darkprince14
  • darkprince14
Prove: If A and B are finite sets than A U B is a finite set and card(AUB) = card(A) + card(B)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Suppose \(A=B\), is it true?
anonymous
  • anonymous
\[ \|A\cup B\| = \|A\| + \|B\| - \|A\cap B\| \]
darkprince14
  • darkprince14
Nope, A is not equal to B

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darkprince14
  • darkprince14
It just said that they're finite..
anonymous
  • anonymous
Two finite sets can be equal. I'm telling you about a simple counter example.
darkprince14
  • darkprince14
Oooh... sorry for that...
zzr0ck3r
  • zzr0ck3r
if they are both finite and nonempty, then there exists a bijection(f) \[f:(1,2....m)\rightarrow A\\and\\g:(1,2,....n)\rightarrow B\] now define a new fuction\[h:(1,2,....,m+n)\rightarrow A\cup B\\where\space h(i)=i,for\space i=1,2,...m\\and\\h(i)=g(i-m)\space for\space i=m+1,m+2,....m+n \] now show that h is surjective.
AkashdeepDeb
  • AkashdeepDeb
Are you talking about there cardinal numbers?
zzr0ck3r
  • zzr0ck3r
yes
zzr0ck3r
  • zzr0ck3r
hmm let A = {1},|A| = 1 let B = {1},|B| = 1 AUB = {1} but|A|+|B| = 2
zzr0ck3r
  • zzr0ck3r
@darkprince14 you understand? I dont think you can prove that last part.
terenzreignz
  • terenzreignz
Maybe this instead... \[\Large |A \cup B | \le |A| +|B|\]
zzr0ck3r
  • zzr0ck3r
it must be
darkprince14
  • darkprince14
@satellite73 please help..
zzr0ck3r
  • zzr0ck3r
what do you need help with?

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