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Union Set and Correspondence In $R_+^2$

2

1

$begingroup$

I have these sets: $q = [0,1]$ and $Z=1,2$, Let $mathbbY$ be the set combination $(q,z)$ .

Let Y and Y' elements of $mathbbY$. $Y = [0,1] times 1 cup (0,2)$ and $Y' = [0,1] times 2 cup (0,1)$

My questions:

1) Y is a vertical line on 1? And Y' is a vertical line on 2?

2) It is clear that there is no relation such $subseteq$ or $supseteq$ between $Y'$ and $Y$. Right?

3) What Is the joint set: $Y cup Y'$ and the intersection set: $Y cap Y'$?

What I did:

1) I believe it is right that there is no relation $cup$ or $cap$ between $Y'$ and $Y$.

2) This what I did:

$Y cup Y' = [0,1] cup 1,2$ It would be a vertical line on 1 and a vertical lie on 2?

elementary-set-theory

share|cite|improve this question

edited Mar 24 at 20:38

Andrés E. Caicedo

65.9k8160252

asked Mar 24 at 17:54

LauraLaura

3708

$endgroup$

add a comment | 

2

1

$begingroup$

I have these sets: $q = [0,1]$ and $Z=1,2$, Let $mathbbY$ be the set combination $(q,z)$ .

Let Y and Y' elements of $mathbbY$. $Y = [0,1] times 1 cup (0,2)$ and $Y' = [0,1] times 2 cup (0,1)$

My questions:

1) Y is a vertical line on 1? And Y' is a vertical line on 2?

2) It is clear that there is no relation such $subseteq$ or $supseteq$ between $Y'$ and $Y$. Right?

3) What Is the joint set: $Y cup Y'$ and the intersection set: $Y cap Y'$?

What I did:

1) I believe it is right that there is no relation $cup$ or $cap$ between $Y'$ and $Y$.

2) This what I did:

$Y cup Y' = [0,1] cup 1,2$ It would be a vertical line on 1 and a vertical lie on 2?

elementary-set-theory

share|cite|improve this question

edited Mar 24 at 20:38

Andrés E. Caicedo

65.9k8160252

asked Mar 24 at 17:54

LauraLaura

3708

$endgroup$

add a comment | 

$begingroup$

I have these sets: $q = [0,1]$ and $Z=1,2$, Let $mathbbY$ be the set combination $(q,z)$ .

Let Y and Y' elements of $mathbbY$. $Y = [0,1] times 1 cup (0,2)$ and $Y' = [0,1] times 2 cup (0,1)$

My questions:

1) Y is a vertical line on 1? And Y' is a vertical line on 2?

2) It is clear that there is no relation such $subseteq$ or $supseteq$ between $Y'$ and $Y$. Right?

3) What Is the joint set: $Y cup Y'$ and the intersection set: $Y cap Y'$?

What I did:

1) I believe it is right that there is no relation $cup$ or $cap$ between $Y'$ and $Y$.

2) This what I did:

$Y cup Y' = [0,1] cup 1,2$ It would be a vertical line on 1 and a vertical lie on 2?

elementary-set-theory

share|cite|improve this question

edited Mar 24 at 20:38

Andrés E. Caicedo

65.9k8160252

asked Mar 24 at 17:54

LauraLaura

3708

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:38

Andrés E. Caicedo

65.9k8160252

asked Mar 24 at 17:54

LauraLaura

3708

share|cite|improve this question

edited Mar 24 at 20:38

Andrés E. Caicedo

65.9k8160252

asked Mar 24 at 17:54

LauraLaura

3708

edited Mar 24 at 20:38

Andrés E. Caicedo

65.9k8160252

edited Mar 24 at 20:38

Andrés E. Caicedo

65.9k8160252

asked Mar 24 at 17:54

LauraLaura

3708

asked Mar 24 at 17:54

LauraLaura

3708

1 Answer
1

active

oldest

votes

1

$begingroup$

As it stands, there are a few things that don't quite make sense about your post, but I think I can tell what you're getting at. It seems that you mean the following:

• 
  $Q$ is the real interval from $0$ to $1,$ inclusive.


• 
  $Z$ is the set whose elements are just $1$ and $2.$
  


• $Bbb Y=Qtimes Z.$

Now, it looks like we have that

• 
  $Y$ is the set comprised of the ordered pair $langle 0,2rangle$ together with the elements of $Qtimes1.$
  


• 
  $Y'$ is the set comprised of the ordered pair $langle 0,1rangle$ together with the elements of $Qtimes2.$
  

If I'm correct about what you mean by $Bbb Y,$ this would make $Y$ and $Y'$ subsets of $Bbb Y,$ rather than elements of $Bbb Y.$ I can't think of an interpretation of $Bbb Y$ that would make $Y$ and $Y'$ elements of it, but there may be something I'm missing.

So, we could think of $Y$ as a vertical segment on $1,$ together with a single point off that segment. We can think of $Y'$ similarly. It's quite correct that $Ynotsubseteq Y'$ and $Ynotsupseteq Y'.$ $Ycup Y'$ would consist of a vertical segment on both $1$ and $2.$ Moreover, if I'm correct about what you mean by $Bbb Y,$ we in fact have that $Ycup Y'=Bbb Y.$

share|cite|improve this answer

answered Mar 24 at 19:37

Cameron BuieCameron Buie

86.8k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for you effort in understand my question You are right about the $Q$, $Z$ and$ mathbbY$ sets. And what about the intersection? I think it is: $(0,1), (0,2)$. Only two points. Am I right?
  $endgroup$
  – Laura
  Mar 24 at 22:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, indeed! Nicely done, Laura.
  $endgroup$
  – Cameron Buie
  Mar 25 at 0:48
  

  
  
  
  

add a comment | 

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1

$begingroup$

As it stands, there are a few things that don't quite make sense about your post, but I think I can tell what you're getting at. It seems that you mean the following:

• 
  $Q$ is the real interval from $0$ to $1,$ inclusive.


• 
  $Z$ is the set whose elements are just $1$ and $2.$
  


• $Bbb Y=Qtimes Z.$

Now, it looks like we have that

• 
  $Y$ is the set comprised of the ordered pair $langle 0,2rangle$ together with the elements of $Qtimes1.$
  


• 
  $Y'$ is the set comprised of the ordered pair $langle 0,1rangle$ together with the elements of $Qtimes2.$
  

If I'm correct about what you mean by $Bbb Y,$ this would make $Y$ and $Y'$ subsets of $Bbb Y,$ rather than elements of $Bbb Y.$ I can't think of an interpretation of $Bbb Y$ that would make $Y$ and $Y'$ elements of it, but there may be something I'm missing.

So, we could think of $Y$ as a vertical segment on $1,$ together with a single point off that segment. We can think of $Y'$ similarly. It's quite correct that $Ynotsubseteq Y'$ and $Ynotsupseteq Y'.$ $Ycup Y'$ would consist of a vertical segment on both $1$ and $2.$ Moreover, if I'm correct about what you mean by $Bbb Y,$ we in fact have that $Ycup Y'=Bbb Y.$

share|cite|improve this answer

answered Mar 24 at 19:37

Cameron BuieCameron Buie

86.8k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for you effort in understand my question You are right about the $Q$, $Z$ and$ mathbbY$ sets. And what about the intersection? I think it is: $(0,1), (0,2)$. Only two points. Am I right?
  $endgroup$
  – Laura
  Mar 24 at 22:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, indeed! Nicely done, Laura.
  $endgroup$
  – Cameron Buie
  Mar 25 at 0:48
  

  
  
  
  

add a comment | 

1

$begingroup$

As it stands, there are a few things that don't quite make sense about your post, but I think I can tell what you're getting at. It seems that you mean the following:

• 
  $Q$ is the real interval from $0$ to $1,$ inclusive.


• 
  $Z$ is the set whose elements are just $1$ and $2.$
  


• $Bbb Y=Qtimes Z.$

Now, it looks like we have that

• 
  $Y$ is the set comprised of the ordered pair $langle 0,2rangle$ together with the elements of $Qtimes1.$
  


• 
  $Y'$ is the set comprised of the ordered pair $langle 0,1rangle$ together with the elements of $Qtimes2.$
  

If I'm correct about what you mean by $Bbb Y,$ this would make $Y$ and $Y'$ subsets of $Bbb Y,$ rather than elements of $Bbb Y.$ I can't think of an interpretation of $Bbb Y$ that would make $Y$ and $Y'$ elements of it, but there may be something I'm missing.

So, we could think of $Y$ as a vertical segment on $1,$ together with a single point off that segment. We can think of $Y'$ similarly. It's quite correct that $Ynotsubseteq Y'$ and $Ynotsupseteq Y'.$ $Ycup Y'$ would consist of a vertical segment on both $1$ and $2.$ Moreover, if I'm correct about what you mean by $Bbb Y,$ we in fact have that $Ycup Y'=Bbb Y.$

share|cite|improve this answer

answered Mar 24 at 19:37

Cameron BuieCameron Buie

86.8k773161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for you effort in understand my question You are right about the $Q$, $Z$ and$ mathbbY$ sets. And what about the intersection? I think it is: $(0,1), (0,2)$. Only two points. Am I right?
  $endgroup$
  – Laura
  Mar 24 at 22:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, indeed! Nicely done, Laura.
  $endgroup$
  – Cameron Buie
  Mar 25 at 0:48
  

  
  
  
  

add a comment | 

$begingroup$

As it stands, there are a few things that don't quite make sense about your post, but I think I can tell what you're getting at. It seems that you mean the following:

• 
  $Q$ is the real interval from $0$ to $1,$ inclusive.


• 
  $Z$ is the set whose elements are just $1$ and $2.$
  


• $Bbb Y=Qtimes Z.$

Now, it looks like we have that

• 
  $Y$ is the set comprised of the ordered pair $langle 0,2rangle$ together with the elements of $Qtimes1.$
  


• 
  $Y'$ is the set comprised of the ordered pair $langle 0,1rangle$ together with the elements of $Qtimes2.$
  

If I'm correct about what you mean by $Bbb Y,$ this would make $Y$ and $Y'$ subsets of $Bbb Y,$ rather than elements of $Bbb Y.$ I can't think of an interpretation of $Bbb Y$ that would make $Y$ and $Y'$ elements of it, but there may be something I'm missing.

So, we could think of $Y$ as a vertical segment on $1,$ together with a single point off that segment. We can think of $Y'$ similarly. It's quite correct that $Ynotsubseteq Y'$ and $Ynotsupseteq Y'.$ $Ycup Y'$ would consist of a vertical segment on both $1$ and $2.$ Moreover, if I'm correct about what you mean by $Bbb Y,$ we in fact have that $Ycup Y'=Bbb Y.$

share|cite|improve this answer

answered Mar 24 at 19:37

Cameron BuieCameron Buie

86.8k773161

$endgroup$

share|cite|improve this answer

answered Mar 24 at 19:37

Cameron BuieCameron Buie

86.8k773161

answered Mar 24 at 19:37

Cameron BuieCameron Buie

86.8k773161

answered Mar 24 at 19:37

Cameron BuieCameron Buie

86.8k773161