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bergausstein

Active member
Jul 30, 2013
191
please correct my mistakes in this solution in 1 and help me solve 2.

1. show that if $A\cap B\,=\,\emptyset$, then $\left(A\,x\,C\right)\cap \left(B\,x\,C\right)\,=\,\emptyset$ that is if A and B have no elements in common, then there can be no ordered pair $\left(x,c\right)$ which is in both AxB and BxC.

my solution,

$A\,x\,C\,=\,\{\left(a,c\right)|a\in A\,\wedge\,c\in C\}$
$B\,x\,C\,=\,\{\left(b,c\right)|b\in B\,\wedge\,c\in C\}$

since AxC and BxC have no elements in common $A\cap B\,=\,\emptyset$

2. explain why $\left(A\cap B\right)\,x\,C\,=\,\left(A\,x\,C \right)\cap \left(B\,x\,C\right)$, that is if an ordered pair (x,c) is in AxC and in BxC, then (x,c) must be in $\left(A\cap B\right)\,x\,C$ and conversely.

thanks!
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
please correct my mistakes in this solution in 1 and help me solve 2.

1. show that if $A\cap B\,=\,\emptyset$, then $\left(A\,x\,C\right)\cap \left(B\,x\,C\right)\,=\,\emptyset$ that is if A and B have no elements in common, then there can be no ordered pair $\left(x,c\right)$ which is in both AxB and BxC.

my solution,

$A\,x\,C\,=\,\{\left(a,c\right)|a\in A\,\wedge\,c\in C\}$
$B\,x\,C\,=\,\{\left(b,c\right)|b\in B\,\wedge\,c\in C\}$

since AxC and BxC have no elements in common $A\cap B\,=\,\emptyset$

2. explain why $\left(A\cap B\right)\,x\,C\,=\,\left(A\,x\,C \right)\cap \left(B\,x\,C\right)$, that is if an ordered pair (x,c) is in AxC and in BxC, then (x,c) must be in $\left(A\cap B\right)\,x\,C$ and conversely.

thanks!
Hey Bergausstein. You sure have a daunting nick right there!! :)

In the first one you are doing the opposite of what you are required to do. You need to show the emptiness of $(A\times B)\cap(B\times C)$ by assuming that $A\cap B=\emptyset$.. not the other way round.

For the second one let $(x,y)\in (A\cap B)\times C$. Then $x\in A\cap B$ and $y\in C$. Thus $(x,y)\in A\times C$ and $(x,y)\in B\times C$. Therefore $(x,y)\in (A\times C)\cap (B\times C)$. For the reverse containment take $(x,y)\in (A\times C)\cap (B\times C)$. Thus $(x,y)\in A\times C$ and $(x,y)\in B\times C$. This means $x\in A\cap B$ and $y\in C$ and therefore $(x,y)\in (A\cap B)\times C$ and we are done.

Tell me if you have further doubts.
 

paulmdrdo

Active member
May 13, 2013
386
can you show me how will you answer 1.?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
can you show me how will you answer 1.?
caffeinemachine has already showed that if $(x,y)\in (A\times C)\cap (B\times C)$ for some $x,y$, i.e., if $(A\times C)\cap (B\times C)$ is nonempty, then $x\in A\cap B$, i.e., $A\cap B$ is nonempty.
 

paulmdrdo

Active member
May 13, 2013
386
but the problem is asking to show the emptiness of $A\cap B$ not its non emptiness.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
1. show that if $A\cap B\,=\,\emptyset$, then $\left(A\,\times\,C\right)\cap \left(B\,\times\,C\right)\,=\,\emptyset$
but the problem is asking to show the emptiness of $A\cap B$ not its non emptiness.
Really? Does the problem ask you to show $A\cap B=\emptyset$? Under which assumption?

LaTeX hints: Use \times for Cartesian product. There is no need to insert thin spaces; LaTeX is designed to insert correct spaces around operators and relations. Also, there is no need to write \left and \right unless the content between the parentheses is taller than them, e.g., when parentheses surround a fraction.
 

bergausstein

Active member
Jul 30, 2013
191
guys i'm still confused.

in my solution to prob. 1 I showed that,

$A\,\times\,C\,=\,\{\left(a,c\right)|a\in A\,\wedge\,c\in C\}$
$B\,\times\,C\,=\,\{\left(b,c\right)|b\in B\,\wedge\,c\in C\}$

there's no ordered pair common to both set. hence, $\left(A\,\times\,C\right)\cap \left(B\,\times C\right)\,=\,\emptyset$. how come i'm doing the other way around? please help me. can you give an example analogous to this problem. :(
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
guys i'm still confused.

in my solution to prob. 1 I showed that,

$A\,\times\,C\,=\,\{\left(a,c\right)|a\in A\,\wedge\,c\in C\}$
$B\,\times\,C\,=\,\{\left(b,c\right)|b\in B\,\wedge\,c\in C\}$

there's no ordered pair common to both set. hence, $\left(A\,\times\,C\right)\cap \left(B\,\times C\right)\,=\,\emptyset$. how come i'm doing the other way around? please help me. can you give an example analogous to this problem. :(
In your first post you have
"since AxC and BxC have no elements in common [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]∩[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]∅[/FONT]"
This sounds like you are trying to show $(A\times C)\cap (B\times C)=\emptyset ~\Rightarrow ~A\cap B=\emptyset$.. don't you think?
 

LATEBLOOMER

New member
Aug 7, 2013
21
if you want example consider this

Say $A=\{1,2\}$ , $B=\{3,4\}$, $C=\{s,t\}$ you see that $A\cap B =\emptyset$

now if we take the product of A and C we have,
$A\times C = \{(1,s),\,(1,t),\,(2,s),\,(2,t)\}$
the product of B and C we have,
$B\times C = \{(3,s),\,(3,t),\,(4,s),\,(4,t)\}$

now if we take $(A\times C)\cap (B\times C)$ it is empty since they don't have elements in common.
now we can say that $(A\times C)\cap (B\times C)=\emptyset$ Because in the first place $A\cap B=\emptyset$.

hope this would help! :)
 

bergausstein

Active member
Jul 30, 2013
191
now I understand it! thanks latebloomer.