Given $B subseteq A$ and $x in A$, x could also be in B [closed]












2












$begingroup$


This statement should be true right? If $B subseteq A$ and $x in A$, $exists x(x in B)$. I was trying to prove this statement but my previous approach was wrong.



My previous approach involved misstating the property if $B subseteq A$ then $B land A = A$. This obviously is wrong as it should be $B land A = B$.



So this is my new proof.



Given $B subseteq A$, $B lor A = A$. This means:



$forall x(x in (B lor A) leftrightarrow x in A)$



The statement of interest is:



$forall x(x in A rightarrow x in (B lor A)$



Thus if $x in A$, x could be in B. Or that $exists x(x in B)$.






proof-verification







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by lulu, max_zorn, Riccardo.Alestra, José Carlos Santos, Joel Reyes Noche Jan 23 at 12:34


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 2




    $begingroup$
    The fact that $Bsubseteq A$ tells us that $Bcap A=B$, not $A$.
    $endgroup$
    – lulu
    Jan 22 at 22:12










  • $begingroup$
    Note: the claim in the header is clearly false...did you mean to write $Asubseteq B$?
    $endgroup$
    – lulu
    Jan 22 at 22:13






  • 1




    $begingroup$
    This is false. You cannot prove it.
    $endgroup$
    – Ben W
    Jan 22 at 22:14










  • $begingroup$
    Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped?
    $endgroup$
    – lulu
    Jan 22 at 22:18










  • $begingroup$
    Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again.
    $endgroup$
    – fesodes
    Jan 22 at 22:20
















2












$begingroup$


This statement should be true right? If $B subseteq A$ and $x in A$, $exists x(x in B)$. I was trying to prove this statement but my previous approach was wrong.



My previous approach involved misstating the property if $B subseteq A$ then $B land A = A$. This obviously is wrong as it should be $B land A = B$.



So this is my new proof.



Given $B subseteq A$, $B lor A = A$. This means:



$forall x(x in (B lor A) leftrightarrow x in A)$



The statement of interest is:



$forall x(x in A rightarrow x in (B lor A)$



Thus if $x in A$, x could be in B. Or that $exists x(x in B)$.






proof-verification







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by lulu, max_zorn, Riccardo.Alestra, José Carlos Santos, Joel Reyes Noche Jan 23 at 12:34


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 2




    $begingroup$
    The fact that $Bsubseteq A$ tells us that $Bcap A=B$, not $A$.
    $endgroup$
    – lulu
    Jan 22 at 22:12










  • $begingroup$
    Note: the claim in the header is clearly false...did you mean to write $Asubseteq B$?
    $endgroup$
    – lulu
    Jan 22 at 22:13






  • 1




    $begingroup$
    This is false. You cannot prove it.
    $endgroup$
    – Ben W
    Jan 22 at 22:14










  • $begingroup$
    Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped?
    $endgroup$
    – lulu
    Jan 22 at 22:18










  • $begingroup$
    Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again.
    $endgroup$
    – fesodes
    Jan 22 at 22:20














2












2








2





$begingroup$


This statement should be true right? If $B subseteq A$ and $x in A$, $exists x(x in B)$. I was trying to prove this statement but my previous approach was wrong.



My previous approach involved misstating the property if $B subseteq A$ then $B land A = A$. This obviously is wrong as it should be $B land A = B$.



So this is my new proof.



Given $B subseteq A$, $B lor A = A$. This means:



$forall x(x in (B lor A) leftrightarrow x in A)$



The statement of interest is:



$forall x(x in A rightarrow x in (B lor A)$



Thus if $x in A$, x could be in B. Or that $exists x(x in B)$.






proof-verification







share|cite|improve this question











$endgroup$




This statement should be true right? If $B subseteq A$ and $x in A$, $exists x(x in B)$. I was trying to prove this statement but my previous approach was wrong.



My previous approach involved misstating the property if $B subseteq A$ then $B land A = A$. This obviously is wrong as it should be $B land A = B$.



So this is my new proof.



Given $B subseteq A$, $B lor A = A$. This means:



$forall x(x in (B lor A) leftrightarrow x in A)$



The statement of interest is:



$forall x(x in A rightarrow x in (B lor A)$



Thus if $x in A$, x could be in B. Or that $exists x(x in B)$.






proof-verification




proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 22:55







fesodes

















asked Jan 22 at 22:11









fesodesfesodes

262




262




closed as unclear what you're asking by lulu, max_zorn, Riccardo.Alestra, José Carlos Santos, Joel Reyes Noche Jan 23 at 12:34


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by lulu, max_zorn, Riccardo.Alestra, José Carlos Santos, Joel Reyes Noche Jan 23 at 12:34


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 2




    $begingroup$
    The fact that $Bsubseteq A$ tells us that $Bcap A=B$, not $A$.
    $endgroup$
    – lulu
    Jan 22 at 22:12










  • $begingroup$
    Note: the claim in the header is clearly false...did you mean to write $Asubseteq B$?
    $endgroup$
    – lulu
    Jan 22 at 22:13






  • 1




    $begingroup$
    This is false. You cannot prove it.
    $endgroup$
    – Ben W
    Jan 22 at 22:14










  • $begingroup$
    Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped?
    $endgroup$
    – lulu
    Jan 22 at 22:18










  • $begingroup$
    Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again.
    $endgroup$
    – fesodes
    Jan 22 at 22:20














  • 2




    $begingroup$
    The fact that $Bsubseteq A$ tells us that $Bcap A=B$, not $A$.
    $endgroup$
    – lulu
    Jan 22 at 22:12










  • $begingroup$
    Note: the claim in the header is clearly false...did you mean to write $Asubseteq B$?
    $endgroup$
    – lulu
    Jan 22 at 22:13






  • 1




    $begingroup$
    This is false. You cannot prove it.
    $endgroup$
    – Ben W
    Jan 22 at 22:14










  • $begingroup$
    Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped?
    $endgroup$
    – lulu
    Jan 22 at 22:18










  • $begingroup$
    Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again.
    $endgroup$
    – fesodes
    Jan 22 at 22:20








2




2




$begingroup$
The fact that $Bsubseteq A$ tells us that $Bcap A=B$, not $A$.
$endgroup$
– lulu
Jan 22 at 22:12




$begingroup$
The fact that $Bsubseteq A$ tells us that $Bcap A=B$, not $A$.
$endgroup$
– lulu
Jan 22 at 22:12












$begingroup$
Note: the claim in the header is clearly false...did you mean to write $Asubseteq B$?
$endgroup$
– lulu
Jan 22 at 22:13




$begingroup$
Note: the claim in the header is clearly false...did you mean to write $Asubseteq B$?
$endgroup$
– lulu
Jan 22 at 22:13




1




1




$begingroup$
This is false. You cannot prove it.
$endgroup$
– Ben W
Jan 22 at 22:14




$begingroup$
This is false. You cannot prove it.
$endgroup$
– Ben W
Jan 22 at 22:14












$begingroup$
Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped?
$endgroup$
– lulu
Jan 22 at 22:18




$begingroup$
Can you clarify your question? As stated, the statement you seek to prove is clearly wrong. Perhaps you mistyped?
$endgroup$
– lulu
Jan 22 at 22:18












$begingroup$
Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again.
$endgroup$
– fesodes
Jan 22 at 22:20




$begingroup$
Ah thanks for pointing out the mistake. Let me quickly review the question and my proof again.
$endgroup$
– fesodes
Jan 22 at 22:20










1 Answer
1






active

oldest

votes


















0












$begingroup$

Sometimes a figure is worth 1000 words.



enter image description here



By the way, why do you need "and $x in A$" in your title?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if B is the empty set?
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:23










  • $begingroup$
    @JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid.
    $endgroup$
    – David G. Stork
    Jan 22 at 22:25












  • $begingroup$
    Yeah, that's what makes me wonder if he meant $x in B$.
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:26












  • $begingroup$
    Me too.........
    $endgroup$
    – David G. Stork
    Jan 22 at 22:26










  • $begingroup$
    I realized that the property I used was wrong. $B subseteq A$ means $B land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B.
    $endgroup$
    – fesodes
    Jan 22 at 22:36




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Sometimes a figure is worth 1000 words.



enter image description here



By the way, why do you need "and $x in A$" in your title?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if B is the empty set?
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:23










  • $begingroup$
    @JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid.
    $endgroup$
    – David G. Stork
    Jan 22 at 22:25












  • $begingroup$
    Yeah, that's what makes me wonder if he meant $x in B$.
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:26












  • $begingroup$
    Me too.........
    $endgroup$
    – David G. Stork
    Jan 22 at 22:26










  • $begingroup$
    I realized that the property I used was wrong. $B subseteq A$ means $B land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B.
    $endgroup$
    – fesodes
    Jan 22 at 22:36


















0












$begingroup$

Sometimes a figure is worth 1000 words.



enter image description here



By the way, why do you need "and $x in A$" in your title?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if B is the empty set?
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:23










  • $begingroup$
    @JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid.
    $endgroup$
    – David G. Stork
    Jan 22 at 22:25












  • $begingroup$
    Yeah, that's what makes me wonder if he meant $x in B$.
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:26












  • $begingroup$
    Me too.........
    $endgroup$
    – David G. Stork
    Jan 22 at 22:26










  • $begingroup$
    I realized that the property I used was wrong. $B subseteq A$ means $B land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B.
    $endgroup$
    – fesodes
    Jan 22 at 22:36
















0












0








0





$begingroup$

Sometimes a figure is worth 1000 words.



enter image description here



By the way, why do you need "and $x in A$" in your title?






share|cite|improve this answer









$endgroup$



Sometimes a figure is worth 1000 words.



enter image description here



By the way, why do you need "and $x in A$" in your title?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 22 at 22:20









David G. StorkDavid G. Stork

11k41432




11k41432












  • $begingroup$
    What if B is the empty set?
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:23










  • $begingroup$
    @JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid.
    $endgroup$
    – David G. Stork
    Jan 22 at 22:25












  • $begingroup$
    Yeah, that's what makes me wonder if he meant $x in B$.
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:26












  • $begingroup$
    Me too.........
    $endgroup$
    – David G. Stork
    Jan 22 at 22:26










  • $begingroup$
    I realized that the property I used was wrong. $B subseteq A$ means $B land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B.
    $endgroup$
    – fesodes
    Jan 22 at 22:36




















  • $begingroup$
    What if B is the empty set?
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:23










  • $begingroup$
    @JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid.
    $endgroup$
    – David G. Stork
    Jan 22 at 22:25












  • $begingroup$
    Yeah, that's what makes me wonder if he meant $x in B$.
    $endgroup$
    – Jack Pfaffinger
    Jan 22 at 22:26












  • $begingroup$
    Me too.........
    $endgroup$
    – David G. Stork
    Jan 22 at 22:26










  • $begingroup$
    I realized that the property I used was wrong. $B subseteq A$ means $B land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B.
    $endgroup$
    – fesodes
    Jan 22 at 22:36


















$begingroup$
What if B is the empty set?
$endgroup$
– Jack Pfaffinger
Jan 22 at 22:23




$begingroup$
What if B is the empty set?
$endgroup$
– Jack Pfaffinger
Jan 22 at 22:23












$begingroup$
@JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid.
$endgroup$
– David G. Stork
Jan 22 at 22:25






$begingroup$
@JackPfaffinger: Then the statement is false and any proof that doesn't exclude the empty set is invalid.
$endgroup$
– David G. Stork
Jan 22 at 22:25














$begingroup$
Yeah, that's what makes me wonder if he meant $x in B$.
$endgroup$
– Jack Pfaffinger
Jan 22 at 22:26






$begingroup$
Yeah, that's what makes me wonder if he meant $x in B$.
$endgroup$
– Jack Pfaffinger
Jan 22 at 22:26














$begingroup$
Me too.........
$endgroup$
– David G. Stork
Jan 22 at 22:26




$begingroup$
Me too.........
$endgroup$
– David G. Stork
Jan 22 at 22:26












$begingroup$
I realized that the property I used was wrong. $B subseteq A$ means $B land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B.
$endgroup$
– fesodes
Jan 22 at 22:36






$begingroup$
I realized that the property I used was wrong. $B subseteq A$ means $B land A = B$. With that said, the question should still be valid right? As by the picture provided by David, there should at least be an x that is in A and B.
$endgroup$
– fesodes
Jan 22 at 22:36





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