If ∩F and ∩G are disjoint, then for some A ∈ F and B ∈ G, A and B are disjoint.











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If ∩F and ∩G are disjoint, then for some A ∈ F and B ∈ G, A and B are disjoint.
Give a proof or counter-example.



I'm like 99 percent the theorem is true, can't seem to prove it. Have tried a few methods, closest I have got is trying to prove the contra-positive. Any help with this problem would be extremely appreciated, thanks in advance!






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  • Have you looked for a counterexample?
    – Lord Shark the Unknown
    Nov 21 at 7:09






  • 3




    The above counterexample does not work, since $A cap D$ and $C cap D$ are both empty. However, by taking $D = {1,6,5}$ I think we are done : although $A cap C = {3}$ and $B cap D = {6}$ are disjoint, we see that $3 in A cap B, 1 in A cap D, 3 in C cap B$ and $5 in C cap D$ are true. So $F = {A,C}$ and $G = {B,D}$ serves as a counterexample.
    – астон вілла олоф мэллбэрг
    Nov 21 at 7:22








  • 1




    @астонвіллаолофмэллбэрг, your comment qualifies as an actual answer to the OP. Please post it as an actual answer, so that the question does not remain unanswered.
    – Jose Arnaldo Bebita Dris
    Nov 21 at 8:08






  • 1




    karagila.org/2015/how-to-solve-your-problems
    – Asaf Karagila♦
    Nov 21 at 8:16






  • 1




    @JoseArnaldoBebitaDris Thank you, I have done it below.
    – астон вілла олоф мэллбэрг
    Nov 21 at 8:25















up vote
0
down vote

favorite












If ∩F and ∩G are disjoint, then for some A ∈ F and B ∈ G, A and B are disjoint.
Give a proof or counter-example.



I'm like 99 percent the theorem is true, can't seem to prove it. Have tried a few methods, closest I have got is trying to prove the contra-positive. Any help with this problem would be extremely appreciated, thanks in advance!






elementary-set-theory proof-writing







share|cite|improve this question
























  • Have you looked for a counterexample?
    – Lord Shark the Unknown
    Nov 21 at 7:09






  • 3




    The above counterexample does not work, since $A cap D$ and $C cap D$ are both empty. However, by taking $D = {1,6,5}$ I think we are done : although $A cap C = {3}$ and $B cap D = {6}$ are disjoint, we see that $3 in A cap B, 1 in A cap D, 3 in C cap B$ and $5 in C cap D$ are true. So $F = {A,C}$ and $G = {B,D}$ serves as a counterexample.
    – астон вілла олоф мэллбэрг
    Nov 21 at 7:22








  • 1




    @астонвіллаолофмэллбэрг, your comment qualifies as an actual answer to the OP. Please post it as an actual answer, so that the question does not remain unanswered.
    – Jose Arnaldo Bebita Dris
    Nov 21 at 8:08






  • 1




    karagila.org/2015/how-to-solve-your-problems
    – Asaf Karagila♦
    Nov 21 at 8:16






  • 1




    @JoseArnaldoBebitaDris Thank you, I have done it below.
    – астон вілла олоф мэллбэрг
    Nov 21 at 8:25













up vote
0
down vote

favorite









up vote
0
down vote

favorite











If ∩F and ∩G are disjoint, then for some A ∈ F and B ∈ G, A and B are disjoint.
Give a proof or counter-example.



I'm like 99 percent the theorem is true, can't seem to prove it. Have tried a few methods, closest I have got is trying to prove the contra-positive. Any help with this problem would be extremely appreciated, thanks in advance!






elementary-set-theory proof-writing







share|cite|improve this question















If ∩F and ∩G are disjoint, then for some A ∈ F and B ∈ G, A and B are disjoint.
Give a proof or counter-example.



I'm like 99 percent the theorem is true, can't seem to prove it. Have tried a few methods, closest I have got is trying to prove the contra-positive. Any help with this problem would be extremely appreciated, thanks in advance!






elementary-set-theory proof-writing




elementary-set-theory proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 7:39









астон вілла олоф мэллбэрг

37k33376




37k33376










asked Nov 21 at 7:07









Anon96

1




1












  • Have you looked for a counterexample?
    – Lord Shark the Unknown
    Nov 21 at 7:09






  • 3




    The above counterexample does not work, since $A cap D$ and $C cap D$ are both empty. However, by taking $D = {1,6,5}$ I think we are done : although $A cap C = {3}$ and $B cap D = {6}$ are disjoint, we see that $3 in A cap B, 1 in A cap D, 3 in C cap B$ and $5 in C cap D$ are true. So $F = {A,C}$ and $G = {B,D}$ serves as a counterexample.
    – астон вілла олоф мэллбэрг
    Nov 21 at 7:22








  • 1




    @астонвіллаолофмэллбэрг, your comment qualifies as an actual answer to the OP. Please post it as an actual answer, so that the question does not remain unanswered.
    – Jose Arnaldo Bebita Dris
    Nov 21 at 8:08






  • 1




    karagila.org/2015/how-to-solve-your-problems
    – Asaf Karagila♦
    Nov 21 at 8:16






  • 1




    @JoseArnaldoBebitaDris Thank you, I have done it below.
    – астон вілла олоф мэллбэрг
    Nov 21 at 8:25


















  • Have you looked for a counterexample?
    – Lord Shark the Unknown
    Nov 21 at 7:09






  • 3




    The above counterexample does not work, since $A cap D$ and $C cap D$ are both empty. However, by taking $D = {1,6,5}$ I think we are done : although $A cap C = {3}$ and $B cap D = {6}$ are disjoint, we see that $3 in A cap B, 1 in A cap D, 3 in C cap B$ and $5 in C cap D$ are true. So $F = {A,C}$ and $G = {B,D}$ serves as a counterexample.
    – астон вілла олоф мэллбэрг
    Nov 21 at 7:22








  • 1




    @астонвіллаолофмэллбэрг, your comment qualifies as an actual answer to the OP. Please post it as an actual answer, so that the question does not remain unanswered.
    – Jose Arnaldo Bebita Dris
    Nov 21 at 8:08






  • 1




    karagila.org/2015/how-to-solve-your-problems
    – Asaf Karagila♦
    Nov 21 at 8:16






  • 1




    @JoseArnaldoBebitaDris Thank you, I have done it below.
    – астон вілла олоф мэллбэрг
    Nov 21 at 8:25
















Have you looked for a counterexample?
– Lord Shark the Unknown
Nov 21 at 7:09




Have you looked for a counterexample?
– Lord Shark the Unknown
Nov 21 at 7:09




3




3




The above counterexample does not work, since $A cap D$ and $C cap D$ are both empty. However, by taking $D = {1,6,5}$ I think we are done : although $A cap C = {3}$ and $B cap D = {6}$ are disjoint, we see that $3 in A cap B, 1 in A cap D, 3 in C cap B$ and $5 in C cap D$ are true. So $F = {A,C}$ and $G = {B,D}$ serves as a counterexample.
– астон вілла олоф мэллбэрг
Nov 21 at 7:22






The above counterexample does not work, since $A cap D$ and $C cap D$ are both empty. However, by taking $D = {1,6,5}$ I think we are done : although $A cap C = {3}$ and $B cap D = {6}$ are disjoint, we see that $3 in A cap B, 1 in A cap D, 3 in C cap B$ and $5 in C cap D$ are true. So $F = {A,C}$ and $G = {B,D}$ serves as a counterexample.
– астон вілла олоф мэллбэрг
Nov 21 at 7:22






1




1




@астонвіллаолофмэллбэрг, your comment qualifies as an actual answer to the OP. Please post it as an actual answer, so that the question does not remain unanswered.
– Jose Arnaldo Bebita Dris
Nov 21 at 8:08




@астонвіллаолофмэллбэрг, your comment qualifies as an actual answer to the OP. Please post it as an actual answer, so that the question does not remain unanswered.
– Jose Arnaldo Bebita Dris
Nov 21 at 8:08




1




1




karagila.org/2015/how-to-solve-your-problems
– Asaf Karagila♦
Nov 21 at 8:16




karagila.org/2015/how-to-solve-your-problems
– Asaf Karagila♦
Nov 21 at 8:16




1




1




@JoseArnaldoBebitaDris Thank you, I have done it below.
– астон вілла олоф мэллбэрг
Nov 21 at 8:25




@JoseArnaldoBebitaDris Thank you, I have done it below.
– астон вілла олоф мэллбэрг
Nov 21 at 8:25










2 Answers
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Counterexample : $A = {1,2,3}$, $C = {3,4,5}$, $B = {3,6,7}$, $D = {1,5,6}$ with $F = {A,C}$ and $G = {B,D}$.



In particular, $3 in A cap B$, $3 in C cap B$, $5 in C cap D$ , $1 in A cap D$. However, $cap F = {3}$ and $cap G = {6}$ are disjoint.






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    up vote
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    Think small: $mathcal{F}=bigl{{1,2}bigr}$, $mathcal{G}=bigl{{1},{2}bigr}$.



    If you don't like that the sets have empty intersection, use $mathcal{G}=bigl{{1,3},{2,3}bigr}$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      Counterexample : $A = {1,2,3}$, $C = {3,4,5}$, $B = {3,6,7}$, $D = {1,5,6}$ with $F = {A,C}$ and $G = {B,D}$.



      In particular, $3 in A cap B$, $3 in C cap B$, $5 in C cap D$ , $1 in A cap D$. However, $cap F = {3}$ and $cap G = {6}$ are disjoint.






      share|cite|improve this answer

























        up vote
        1
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        Counterexample : $A = {1,2,3}$, $C = {3,4,5}$, $B = {3,6,7}$, $D = {1,5,6}$ with $F = {A,C}$ and $G = {B,D}$.



        In particular, $3 in A cap B$, $3 in C cap B$, $5 in C cap D$ , $1 in A cap D$. However, $cap F = {3}$ and $cap G = {6}$ are disjoint.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Counterexample : $A = {1,2,3}$, $C = {3,4,5}$, $B = {3,6,7}$, $D = {1,5,6}$ with $F = {A,C}$ and $G = {B,D}$.



          In particular, $3 in A cap B$, $3 in C cap B$, $5 in C cap D$ , $1 in A cap D$. However, $cap F = {3}$ and $cap G = {6}$ are disjoint.






          share|cite|improve this answer












          Counterexample : $A = {1,2,3}$, $C = {3,4,5}$, $B = {3,6,7}$, $D = {1,5,6}$ with $F = {A,C}$ and $G = {B,D}$.



          In particular, $3 in A cap B$, $3 in C cap B$, $5 in C cap D$ , $1 in A cap D$. However, $cap F = {3}$ and $cap G = {6}$ are disjoint.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 at 8:25









          астон вілла олоф мэллбэрг

          37k33376




          37k33376






















              up vote
              1
              down vote













              Think small: $mathcal{F}=bigl{{1,2}bigr}$, $mathcal{G}=bigl{{1},{2}bigr}$.



              If you don't like that the sets have empty intersection, use $mathcal{G}=bigl{{1,3},{2,3}bigr}$






              share|cite|improve this answer

























                up vote
                1
                down vote













                Think small: $mathcal{F}=bigl{{1,2}bigr}$, $mathcal{G}=bigl{{1},{2}bigr}$.



                If you don't like that the sets have empty intersection, use $mathcal{G}=bigl{{1,3},{2,3}bigr}$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Think small: $mathcal{F}=bigl{{1,2}bigr}$, $mathcal{G}=bigl{{1},{2}bigr}$.



                  If you don't like that the sets have empty intersection, use $mathcal{G}=bigl{{1,3},{2,3}bigr}$






                  share|cite|improve this answer












                  Think small: $mathcal{F}=bigl{{1,2}bigr}$, $mathcal{G}=bigl{{1},{2}bigr}$.



                  If you don't like that the sets have empty intersection, use $mathcal{G}=bigl{{1,3},{2,3}bigr}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 at 8:57









                  egreg

                  176k1384198




                  176k1384198






























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