If for some sets $A, B$ we have their symmetric difference $(A-B) cup (B-A) =B$. Then does that imply $A=emptyset$

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If for some sets $A, B$ we have their symmetric difference $(A-B) cup (B-A) =B$. Then does that imply $A=emptyset$

My Attempt

If $xin B$ then

$$ x in (A-B) cup (B-A) $$

Then

$$x in (A-B) $$ or $$ xin (B-A) $$

Note that $x in (A-B)$ is impossible since $xin B$

We must have $$ xin (B-A) $$ such implies that $ x not in A$ hence $A cap B=emptyset$ (Hope I am correct upto this.)

If I further show $ A cup B=B$ then I can conclude that $A= emptyset$, this solution is one of my classmates. But can there be a better proof?

But can someone help me with an elementary proof.

elementary-set-theory

share|cite|improve this question

edited Jul 30 at 2:05

Andrés E. Caicedo

63.1k7151235

asked Jul 30 at 0:57

Rakesh Bhatt

553112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kaynex the property you mention is exactly the property needing to be proven here. You can't use a result to prove itself, that is circular reasoning.
  – JMoravitz
  Jul 30 at 1:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As for your attempt at a proof @ Rakesh, it is fine. You showed that $Acap B=emptyset$ well, but rather than showing $A=emptyset$ via showing $Acup B=B$., I would instead approach by contrapositive/contradiction. Supposing that $Acap B=emptyset$ and $A$ is nonempty, then there should be some element $ain A-B$, but then that would have meant ______... (also, as a typesetting note, emptyset is more common than phi)
  – JMoravitz
  Jul 30 at 1:05
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

If for some sets $A, B$ we have their symmetric difference $(A-B) cup (B-A) =B$. Then does that imply $A=emptyset$

My Attempt

If $xin B$ then

$$ x in (A-B) cup (B-A) $$

Then

$$x in (A-B) $$ or $$ xin (B-A) $$

Note that $x in (A-B)$ is impossible since $xin B$

We must have $$ xin (B-A) $$ such implies that $ x not in A$ hence $A cap B=emptyset$ (Hope I am correct upto this.)

If I further show $ A cup B=B$ then I can conclude that $A= emptyset$, this solution is one of my classmates. But can there be a better proof?

But can someone help me with an elementary proof.

elementary-set-theory

share|cite|improve this question

edited Jul 30 at 2:05

Andrés E. Caicedo

63.1k7151235

asked Jul 30 at 0:57

Rakesh Bhatt

553112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kaynex the property you mention is exactly the property needing to be proven here. You can't use a result to prove itself, that is circular reasoning.
  – JMoravitz
  Jul 30 at 1:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As for your attempt at a proof @ Rakesh, it is fine. You showed that $Acap B=emptyset$ well, but rather than showing $A=emptyset$ via showing $Acup B=B$., I would instead approach by contrapositive/contradiction. Supposing that $Acap B=emptyset$ and $A$ is nonempty, then there should be some element $ain A-B$, but then that would have meant ______... (also, as a typesetting note, emptyset is more common than phi)
  – JMoravitz
  Jul 30 at 1:05
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If for some sets $A, B$ we have their symmetric difference $(A-B) cup (B-A) =B$. Then does that imply $A=emptyset$

My Attempt

If $xin B$ then

$$ x in (A-B) cup (B-A) $$

Then

$$x in (A-B) $$ or $$ xin (B-A) $$

Note that $x in (A-B)$ is impossible since $xin B$

We must have $$ xin (B-A) $$ such implies that $ x not in A$ hence $A cap B=emptyset$ (Hope I am correct upto this.)

If I further show $ A cup B=B$ then I can conclude that $A= emptyset$, this solution is one of my classmates. But can there be a better proof?

But can someone help me with an elementary proof.

elementary-set-theory

share|cite|improve this question

edited Jul 30 at 2:05

Andrés E. Caicedo

63.1k7151235

asked Jul 30 at 0:57

Rakesh Bhatt

553112

If for some sets $A, B$ we have their symmetric difference $(A-B) cup (B-A) =B$. Then does that imply $A=emptyset$

My Attempt

If $xin B$ then

$$ x in (A-B) cup (B-A) $$

Then

$$x in (A-B) $$ or $$ xin (B-A) $$

Note that $x in (A-B)$ is impossible since $xin B$

We must have $$ xin (B-A) $$ such implies that $ x not in A$ hence $A cap B=emptyset$ (Hope I am correct upto this.)

If I further show $ A cup B=B$ then I can conclude that $A= emptyset$, this solution is one of my classmates. But can there be a better proof?

But can someone help me with an elementary proof.

elementary-set-theory

share|cite|improve this question

edited Jul 30 at 2:05

Andrés E. Caicedo

63.1k7151235

asked Jul 30 at 0:57

Rakesh Bhatt

553112

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 2:05

Andrés E. Caicedo

63.1k7151235

edited Jul 30 at 2:05

Andrés E. Caicedo

63.1k7151235

edited Jul 30 at 2:05

Andrés E. Caicedo

63.1k7151235

63.1k7151235

asked Jul 30 at 0:57

Rakesh Bhatt

553112

asked Jul 30 at 0:57

Rakesh Bhatt

553112

asked Jul 30 at 0:57

Rakesh Bhatt

553112

553112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kaynex the property you mention is exactly the property needing to be proven here. You can't use a result to prove itself, that is circular reasoning.
  – JMoravitz
  Jul 30 at 1:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As for your attempt at a proof @ Rakesh, it is fine. You showed that $Acap B=emptyset$ well, but rather than showing $A=emptyset$ via showing $Acup B=B$., I would instead approach by contrapositive/contradiction. Supposing that $Acap B=emptyset$ and $A$ is nonempty, then there should be some element $ain A-B$, but then that would have meant ______... (also, as a typesetting note, emptyset is more common than phi)
  – JMoravitz
  Jul 30 at 1:05
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Kaynex the property you mention is exactly the property needing to be proven here. You can't use a result to prove itself, that is circular reasoning.
  – JMoravitz
  Jul 30 at 1:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As for your attempt at a proof @ Rakesh, it is fine. You showed that $Acap B=emptyset$ well, but rather than showing $A=emptyset$ via showing $Acup B=B$., I would instead approach by contrapositive/contradiction. Supposing that $Acap B=emptyset$ and $A$ is nonempty, then there should be some element $ain A-B$, but then that would have meant ______... (also, as a typesetting note, emptyset is more common than phi)
  – JMoravitz
  Jul 30 at 1:05
  
  

  
  
  
  

@Kaynex the property you mention is exactly the property needing to be proven here. You can't use a result to prove itself, that is circular reasoning.
– JMoravitz
Jul 30 at 1:02

@Kaynex the property you mention is exactly the property needing to be proven here. You can't use a result to prove itself, that is circular reasoning.
– JMoravitz
Jul 30 at 1:02

1

1

As for your attempt at a proof @ Rakesh, it is fine. You showed that $Acap B=emptyset$ well, but rather than showing $A=emptyset$ via showing $Acup B=B$., I would instead approach by contrapositive/contradiction. Supposing that $Acap B=emptyset$ and $A$ is nonempty, then there should be some element $ain A-B$, but then that would have meant ______... (also, as a typesetting note, emptyset is more common than phi)
– JMoravitz
Jul 30 at 1:05

As for your attempt at a proof @ Rakesh, it is fine. You showed that $Acap B=emptyset$ well, but rather than showing $A=emptyset$ via showing $Acup B=B$., I would instead approach by contrapositive/contradiction. Supposing that $Acap B=emptyset$ and $A$ is nonempty, then there should be some element $ain A-B$, but then that would have meant ______... (also, as a typesetting note, emptyset is more common than phi)
– JMoravitz
Jul 30 at 1:05

add a comment | 

1 Answer
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up vote
2
down vote



accepted

If $ain A$ then

(i). If $;ain B$ then $ain Acap B=Acap (ADelta B)=A$ $B$ so $anot in B, $ a contradiction.

(ii). If $; anot in B$ then $ain A$ $ B subset ADelta B=B$ so $ ain B, $ a contradiction.

So no $a$ belongs to $A.$

share|cite|improve this answer

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a direct proof: Since $A=(A$ $B ) cup (Acap B),$ show first that $Acap B=A$ $B$ and show second that $A backslash B =phi.$ First, $Acap B=A cap (ADelta B)=A$ $ B.$ Second, $A$ $ B= (Abackslash B)cap (ADelta B)=(A backslash B)cap B=phi.$
  – DanielWainfleet
  Jul 30 at 18:59
  
  

  
  
  
  

add a comment | 

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

If $ain A$ then

(i). If $;ain B$ then $ain Acap B=Acap (ADelta B)=A$ $B$ so $anot in B, $ a contradiction.

(ii). If $; anot in B$ then $ain A$ $ B subset ADelta B=B$ so $ ain B, $ a contradiction.

So no $a$ belongs to $A.$

share|cite|improve this answer

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a direct proof: Since $A=(A$ $B ) cup (Acap B),$ show first that $Acap B=A$ $B$ and show second that $A backslash B =phi.$ First, $Acap B=A cap (ADelta B)=A$ $ B.$ Second, $A$ $ B= (Abackslash B)cap (ADelta B)=(A backslash B)cap B=phi.$
  – DanielWainfleet
  Jul 30 at 18:59
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

If $ain A$ then

(i). If $;ain B$ then $ain Acap B=Acap (ADelta B)=A$ $B$ so $anot in B, $ a contradiction.

(ii). If $; anot in B$ then $ain A$ $ B subset ADelta B=B$ so $ ain B, $ a contradiction.

So no $a$ belongs to $A.$

share|cite|improve this answer

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a direct proof: Since $A=(A$ $B ) cup (Acap B),$ show first that $Acap B=A$ $B$ and show second that $A backslash B =phi.$ First, $Acap B=A cap (ADelta B)=A$ $ B.$ Second, $A$ $ B= (Abackslash B)cap (ADelta B)=(A backslash B)cap B=phi.$
  – DanielWainfleet
  Jul 30 at 18:59
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

If $ain A$ then

(i). If $;ain B$ then $ain Acap B=Acap (ADelta B)=A$ $B$ so $anot in B, $ a contradiction.

(ii). If $; anot in B$ then $ain A$ $ B subset ADelta B=B$ so $ ain B, $ a contradiction.

So no $a$ belongs to $A.$

share|cite|improve this answer

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

If $ain A$ then

(i). If $;ain B$ then $ain Acap B=Acap (ADelta B)=A$ $B$ so $anot in B, $ a contradiction.

(ii). If $; anot in B$ then $ain A$ $ B subset ADelta B=B$ so $ ain B, $ a contradiction.

So no $a$ belongs to $A.$

share|cite|improve this answer

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

answered Jul 30 at 2:03

DanielWainfleet

31.4k31542

31.4k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a direct proof: Since $A=(A$ $B ) cup (Acap B),$ show first that $Acap B=A$ $B$ and show second that $A backslash B =phi.$ First, $Acap B=A cap (ADelta B)=A$ $ B.$ Second, $A$ $ B= (Abackslash B)cap (ADelta B)=(A backslash B)cap B=phi.$
  – DanielWainfleet
  Jul 30 at 18:59
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  For a direct proof: Since $A=(A$ $B ) cup (Acap B),$ show first that $Acap B=A$ $B$ and show second that $A backslash B =phi.$ First, $Acap B=A cap (ADelta B)=A$ $ B.$ Second, $A$ $ B= (Abackslash B)cap (ADelta B)=(A backslash B)cap B=phi.$
  – DanielWainfleet
  Jul 30 at 18:59
  
  

  
  
  
  

For a direct proof: Since $A=(A$ $B ) cup (Acap B),$ show first that $Acap B=A$ $B$ and show second that $A backslash B =phi.$ First, $Acap B=A cap (ADelta B)=A$ $ B.$ Second, $A$ $ B= (Abackslash B)cap (ADelta B)=(A backslash B)cap B=phi.$
– DanielWainfleet
Jul 30 at 18:59

For a direct proof: Since $A=(A$ $B ) cup (Acap B),$ show first that $Acap B=A$ $B$ and show second that $A backslash B =phi.$ First, $Acap B=A cap (ADelta B)=A$ $ B.$ Second, $A$ $ B= (Abackslash B)cap (ADelta B)=(A backslash B)cap B=phi.$
– DanielWainfleet
Jul 30 at 18:59

add a comment | 

 

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