darkprince14
  • darkprince14
Prove: If A and B are finite sets then A X B is a finite set and card(AXB) = card(A) * card(B)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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darkprince14
  • darkprince14
@zzr0ck3r please help me on this one..
zzr0ck3r
  • zzr0ck3r
give me a few im helping someone else on something, we are almost done.
darkprince14
  • darkprince14
thanks :))

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zzr0ck3r
  • zzr0ck3r
\[\text{let a be fixed in A}\\\text{define}\space B_a=\{(a,b)\in A\space \text{x}\space B\space|\space b\in B\}\\\text{now}\space B_a \space \text{is finite because B is finite}\\A\space \text{x}\space B=\large\cup_{a\in A}\normalsize B_a\text{ is the union of countable sets, and thus countable.}\\\text{-qed}\]
zzr0ck3r
  • zzr0ck3r
do you need the proof of the fact that the union of countable sets is countable?
darkprince14
  • darkprince14
nope.. what is \[B _{a}\]? and what does fixing a in A means? sorry, i'm kinda lost since all of these are new to me..
zzr0ck3r
  • zzr0ck3r
a is fixed means that a is one of the elements, it is not a place holder for all of them B_a is the set as defined example say we have A={a_0,a_1,a_2} B = {1,2,3,4,} when i say let a be fixed in A I mean choose one of the elements say we choose a_0=a then B_a= the cross product of that one element of A with all of b B_a={(a_0,x),(a_0,y),(a_0,z)} now we do it with a = a_1, then a_2....
zzr0ck3r
  • zzr0ck3r
so at the end of the day we have the cross product of one of the elements with all of B then we have the cross product of a different element in A and all of B and then we have the cross product yet a different element of A and all of B when we are done, if we take the union of all these then we have AxB
zzr0ck3r
  • zzr0ck3r
that union is U_{a in A} B_a
darkprince14
  • darkprince14
what about the cardinality of AXB? how can I prove that | AXB | = |A|*|B|?
darkprince14
  • darkprince14
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
sec making dinner
darkprince14
  • darkprince14
oops, sorry.. i'll wait..
zzr0ck3r
  • zzr0ck3r
what do you have? think of your union of B_a this will have |B| elements for every element in A and the union is AxB |UB_a| = |AxB| = |A||B|
darkprince14
  • darkprince14
thanks:)) Last one, how can I prove \[A X \left\{ b \right\}\approx A\] the curl equal was supposed to be the symbol for equinumerous
zzr0ck3r
  • zzr0ck3r
|AXB| = |A|*|b| so |Ax{b}| = |A|*1=|A| so they have the same amount of elements so contruct a bijection
zzr0ck3r
  • zzr0ck3r
f:AXB->A just assign each element a in A to the ordered pair (a,b) then we have a bijection
zzr0ck3r
  • zzr0ck3r
\[(a_i,b)\rightarrow a_i,\space i\in N\]... make sense @darkprince14 ?

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