Question

Prove: If A and B are finite sets then A X B is a finite set and card(AXB) = card(A) * card(B)

Answer 1

@zzr0ck3r please help me on this one..

Answer 2

give me a few im helping someone else on something, we are almost done.

Answer 3

thanks :))

Answer 4

\[\text{let a be fixed in A}\\\text{define}\space B_a=\{(a,b)\in A\space \text{x}\space B\space|\space b\in B\}\\\text{now}\space B_a \space \text{is finite because B is finite}\\A\space \text{x}\space B=\large\cup_{a\in A}\normalsize B_a\text{ is the union of countable sets, and thus countable.}\\\text{-qed}\]

Answer 5

do you need the proof of the fact that the union of countable sets is countable?

Answer 6

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..

Answer 7

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....

Answer 8

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

Answer 9

that union is
U_{a in A} B_a

Answer 10

what about the cardinality of AXB?

how can I prove that | AXB | = |A|*|B|?

Answer 11

@zzr0ck3r

Answer 12

sec making dinner

Answer 13

oops, sorry.. i'll wait..

Answer 14

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|

Answer 15

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

Answer 16

|AXB| = |A|*|b|
so
|Ax{b}| = |A|*1=|A| so they have the same amount of elements
so contruct a bijection

Answer 17

f:AXB->A
just assign each element a in A to the ordered pair (a,b)
then we have a bijection

Answer 18

\[(a_i,b)\rightarrow a_i,\space i\in N\]...
make sense @darkprince14 ?