express   [  x      ]          ∫       in terms of the function   f  :  [  x      ]          ∫        =  {      x    ′    ∈  X  :  f  (      x    ′    )  .  .  .  ?  ?  .  .  .  } (I do not want to see   &quot;  [  x      ]          ∫        =  {      x    ′    ∈  X  :      x    ′    ∼  x  }  &quot;  .  )

Andres Barrett · Accepted Answer

Step 1Given x, y be two non empty sets and let   f  :  x  →  yFor   a  ,  b  ∈  x  ,  a  ∼  b iff   f  (  a  )  =  f  (  b  ).a) To prove:   ∼ is equivalence relation on xProof:   ∼ is reflexiveFor any   a  ∼  a  ,  ∀  a  ∈  xhence   ∼ is reflexive  ∼ is symmetricFor   a  ,  b  ∈  x     let     a  ∼  b  f  (  a  )  =  f  (  b  )  f  (  b  )  =  f  (  a  )  b  ∼  ahence   ∼ is symmetric  ∼ is transitiveStep 2Let   a  ,  b  ,  c  ∈  x such that   a  ∼  b and   b  ∼  c  f  (  a  )  =  f  (  b  )     and     f  (  b  )  =  f  (  c  )  f  (  a  )  =  f  (  c  )  a  ∼  c  ⇒∼ is TransitiveHence   ∼ is equivalence relation  [  x      ]    f    =  {      x    ′    ∈  x      |    f  (  x  )  =  f  (      x    ′    )  }

Let X, Y be two nonempty sets and let f:X rightarrow Y. For a, b in X we write a ~ b iff f(a)=f(b).

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