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Find the interior and boundary of a set.
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I need to find the interior and boundary of this set:
$$A=(x,y,z)in R^3 : 0leq xleq1 ,0 leq yleq2, 0leq zlt3 setminus(0,0,0). $$
We defined the interior as the set of all interior points, where we defined an interior point as:
point $ain R^n$ is interior for $A subseteq R^n$ if $exists r>0 $ so that $K(a,r) subseteq A$. (K being an open ball with a centre in a and a radius of r).
I understand the definitions in a logical sense but don't know how to apply the "ball condition" to a real example. I also don't understand how the different boundaries ($<, leq$) impact it.
real-analysis general-topology
asked Jul 30 at 14:49
TheOakDwarf
31
 |Â
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up vote
0
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favorite
I need to find the interior and boundary of this set:
$$A=(x,y,z)in R^3 : 0leq xleq1 ,0 leq yleq2, 0leq zlt3 setminus(0,0,0). $$
We defined the interior as the set of all interior points, where we defined an interior point as:
point $ain R^n$ is interior for $A subseteq R^n$ if $exists r>0 $ so that $K(a,r) subseteq A$. (K being an open ball with a centre in a and a radius of r).
I understand the definitions in a logical sense but don't know how to apply the "ball condition" to a real example. I also don't understand how the different boundaries ($<, leq$) impact it.
real-analysis general-topology
asked Jul 30 at 14:49
TheOakDwarf
31
First guess it, then prove it by using the ball condition.
– xbh
Jul 30 at 14:53
Try answering this question on the following sets: $X = xinBbbR: 0le x le 1$ and $Y = (x,y)in BbbR^2: 0le x le 1; 0 le y le 2$. Draw them out, pick random points, and ask if they are in the interior, or the boundary.
– Santana Afton
Jul 30 at 14:53
An equivalent definition is that the interior of $A$ is the largest open set contained in $A$. It's not hard to figure out the candidate to be interior.
– Yagger
Jul 30 at 14:55
If i were to guess, I'd say the interior for x would be $x in (0,1)$ , $y in (0,2) $ but i don't know how < impacts z. Is it also (0,3) even though it has < instead of $leq$ ?
– TheOakDwarf
Jul 30 at 15:11
@TheOakDwarf Yeah, the $leqslant$ does not make much difference. Now combine them together to derive the interior for the solid in 3-d space.
– xbh
Jul 30 at 15:18
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to find the interior and boundary of this set:
$$A=(x,y,z)in R^3 : 0leq xleq1 ,0 leq yleq2, 0leq zlt3 setminus(0,0,0). $$
We defined the interior as the set of all interior points, where we defined an interior point as:
point $ain R^n$ is interior for $A subseteq R^n$ if $exists r>0 $ so that $K(a,r) subseteq A$. (K being an open ball with a centre in a and a radius of r).
I understand the definitions in a logical sense but don't know how to apply the "ball condition" to a real example. I also don't understand how the different boundaries ($<, leq$) impact it.
real-analysis general-topology
asked Jul 30 at 14:49
TheOakDwarf
31
I need to find the interior and boundary of this set:
$$A=(x,y,z)in R^3 : 0leq xleq1 ,0 leq yleq2, 0leq zlt3 setminus(0,0,0). $$
We defined the interior as the set of all interior points, where we defined an interior point as:
point $ain R^n$ is interior for $A subseteq R^n$ if $exists r>0 $ so that $K(a,r) subseteq A$. (K being an open ball with a centre in a and a radius of r).
I understand the definitions in a logical sense but don't know how to apply the "ball condition" to a real example. I also don't understand how the different boundaries ($<, leq$) impact it.
real-analysis general-topology
asked Jul 30 at 14:49
TheOakDwarf
31
asked Jul 30 at 14:49
TheOakDwarf
31
asked Jul 30 at 14:49
TheOakDwarf
31
asked Jul 30 at 14:49
TheOakDwarf
31
31
First guess it, then prove it by using the ball condition.
– xbh
Jul 30 at 14:53
Try answering this question on the following sets: $X = xinBbbR: 0le x le 1$ and $Y = (x,y)in BbbR^2: 0le x le 1; 0 le y le 2$. Draw them out, pick random points, and ask if they are in the interior, or the boundary.
– Santana Afton
Jul 30 at 14:53
An equivalent definition is that the interior of $A$ is the largest open set contained in $A$. It's not hard to figure out the candidate to be interior.
– Yagger
Jul 30 at 14:55
If i were to guess, I'd say the interior for x would be $x in (0,1)$ , $y in (0,2) $ but i don't know how < impacts z. Is it also (0,3) even though it has < instead of $leq$ ?
– TheOakDwarf
Jul 30 at 15:11
@TheOakDwarf Yeah, the $leqslant$ does not make much difference. Now combine them together to derive the interior for the solid in 3-d space.
– xbh
Jul 30 at 15:18
 |Â
show 5 more comments
First guess it, then prove it by using the ball condition.
– xbh
Jul 30 at 14:53
Try answering this question on the following sets: $X = xinBbbR: 0le x le 1$ and $Y = (x,y)in BbbR^2: 0le x le 1; 0 le y le 2$. Draw them out, pick random points, and ask if they are in the interior, or the boundary.
– Santana Afton
Jul 30 at 14:53
An equivalent definition is that the interior of $A$ is the largest open set contained in $A$. It's not hard to figure out the candidate to be interior.
– Yagger
Jul 30 at 14:55
If i were to guess, I'd say the interior for x would be $x in (0,1)$ , $y in (0,2) $ but i don't know how < impacts z. Is it also (0,3) even though it has < instead of $leq$ ?
– TheOakDwarf
Jul 30 at 15:11
@TheOakDwarf Yeah, the $leqslant$ does not make much difference. Now combine them together to derive the interior for the solid in 3-d space.
– xbh
Jul 30 at 15:18
First guess it, then prove it by using the ball condition.
– xbh
Jul 30 at 14:53
First guess it, then prove it by using the ball condition.
– xbh
Jul 30 at 14:53
Try answering this question on the following sets: $X = xinBbbR: 0le x le 1$ and $Y = (x,y)in BbbR^2: 0le x le 1; 0 le y le 2$. Draw them out, pick random points, and ask if they are in the interior, or the boundary.
– Santana Afton
Jul 30 at 14:53
Try answering this question on the following sets: $X = xinBbbR: 0le x le 1$ and $Y = (x,y)in BbbR^2: 0le x le 1; 0 le y le 2$. Draw them out, pick random points, and ask if they are in the interior, or the boundary.
– Santana Afton
Jul 30 at 14:53
An equivalent definition is that the interior of $A$ is the largest open set contained in $A$. It's not hard to figure out the candidate to be interior.
– Yagger
Jul 30 at 14:55
An equivalent definition is that the interior of $A$ is the largest open set contained in $A$. It's not hard to figure out the candidate to be interior.
– Yagger
Jul 30 at 14:55
If i were to guess, I'd say the interior for x would be $x in (0,1)$ , $y in (0,2) $ but i don't know how < impacts z. Is it also (0,3) even though it has < instead of $leq$ ?
– TheOakDwarf
Jul 30 at 15:11
If i were to guess, I'd say the interior for x would be $x in (0,1)$ , $y in (0,2) $ but i don't know how < impacts z. Is it also (0,3) even though it has < instead of $leq$ ?
– TheOakDwarf
Jul 30 at 15:11
@TheOakDwarf Yeah, the $leqslant$ does not make much difference. Now combine them together to derive the interior for the solid in 3-d space.
– xbh
Jul 30 at 15:18
@TheOakDwarf Yeah, the $leqslant$ does not make much difference. Now combine them together to derive the interior for the solid in 3-d space.
– xbh
Jul 30 at 15:18
 |Â
show 5 more comments
2 Answers
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When applying the "ball condition" it is important to remember that you can choose a radius $r$ that works.
Just working in one dimension, we have
$$ beginalign
A &=x in mathbbR : 0leq xleq1 setminus0\ &=(0,1]
endalign$$
In order to show that $mathrmint(A)=(0,1)$ ... take $a in (0,1)$ and define $r:=mathrmminfraca2,frac1-a2$. Then (perhaps with the help of a sketch) you can show that $K(a,r) subset (0,1)$.
Can you complete this proof, then do similar in $mathbbR^2$ and $mathbbR^3$?
answered Jul 30 at 15:17
Malkin
915421
What i don't understand is how 1 can be part of the interior. If you take 1, you can't find a ball around it that wouldn't be outside of A, right?
– TheOakDwarf
Jul 30 at 15:24
1 isn't part of the interior. $(0,1)$ is the set whose points that satisfy $0 < x < 1$
– JuliusL33t
Jul 30 at 15:32
Yes, but at the start he says (0,1].
– TheOakDwarf
Jul 30 at 15:35
Yeah, he started with [0,1] and cut out 0. The interior of the resulting set is (0,1). This is analogous to your care, where 0 is cut out.
– JuliusL33t
Jul 30 at 15:37
Yeah, have a read of it again. I say that $A=(0,1]$ and that $mathrmint(A)=(0,1)$. You are correct that $1$ can't be part of the interior. In fact, $1$ is the boundary in this case.
– Malkin
Jul 30 at 15:40
 |Â
show 1 more comment
up vote
0
down vote
If $a=(x,y,z)in A$ and $0in x,y,z$ and $r>0$ then the ball $K(a,r)$ will contain a point with a negative co-ordinate, so $K(a,r) not subset A.$ Because $(x-r/2, y-r/2, z-r/2)in K(a,r),$ and if at least one of $x,y,z$ is $0$ then at least one of $x-r/2,;y-r/2,;z-r/2$ is negative.
Similarly if $ain A$ and $(x=1lor y=2lor z=3)$ and $r>0$ then $(x+r/2,y+r/2,z+r/2)in K(a,r)$ but $(x+r/2>1lor y+r/2>2 lor z+r/2>3)$ so $K(a,r)not subset A.$
Now $[;a=(x,y,z)in A$ $ land 0not in x,y,z)land (xne 1land yne 2 land zne 3);]$ if and only if $ain B=(0,1) times (0,2)times (0,3).$.. (where $(u,v)$ denotes an open interval in $Bbb R$.) .... If $ain B$ let $r=min x,;1-x,;y,;2-y,;z,;3-z$. Then $r>0.$ And $(x',y',z')in K(a,r)implies max (|x'-x|, |y'-y|,|z'-z|)<r$ $ implies (x',y',z')in B.$.. So $K(a,r)subset A.$
Similarly to $Bbb R^2$ and to $Bbb R,$ if $(x,y,z)in Ssubset Bbb R^3$ and if either (i).. $(x,y,z') not in S$ whenever $z'>z$ or (ii).. $(x,y,z')not in S$ whenever $z'<z,;$... then $(x,y,z) not in $ int$(S)$.
Remarks. The phrase $0not in x,y,zland (xne 1land yne 2 land zne 3)$ can be written $0not in x,y,z,1-x,2-y,3-z.$.. The most common notation for an open ball $K(a,r)$ is $B(a,r)$...("$B$" for ball).
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
When applying the "ball condition" it is important to remember that you can choose a radius $r$ that works.
Just working in one dimension, we have
$$ beginalign
A &=x in mathbbR : 0leq xleq1 setminus0\ &=(0,1]
endalign$$
In order to show that $mathrmint(A)=(0,1)$ ... take $a in (0,1)$ and define $r:=mathrmminfraca2,frac1-a2$. Then (perhaps with the help of a sketch) you can show that $K(a,r) subset (0,1)$.
Can you complete this proof, then do similar in $mathbbR^2$ and $mathbbR^3$?
answered Jul 30 at 15:17
Malkin
915421
What i don't understand is how 1 can be part of the interior. If you take 1, you can't find a ball around it that wouldn't be outside of A, right?
– TheOakDwarf
Jul 30 at 15:24
1 isn't part of the interior. $(0,1)$ is the set whose points that satisfy $0 < x < 1$
– JuliusL33t
Jul 30 at 15:32
Yes, but at the start he says (0,1].
– TheOakDwarf
Jul 30 at 15:35
Yeah, he started with [0,1] and cut out 0. The interior of the resulting set is (0,1). This is analogous to your care, where 0 is cut out.
– JuliusL33t
Jul 30 at 15:37
Yeah, have a read of it again. I say that $A=(0,1]$ and that $mathrmint(A)=(0,1)$. You are correct that $1$ can't be part of the interior. In fact, $1$ is the boundary in this case.
– Malkin
Jul 30 at 15:40
 |Â
show 1 more comment
up vote
0
down vote
accepted
When applying the "ball condition" it is important to remember that you can choose a radius $r$ that works.
Just working in one dimension, we have
$$ beginalign
A &=x in mathbbR : 0leq xleq1 setminus0\ &=(0,1]
endalign$$
In order to show that $mathrmint(A)=(0,1)$ ... take $a in (0,1)$ and define $r:=mathrmminfraca2,frac1-a2$. Then (perhaps with the help of a sketch) you can show that $K(a,r) subset (0,1)$.
Can you complete this proof, then do similar in $mathbbR^2$ and $mathbbR^3$?
answered Jul 30 at 15:17
Malkin
915421
What i don't understand is how 1 can be part of the interior. If you take 1, you can't find a ball around it that wouldn't be outside of A, right?
– TheOakDwarf
Jul 30 at 15:24
1 isn't part of the interior. $(0,1)$ is the set whose points that satisfy $0 < x < 1$
– JuliusL33t
Jul 30 at 15:32
Yes, but at the start he says (0,1].
– TheOakDwarf
Jul 30 at 15:35
Yeah, he started with [0,1] and cut out 0. The interior of the resulting set is (0,1). This is analogous to your care, where 0 is cut out.
– JuliusL33t
Jul 30 at 15:37
Yeah, have a read of it again. I say that $A=(0,1]$ and that $mathrmint(A)=(0,1)$. You are correct that $1$ can't be part of the interior. In fact, $1$ is the boundary in this case.
– Malkin
Jul 30 at 15:40
 |Â
show 1 more comment
up vote
0
down vote
accepted
up vote
0
down vote
accepted
When applying the "ball condition" it is important to remember that you can choose a radius $r$ that works.
Just working in one dimension, we have
$$ beginalign
A &=x in mathbbR : 0leq xleq1 setminus0\ &=(0,1]
endalign$$
In order to show that $mathrmint(A)=(0,1)$ ... take $a in (0,1)$ and define $r:=mathrmminfraca2,frac1-a2$. Then (perhaps with the help of a sketch) you can show that $K(a,r) subset (0,1)$.
Can you complete this proof, then do similar in $mathbbR^2$ and $mathbbR^3$?
answered Jul 30 at 15:17
Malkin
915421
When applying the "ball condition" it is important to remember that you can choose a radius $r$ that works.
Just working in one dimension, we have
$$ beginalign
A &=x in mathbbR : 0leq xleq1 setminus0\ &=(0,1]
endalign$$
In order to show that $mathrmint(A)=(0,1)$ ... take $a in (0,1)$ and define $r:=mathrmminfraca2,frac1-a2$. Then (perhaps with the help of a sketch) you can show that $K(a,r) subset (0,1)$.
Can you complete this proof, then do similar in $mathbbR^2$ and $mathbbR^3$?
answered Jul 30 at 15:17
Malkin
915421
answered Jul 30 at 15:17
Malkin
915421
answered Jul 30 at 15:17
Malkin
915421
answered Jul 30 at 15:17
Malkin
915421
915421
What i don't understand is how 1 can be part of the interior. If you take 1, you can't find a ball around it that wouldn't be outside of A, right?
– TheOakDwarf
Jul 30 at 15:24
1 isn't part of the interior. $(0,1)$ is the set whose points that satisfy $0 < x < 1$
– JuliusL33t
Jul 30 at 15:32
Yes, but at the start he says (0,1].
– TheOakDwarf
Jul 30 at 15:35
Yeah, he started with [0,1] and cut out 0. The interior of the resulting set is (0,1). This is analogous to your care, where 0 is cut out.
– JuliusL33t
Jul 30 at 15:37
Yeah, have a read of it again. I say that $A=(0,1]$ and that $mathrmint(A)=(0,1)$. You are correct that $1$ can't be part of the interior. In fact, $1$ is the boundary in this case.
– Malkin
Jul 30 at 15:40
 |Â
show 1 more comment
What i don't understand is how 1 can be part of the interior. If you take 1, you can't find a ball around it that wouldn't be outside of A, right?
– TheOakDwarf
Jul 30 at 15:24
1 isn't part of the interior. $(0,1)$ is the set whose points that satisfy $0 < x < 1$
– JuliusL33t
Jul 30 at 15:32
Yes, but at the start he says (0,1].
– TheOakDwarf
Jul 30 at 15:35
Yeah, he started with [0,1] and cut out 0. The interior of the resulting set is (0,1). This is analogous to your care, where 0 is cut out.
– JuliusL33t
Jul 30 at 15:37
Yeah, have a read of it again. I say that $A=(0,1]$ and that $mathrmint(A)=(0,1)$. You are correct that $1$ can't be part of the interior. In fact, $1$ is the boundary in this case.
– Malkin
Jul 30 at 15:40
What i don't understand is how 1 can be part of the interior. If you take 1, you can't find a ball around it that wouldn't be outside of A, right?
– TheOakDwarf
Jul 30 at 15:24
What i don't understand is how 1 can be part of the interior. If you take 1, you can't find a ball around it that wouldn't be outside of A, right?
– TheOakDwarf
Jul 30 at 15:24
1 isn't part of the interior. $(0,1)$ is the set whose points that satisfy $0 < x < 1$
– JuliusL33t
Jul 30 at 15:32
1 isn't part of the interior. $(0,1)$ is the set whose points that satisfy $0 < x < 1$
– JuliusL33t
Jul 30 at 15:32
Yes, but at the start he says (0,1].
– TheOakDwarf
Jul 30 at 15:35
Yes, but at the start he says (0,1].
– TheOakDwarf
Jul 30 at 15:35
Yeah, he started with [0,1] and cut out 0. The interior of the resulting set is (0,1). This is analogous to your care, where 0 is cut out.
– JuliusL33t
Jul 30 at 15:37
Yeah, he started with [0,1] and cut out 0. The interior of the resulting set is (0,1). This is analogous to your care, where 0 is cut out.
– JuliusL33t
Jul 30 at 15:37
Yeah, have a read of it again. I say that $A=(0,1]$ and that $mathrmint(A)=(0,1)$. You are correct that $1$ can't be part of the interior. In fact, $1$ is the boundary in this case.
– Malkin
Jul 30 at 15:40
Yeah, have a read of it again. I say that $A=(0,1]$ and that $mathrmint(A)=(0,1)$. You are correct that $1$ can't be part of the interior. In fact, $1$ is the boundary in this case.
– Malkin
Jul 30 at 15:40
 |Â
show 1 more comment
up vote
0
down vote
If $a=(x,y,z)in A$ and $0in x,y,z$ and $r>0$ then the ball $K(a,r)$ will contain a point with a negative co-ordinate, so $K(a,r) not subset A.$ Because $(x-r/2, y-r/2, z-r/2)in K(a,r),$ and if at least one of $x,y,z$ is $0$ then at least one of $x-r/2,;y-r/2,;z-r/2$ is negative.
Similarly if $ain A$ and $(x=1lor y=2lor z=3)$ and $r>0$ then $(x+r/2,y+r/2,z+r/2)in K(a,r)$ but $(x+r/2>1lor y+r/2>2 lor z+r/2>3)$ so $K(a,r)not subset A.$
Now $[;a=(x,y,z)in A$ $ land 0not in x,y,z)land (xne 1land yne 2 land zne 3);]$ if and only if $ain B=(0,1) times (0,2)times (0,3).$.. (where $(u,v)$ denotes an open interval in $Bbb R$.) .... If $ain B$ let $r=min x,;1-x,;y,;2-y,;z,;3-z$. Then $r>0.$ And $(x',y',z')in K(a,r)implies max (|x'-x|, |y'-y|,|z'-z|)<r$ $ implies (x',y',z')in B.$.. So $K(a,r)subset A.$
Similarly to $Bbb R^2$ and to $Bbb R,$ if $(x,y,z)in Ssubset Bbb R^3$ and if either (i).. $(x,y,z') not in S$ whenever $z'>z$ or (ii).. $(x,y,z')not in S$ whenever $z'<z,;$... then $(x,y,z) not in $ int$(S)$.
Remarks. The phrase $0not in x,y,zland (xne 1land yne 2 land zne 3)$ can be written $0not in x,y,z,1-x,2-y,3-z.$.. The most common notation for an open ball $K(a,r)$ is $B(a,r)$...("$B$" for ball).
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
add a comment |Â
up vote
0
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If $a=(x,y,z)in A$ and $0in x,y,z$ and $r>0$ then the ball $K(a,r)$ will contain a point with a negative co-ordinate, so $K(a,r) not subset A.$ Because $(x-r/2, y-r/2, z-r/2)in K(a,r),$ and if at least one of $x,y,z$ is $0$ then at least one of $x-r/2,;y-r/2,;z-r/2$ is negative.
Similarly if $ain A$ and $(x=1lor y=2lor z=3)$ and $r>0$ then $(x+r/2,y+r/2,z+r/2)in K(a,r)$ but $(x+r/2>1lor y+r/2>2 lor z+r/2>3)$ so $K(a,r)not subset A.$
Now $[;a=(x,y,z)in A$ $ land 0not in x,y,z)land (xne 1land yne 2 land zne 3);]$ if and only if $ain B=(0,1) times (0,2)times (0,3).$.. (where $(u,v)$ denotes an open interval in $Bbb R$.) .... If $ain B$ let $r=min x,;1-x,;y,;2-y,;z,;3-z$. Then $r>0.$ And $(x',y',z')in K(a,r)implies max (|x'-x|, |y'-y|,|z'-z|)<r$ $ implies (x',y',z')in B.$.. So $K(a,r)subset A.$
Similarly to $Bbb R^2$ and to $Bbb R,$ if $(x,y,z)in Ssubset Bbb R^3$ and if either (i).. $(x,y,z') not in S$ whenever $z'>z$ or (ii).. $(x,y,z')not in S$ whenever $z'<z,;$... then $(x,y,z) not in $ int$(S)$.
Remarks. The phrase $0not in x,y,zland (xne 1land yne 2 land zne 3)$ can be written $0not in x,y,z,1-x,2-y,3-z.$.. The most common notation for an open ball $K(a,r)$ is $B(a,r)$...("$B$" for ball).
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
add a comment |Â
up vote
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down vote
up vote
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If $a=(x,y,z)in A$ and $0in x,y,z$ and $r>0$ then the ball $K(a,r)$ will contain a point with a negative co-ordinate, so $K(a,r) not subset A.$ Because $(x-r/2, y-r/2, z-r/2)in K(a,r),$ and if at least one of $x,y,z$ is $0$ then at least one of $x-r/2,;y-r/2,;z-r/2$ is negative.
Similarly if $ain A$ and $(x=1lor y=2lor z=3)$ and $r>0$ then $(x+r/2,y+r/2,z+r/2)in K(a,r)$ but $(x+r/2>1lor y+r/2>2 lor z+r/2>3)$ so $K(a,r)not subset A.$
Now $[;a=(x,y,z)in A$ $ land 0not in x,y,z)land (xne 1land yne 2 land zne 3);]$ if and only if $ain B=(0,1) times (0,2)times (0,3).$.. (where $(u,v)$ denotes an open interval in $Bbb R$.) .... If $ain B$ let $r=min x,;1-x,;y,;2-y,;z,;3-z$. Then $r>0.$ And $(x',y',z')in K(a,r)implies max (|x'-x|, |y'-y|,|z'-z|)<r$ $ implies (x',y',z')in B.$.. So $K(a,r)subset A.$
Similarly to $Bbb R^2$ and to $Bbb R,$ if $(x,y,z)in Ssubset Bbb R^3$ and if either (i).. $(x,y,z') not in S$ whenever $z'>z$ or (ii).. $(x,y,z')not in S$ whenever $z'<z,;$... then $(x,y,z) not in $ int$(S)$.
Remarks. The phrase $0not in x,y,zland (xne 1land yne 2 land zne 3)$ can be written $0not in x,y,z,1-x,2-y,3-z.$.. The most common notation for an open ball $K(a,r)$ is $B(a,r)$...("$B$" for ball).
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
If $a=(x,y,z)in A$ and $0in x,y,z$ and $r>0$ then the ball $K(a,r)$ will contain a point with a negative co-ordinate, so $K(a,r) not subset A.$ Because $(x-r/2, y-r/2, z-r/2)in K(a,r),$ and if at least one of $x,y,z$ is $0$ then at least one of $x-r/2,;y-r/2,;z-r/2$ is negative.
Similarly if $ain A$ and $(x=1lor y=2lor z=3)$ and $r>0$ then $(x+r/2,y+r/2,z+r/2)in K(a,r)$ but $(x+r/2>1lor y+r/2>2 lor z+r/2>3)$ so $K(a,r)not subset A.$
Now $[;a=(x,y,z)in A$ $ land 0not in x,y,z)land (xne 1land yne 2 land zne 3);]$ if and only if $ain B=(0,1) times (0,2)times (0,3).$.. (where $(u,v)$ denotes an open interval in $Bbb R$.) .... If $ain B$ let $r=min x,;1-x,;y,;2-y,;z,;3-z$. Then $r>0.$ And $(x',y',z')in K(a,r)implies max (|x'-x|, |y'-y|,|z'-z|)<r$ $ implies (x',y',z')in B.$.. So $K(a,r)subset A.$
Similarly to $Bbb R^2$ and to $Bbb R,$ if $(x,y,z)in Ssubset Bbb R^3$ and if either (i).. $(x,y,z') not in S$ whenever $z'>z$ or (ii).. $(x,y,z')not in S$ whenever $z'<z,;$... then $(x,y,z) not in $ int$(S)$.
Remarks. The phrase $0not in x,y,zland (xne 1land yne 2 land zne 3)$ can be written $0not in x,y,z,1-x,2-y,3-z.$.. The most common notation for an open ball $K(a,r)$ is $B(a,r)$...("$B$" for ball).
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
edited Jul 30 at 16:33
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
answered Jul 30 at 16:22
DanielWainfleet
31.4k31542
31.4k31542
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First guess it, then prove it by using the ball condition.
– xbh
Jul 30 at 14:53
Try answering this question on the following sets: $X = xinBbbR: 0le x le 1$ and $Y = (x,y)in BbbR^2: 0le x le 1; 0 le y le 2$. Draw them out, pick random points, and ask if they are in the interior, or the boundary.
– Santana Afton
Jul 30 at 14:53
An equivalent definition is that the interior of $A$ is the largest open set contained in $A$. It's not hard to figure out the candidate to be interior.
– Yagger
Jul 30 at 14:55
If i were to guess, I'd say the interior for x would be $x in (0,1)$ , $y in (0,2) $ but i don't know how < impacts z. Is it also (0,3) even though it has < instead of $leq$ ?
– TheOakDwarf
Jul 30 at 15:11
@TheOakDwarf Yeah, the $leqslant$ does not make much difference. Now combine them together to derive the interior for the solid in 3-d space.
– xbh
Jul 30 at 15:18