Mixing
The union of two subsets of $mathbb R^3$.
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I have the following subsets of $mathbb R^3$:
$$A=left(x,y,z): z_1leq zleq z_2, , x<barx, , y_1leq yleq y_2right$$
$$B=left(x,y,z): z_1leq zleq z_2, , x>barx, , y_3leq yleq y_2right$$
where and $bar x$, $y_1$, $y_2$, $y_3$, $z_1$, $z_2$ are fixed numbers and $y_1leq y_3$.
What is the union of the two sets? I thought:
$$Acup B=z_1leq zleq z_2, , xneqbarx, , y_1leq yleq y_2, .$$
algebra-precalculus elementary-set-theory
asked Jul 16 at 13:57
Mark
3,56251746
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up vote
3
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I have the following subsets of $mathbb R^3$:
$$A=left(x,y,z): z_1leq zleq z_2, , x<barx, , y_1leq yleq y_2right$$
$$B=left(x,y,z): z_1leq zleq z_2, , x>barx, , y_3leq yleq y_2right$$
where and $bar x$, $y_1$, $y_2$, $y_3$, $z_1$, $z_2$ are fixed numbers and $y_1leq y_3$.
What is the union of the two sets? I thought:
$$Acup B=z_1leq zleq z_2, , xneqbarx, , y_1leq yleq y_2, .$$
algebra-precalculus elementary-set-theory
asked Jul 16 at 13:57
Mark
3,56251746
Not quite right. Consider some $z$ between $z_1$ and $z_2$, some $y$ between $y_1$ and $y_3$, and some $xneqbar x$. Then $(x,y,z)$ will never be in $B$ (because $y<y_3$), and it will be in $A$ and therefore in the union only when $x<bar x$. But your formula for $Acup B$ would include $(x,y,z)$ also when $x>bar x$.
– Andreas Blass
Jul 16 at 15:09
@AndreasBlass. Thank you! However, how can I write the union?
– Mark
Jul 16 at 17:17
1
See the corrected answer from gimusi.
– Andreas Blass
Jul 16 at 17:30
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have the following subsets of $mathbb R^3$:
$$A=left(x,y,z): z_1leq zleq z_2, , x<barx, , y_1leq yleq y_2right$$
$$B=left(x,y,z): z_1leq zleq z_2, , x>barx, , y_3leq yleq y_2right$$
where and $bar x$, $y_1$, $y_2$, $y_3$, $z_1$, $z_2$ are fixed numbers and $y_1leq y_3$.
What is the union of the two sets? I thought:
$$Acup B=z_1leq zleq z_2, , xneqbarx, , y_1leq yleq y_2, .$$
algebra-precalculus elementary-set-theory
asked Jul 16 at 13:57
Mark
3,56251746
I have the following subsets of $mathbb R^3$:
$$A=left(x,y,z): z_1leq zleq z_2, , x<barx, , y_1leq yleq y_2right$$
$$B=left(x,y,z): z_1leq zleq z_2, , x>barx, , y_3leq yleq y_2right$$
where and $bar x$, $y_1$, $y_2$, $y_3$, $z_1$, $z_2$ are fixed numbers and $y_1leq y_3$.
What is the union of the two sets? I thought:
$$Acup B=z_1leq zleq z_2, , xneqbarx, , y_1leq yleq y_2, .$$
algebra-precalculus elementary-set-theory
asked Jul 16 at 13:57
Mark
3,56251746
asked Jul 16 at 13:57
Mark
3,56251746
asked Jul 16 at 13:57
Mark
3,56251746
asked Jul 16 at 13:57
Mark
3,56251746
3,56251746
Not quite right. Consider some $z$ between $z_1$ and $z_2$, some $y$ between $y_1$ and $y_3$, and some $xneqbar x$. Then $(x,y,z)$ will never be in $B$ (because $y<y_3$), and it will be in $A$ and therefore in the union only when $x<bar x$. But your formula for $Acup B$ would include $(x,y,z)$ also when $x>bar x$.
– Andreas Blass
Jul 16 at 15:09
@AndreasBlass. Thank you! However, how can I write the union?
– Mark
Jul 16 at 17:17
1
See the corrected answer from gimusi.
– Andreas Blass
Jul 16 at 17:30
add a comment |Â
Not quite right. Consider some $z$ between $z_1$ and $z_2$, some $y$ between $y_1$ and $y_3$, and some $xneqbar x$. Then $(x,y,z)$ will never be in $B$ (because $y<y_3$), and it will be in $A$ and therefore in the union only when $x<bar x$. But your formula for $Acup B$ would include $(x,y,z)$ also when $x>bar x$.
– Andreas Blass
Jul 16 at 15:09
@AndreasBlass. Thank you! However, how can I write the union?
– Mark
Jul 16 at 17:17
1
See the corrected answer from gimusi.
– Andreas Blass
Jul 16 at 17:30
Not quite right. Consider some $z$ between $z_1$ and $z_2$, some $y$ between $y_1$ and $y_3$, and some $xneqbar x$. Then $(x,y,z)$ will never be in $B$ (because $y<y_3$), and it will be in $A$ and therefore in the union only when $x<bar x$. But your formula for $Acup B$ would include $(x,y,z)$ also when $x>bar x$.
– Andreas Blass
Jul 16 at 15:09
Not quite right. Consider some $z$ between $z_1$ and $z_2$, some $y$ between $y_1$ and $y_3$, and some $xneqbar x$. Then $(x,y,z)$ will never be in $B$ (because $y<y_3$), and it will be in $A$ and therefore in the union only when $x<bar x$. But your formula for $Acup B$ would include $(x,y,z)$ also when $x>bar x$.
– Andreas Blass
Jul 16 at 15:09
@AndreasBlass. Thank you! However, how can I write the union?
– Mark
Jul 16 at 17:17
@AndreasBlass. Thank you! However, how can I write the union?
– Mark
Jul 16 at 17:17
1
1
See the corrected answer from gimusi.
– Andreas Blass
Jul 16 at 17:30
See the corrected answer from gimusi.
– Andreas Blass
Jul 16 at 17:30
add a comment |Â
3 Answers
3
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oldest
votes
up vote
3
down vote
accepted
It might help to write the sets as
$$A = [z_1, z_2] times (-infty, overlinex) times [y_1, y_2]$$
$$B = [z_1, z_2] times (overlinex, +infty) times [y_3, y_2]$$
so $$A cup B = [z_1, z_2] times Bigg( Big((-infty, overlinex) times [y_1, y_2]Big) cup Big((overlinex, +infty) times [y_3, y_2]Big)Bigg)$$
It cannot really be simplified any further.
answered Jul 16 at 19:39
mechanodroid
22.3k52041
add a comment |Â
up vote
3
down vote
I thought it might be helpful to have a graphical illustration of what's being joined here:
Both blocks go off infinitely far to the left and right, since there is no lower $x$ bound for the first subset, and no upper $x$ bound for the second subset.
Note the small gap between the blocks at $x = overlinex$; the union does not contain any points for $x = overlinex$.
answered Jul 16 at 20:26
Brian Tung
25.2k32453
add a comment |Â
up vote
1
down vote
As pointed out by Andreas Blass in the comment, your approach isn't correct, indeed it should be
$$Acup B=(x,y,z):z_1leq zleq z_2, , x<barxquad textforquad y_1leq yleq y_3, xneqbarxquad textforquad y_3leq yleq y_2,$$
answered Jul 16 at 13:59
gimusi
65.4k73684
This isn't correct; see my comment on the question.
– Andreas Blass
Jul 16 at 15:10
@AndreasBlass Oh yes of course! I fix, thanks.
– gimusi
Jul 16 at 15:25
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It might help to write the sets as
$$A = [z_1, z_2] times (-infty, overlinex) times [y_1, y_2]$$
$$B = [z_1, z_2] times (overlinex, +infty) times [y_3, y_2]$$
so $$A cup B = [z_1, z_2] times Bigg( Big((-infty, overlinex) times [y_1, y_2]Big) cup Big((overlinex, +infty) times [y_3, y_2]Big)Bigg)$$
It cannot really be simplified any further.
answered Jul 16 at 19:39
mechanodroid
22.3k52041
add a comment |Â
up vote
3
down vote
accepted
It might help to write the sets as
$$A = [z_1, z_2] times (-infty, overlinex) times [y_1, y_2]$$
$$B = [z_1, z_2] times (overlinex, +infty) times [y_3, y_2]$$
so $$A cup B = [z_1, z_2] times Bigg( Big((-infty, overlinex) times [y_1, y_2]Big) cup Big((overlinex, +infty) times [y_3, y_2]Big)Bigg)$$
It cannot really be simplified any further.
answered Jul 16 at 19:39
mechanodroid
22.3k52041
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It might help to write the sets as
$$A = [z_1, z_2] times (-infty, overlinex) times [y_1, y_2]$$
$$B = [z_1, z_2] times (overlinex, +infty) times [y_3, y_2]$$
so $$A cup B = [z_1, z_2] times Bigg( Big((-infty, overlinex) times [y_1, y_2]Big) cup Big((overlinex, +infty) times [y_3, y_2]Big)Bigg)$$
It cannot really be simplified any further.
answered Jul 16 at 19:39
mechanodroid
22.3k52041
It might help to write the sets as
$$A = [z_1, z_2] times (-infty, overlinex) times [y_1, y_2]$$
$$B = [z_1, z_2] times (overlinex, +infty) times [y_3, y_2]$$
so $$A cup B = [z_1, z_2] times Bigg( Big((-infty, overlinex) times [y_1, y_2]Big) cup Big((overlinex, +infty) times [y_3, y_2]Big)Bigg)$$
It cannot really be simplified any further.
answered Jul 16 at 19:39
mechanodroid
22.3k52041
answered Jul 16 at 19:39
mechanodroid
22.3k52041
answered Jul 16 at 19:39
mechanodroid
22.3k52041
answered Jul 16 at 19:39
mechanodroid
22.3k52041
22.3k52041
add a comment |Â
add a comment |Â
up vote
3
down vote
I thought it might be helpful to have a graphical illustration of what's being joined here:
Both blocks go off infinitely far to the left and right, since there is no lower $x$ bound for the first subset, and no upper $x$ bound for the second subset.
Note the small gap between the blocks at $x = overlinex$; the union does not contain any points for $x = overlinex$.
answered Jul 16 at 20:26
Brian Tung
25.2k32453
add a comment |Â
up vote
3
down vote
I thought it might be helpful to have a graphical illustration of what's being joined here:
Both blocks go off infinitely far to the left and right, since there is no lower $x$ bound for the first subset, and no upper $x$ bound for the second subset.
Note the small gap between the blocks at $x = overlinex$; the union does not contain any points for $x = overlinex$.
answered Jul 16 at 20:26
Brian Tung
25.2k32453
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I thought it might be helpful to have a graphical illustration of what's being joined here:
Both blocks go off infinitely far to the left and right, since there is no lower $x$ bound for the first subset, and no upper $x$ bound for the second subset.
Note the small gap between the blocks at $x = overlinex$; the union does not contain any points for $x = overlinex$.
answered Jul 16 at 20:26
Brian Tung
25.2k32453
I thought it might be helpful to have a graphical illustration of what's being joined here:
Both blocks go off infinitely far to the left and right, since there is no lower $x$ bound for the first subset, and no upper $x$ bound for the second subset.
Note the small gap between the blocks at $x = overlinex$; the union does not contain any points for $x = overlinex$.
answered Jul 16 at 20:26
Brian Tung
25.2k32453
answered Jul 16 at 20:26
Brian Tung
25.2k32453
answered Jul 16 at 20:26
Brian Tung
25.2k32453
answered Jul 16 at 20:26
Brian Tung
25.2k32453
25.2k32453
add a comment |Â
add a comment |Â
up vote
1
down vote
As pointed out by Andreas Blass in the comment, your approach isn't correct, indeed it should be
$$Acup B=(x,y,z):z_1leq zleq z_2, , x<barxquad textforquad y_1leq yleq y_3, xneqbarxquad textforquad y_3leq yleq y_2,$$
answered Jul 16 at 13:59
gimusi
65.4k73684
This isn't correct; see my comment on the question.
– Andreas Blass
Jul 16 at 15:10
@AndreasBlass Oh yes of course! I fix, thanks.
– gimusi
Jul 16 at 15:25
add a comment |Â
up vote
1
down vote
As pointed out by Andreas Blass in the comment, your approach isn't correct, indeed it should be
$$Acup B=(x,y,z):z_1leq zleq z_2, , x<barxquad textforquad y_1leq yleq y_3, xneqbarxquad textforquad y_3leq yleq y_2,$$
answered Jul 16 at 13:59
gimusi
65.4k73684
This isn't correct; see my comment on the question.
– Andreas Blass
Jul 16 at 15:10
@AndreasBlass Oh yes of course! I fix, thanks.
– gimusi
Jul 16 at 15:25
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As pointed out by Andreas Blass in the comment, your approach isn't correct, indeed it should be
$$Acup B=(x,y,z):z_1leq zleq z_2, , x<barxquad textforquad y_1leq yleq y_3, xneqbarxquad textforquad y_3leq yleq y_2,$$
answered Jul 16 at 13:59
gimusi
65.4k73684
As pointed out by Andreas Blass in the comment, your approach isn't correct, indeed it should be
$$Acup B=(x,y,z):z_1leq zleq z_2, , x<barxquad textforquad y_1leq yleq y_3, xneqbarxquad textforquad y_3leq yleq y_2,$$
answered Jul 16 at 13:59
gimusi
65.4k73684
edited Jul 16 at 15:28
answered Jul 16 at 13:59
gimusi
65.4k73684
answered Jul 16 at 13:59
gimusi
65.4k73684
answered Jul 16 at 13:59
gimusi
65.4k73684
65.4k73684
This isn't correct; see my comment on the question.
– Andreas Blass
Jul 16 at 15:10
@AndreasBlass Oh yes of course! I fix, thanks.
– gimusi
Jul 16 at 15:25
add a comment |Â
This isn't correct; see my comment on the question.
– Andreas Blass
Jul 16 at 15:10
@AndreasBlass Oh yes of course! I fix, thanks.
– gimusi
Jul 16 at 15:25
This isn't correct; see my comment on the question.
– Andreas Blass
Jul 16 at 15:10
This isn't correct; see my comment on the question.
– Andreas Blass
Jul 16 at 15:10
@AndreasBlass Oh yes of course! I fix, thanks.
– gimusi
Jul 16 at 15:25
@AndreasBlass Oh yes of course! I fix, thanks.
– gimusi
Jul 16 at 15:25
add a comment |Â
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Not quite right. Consider some $z$ between $z_1$ and $z_2$, some $y$ between $y_1$ and $y_3$, and some $xneqbar x$. Then $(x,y,z)$ will never be in $B$ (because $y<y_3$), and it will be in $A$ and therefore in the union only when $x<bar x$. But your formula for $Acup B$ would include $(x,y,z)$ also when $x>bar x$.
– Andreas Blass
Jul 16 at 15:09
@AndreasBlass. Thank you! However, how can I write the union?
– Mark
Jul 16 at 17:17
1
See the corrected answer from gimusi.
– Andreas Blass
Jul 16 at 17:30