Question on what constitutes surfaces in R3
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Our lecturer gave us the following definition of a surface in $mathbb{R}^3$:
$Gamma$ Is a surface in $mathbb{R}^3$ if for all $yin Gamma$ there exists a coordinate patch $sigma : Dsubsetmathbb{R}^2 rightarrow mathbb{R}^3$ such that $y in text{Im}(sigma)subsetGamma$. From this definition, is it not inferred that the union of two surfaces is a surface? I am confused because he went on to mention that the union of the boundary of the unit sphere and the set ${(x,y,0)|x^2+y^2<1}$ is not a surface. Clearly these are both surfaces so why isn’t their union?
differential-geometry surfaces
asked Dec 9 '18 at 22:37
EloElo
83
$endgroup$
add a comment |
$begingroup$
Our lecturer gave us the following definition of a surface in $mathbb{R}^3$:
$Gamma$ Is a surface in $mathbb{R}^3$ if for all $yin Gamma$ there exists a coordinate patch $sigma : Dsubsetmathbb{R}^2 rightarrow mathbb{R}^3$ such that $y in text{Im}(sigma)subsetGamma$. From this definition, is it not inferred that the union of two surfaces is a surface? I am confused because he went on to mention that the union of the boundary of the unit sphere and the set ${(x,y,0)|x^2+y^2<1}$ is not a surface. Clearly these are both surfaces so why isn’t their union?
differential-geometry surfaces
asked Dec 9 '18 at 22:37
EloElo
83
$endgroup$
$begingroup$
The union of two surfaces is hardly ever a surface.
$endgroup$
– Saucy O'Path
Dec 9 '18 at 22:40
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If the two surfaces are disjoint then their union may be a surface. Not connected but still a surface. Did your definition mention connectedness?
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– user25959
Dec 9 '18 at 22:47
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Nothing about connectedness, no
$endgroup$
– Elo
Dec 9 '18 at 22:51
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Depending on what you mean by coordinate patch, this is the definition of an embedded surface, or of an immersed surface, or even something else.
$endgroup$
– Michał Miśkiewicz
Dec 9 '18 at 22:58
add a comment |
$begingroup$
Our lecturer gave us the following definition of a surface in $mathbb{R}^3$:
$Gamma$ Is a surface in $mathbb{R}^3$ if for all $yin Gamma$ there exists a coordinate patch $sigma : Dsubsetmathbb{R}^2 rightarrow mathbb{R}^3$ such that $y in text{Im}(sigma)subsetGamma$. From this definition, is it not inferred that the union of two surfaces is a surface? I am confused because he went on to mention that the union of the boundary of the unit sphere and the set ${(x,y,0)|x^2+y^2<1}$ is not a surface. Clearly these are both surfaces so why isn’t their union?
differential-geometry surfaces
asked Dec 9 '18 at 22:37
EloElo
83
$endgroup$
Our lecturer gave us the following definition of a surface in $mathbb{R}^3$:
$Gamma$ Is a surface in $mathbb{R}^3$ if for all $yin Gamma$ there exists a coordinate patch $sigma : Dsubsetmathbb{R}^2 rightarrow mathbb{R}^3$ such that $y in text{Im}(sigma)subsetGamma$. From this definition, is it not inferred that the union of two surfaces is a surface? I am confused because he went on to mention that the union of the boundary of the unit sphere and the set ${(x,y,0)|x^2+y^2<1}$ is not a surface. Clearly these are both surfaces so why isn’t their union?
differential-geometry surfaces
differential-geometry surfaces
asked Dec 9 '18 at 22:37
EloElo
83
asked Dec 9 '18 at 22:37
EloElo
83
edited Dec 9 '18 at 23:00
Elo
asked Dec 9 '18 at 22:37
EloElo
83
asked Dec 9 '18 at 22:37
EloElo
83
asked Dec 9 '18 at 22:37
EloElo
83
83
$begingroup$
The union of two surfaces is hardly ever a surface.
$endgroup$
– Saucy O'Path
Dec 9 '18 at 22:40
$begingroup$
If the two surfaces are disjoint then their union may be a surface. Not connected but still a surface. Did your definition mention connectedness?
$endgroup$
– user25959
Dec 9 '18 at 22:47
$begingroup$
Nothing about connectedness, no
$endgroup$
– Elo
Dec 9 '18 at 22:51
$begingroup$
Depending on what you mean by coordinate patch, this is the definition of an embedded surface, or of an immersed surface, or even something else.
$endgroup$
– Michał Miśkiewicz
Dec 9 '18 at 22:58
add a comment |
$begingroup$
The union of two surfaces is hardly ever a surface.
$endgroup$
– Saucy O'Path
Dec 9 '18 at 22:40
$begingroup$
If the two surfaces are disjoint then their union may be a surface. Not connected but still a surface. Did your definition mention connectedness?
$endgroup$
– user25959
Dec 9 '18 at 22:47
$begingroup$
Nothing about connectedness, no
$endgroup$
– Elo
Dec 9 '18 at 22:51
$begingroup$
Depending on what you mean by coordinate patch, this is the definition of an embedded surface, or of an immersed surface, or even something else.
$endgroup$
– Michał Miśkiewicz
Dec 9 '18 at 22:58
$begingroup$
The union of two surfaces is hardly ever a surface.
$endgroup$
– Saucy O'Path
Dec 9 '18 at 22:40
$begingroup$
The union of two surfaces is hardly ever a surface.
$endgroup$
– Saucy O'Path
Dec 9 '18 at 22:40
$begingroup$
If the two surfaces are disjoint then their union may be a surface. Not connected but still a surface. Did your definition mention connectedness?
$endgroup$
– user25959
Dec 9 '18 at 22:47
$begingroup$
If the two surfaces are disjoint then their union may be a surface. Not connected but still a surface. Did your definition mention connectedness?
$endgroup$
– user25959
Dec 9 '18 at 22:47
$begingroup$
Nothing about connectedness, no
$endgroup$
– Elo
Dec 9 '18 at 22:51
$begingroup$
Nothing about connectedness, no
$endgroup$
– Elo
Dec 9 '18 at 22:51
$begingroup$
Depending on what you mean by coordinate patch, this is the definition of an embedded surface, or of an immersed surface, or even something else.
$endgroup$
– Michał Miśkiewicz
Dec 9 '18 at 22:58
$begingroup$
Depending on what you mean by coordinate patch, this is the definition of an embedded surface, or of an immersed surface, or even something else.
$endgroup$
– Michał Miśkiewicz
Dec 9 '18 at 22:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $S$ be the unit sphere ${(x, y, z) mid x^2+y^2+z^2=1}$ and $T = {(x, y, 0) mid x^2 +y^2<1}$ be the two surfaces you mention and let $Gamma = S cup T$. Then the point $e = (1, 0, 0)$ in $Gamma$ has no neighbourhood in $U$ that is homeomorphic with an open set $D$ of $Bbb{R}^2$. To prove this takes a little work, but it should be clear visually: any neighbourhood of $e$ in $Gamma$ looks locally like the union along their edges of three half-planes and no open subset of $Bbb{R}^2$ looks like that.
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
$endgroup$
$begingroup$
So must the definition be ammended so that $text{Im}(sigma)$ must contain a neighbourhood of every point of $Gamma$?
$endgroup$
– Elo
Dec 9 '18 at 23:32
$begingroup$
I've changed the notation in my answer to match yours better. The definitions don't need changing. What I am saying is that there can't be a coordinate patch that maps to an open subset of $Gamma$ containing the point $e$.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:38
$begingroup$
I mean the definition in my question. The condition that it maps to an ‘open subset of’ $Gamma$ isn’t implied by the definition in the original question
$endgroup$
– Elo
Dec 9 '18 at 23:53
$begingroup$
Or maybe I’m misunderstanding the implications of being mapped onto by a coordinate patch
$endgroup$
– Elo
Dec 9 '18 at 23:57
$begingroup$
I was assuming your definition of coordinate patch was a homeomorphism between an open subset of $\Bbb{R}^2$ and a neighbourhood in $Gamma$ (and if you have one of those, you can find a slight smaller coordinate patch that gives a homeomorphism with an open neighbourhood in $Gamma$). To go into this in more detail, you need to tell us exactly what definition of "coordinate patch" you are using.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:59
|
show 4 more comments
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1 Answer
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1 Answer
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$begingroup$
Let $S$ be the unit sphere ${(x, y, z) mid x^2+y^2+z^2=1}$ and $T = {(x, y, 0) mid x^2 +y^2<1}$ be the two surfaces you mention and let $Gamma = S cup T$. Then the point $e = (1, 0, 0)$ in $Gamma$ has no neighbourhood in $U$ that is homeomorphic with an open set $D$ of $Bbb{R}^2$. To prove this takes a little work, but it should be clear visually: any neighbourhood of $e$ in $Gamma$ looks locally like the union along their edges of three half-planes and no open subset of $Bbb{R}^2$ looks like that.
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
$endgroup$
$begingroup$
So must the definition be ammended so that $text{Im}(sigma)$ must contain a neighbourhood of every point of $Gamma$?
$endgroup$
– Elo
Dec 9 '18 at 23:32
$begingroup$
I've changed the notation in my answer to match yours better. The definitions don't need changing. What I am saying is that there can't be a coordinate patch that maps to an open subset of $Gamma$ containing the point $e$.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:38
$begingroup$
I mean the definition in my question. The condition that it maps to an ‘open subset of’ $Gamma$ isn’t implied by the definition in the original question
$endgroup$
– Elo
Dec 9 '18 at 23:53
$begingroup$
Or maybe I’m misunderstanding the implications of being mapped onto by a coordinate patch
$endgroup$
– Elo
Dec 9 '18 at 23:57
$begingroup$
I was assuming your definition of coordinate patch was a homeomorphism between an open subset of $\Bbb{R}^2$ and a neighbourhood in $Gamma$ (and if you have one of those, you can find a slight smaller coordinate patch that gives a homeomorphism with an open neighbourhood in $Gamma$). To go into this in more detail, you need to tell us exactly what definition of "coordinate patch" you are using.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:59
|
show 4 more comments
$begingroup$
Let $S$ be the unit sphere ${(x, y, z) mid x^2+y^2+z^2=1}$ and $T = {(x, y, 0) mid x^2 +y^2<1}$ be the two surfaces you mention and let $Gamma = S cup T$. Then the point $e = (1, 0, 0)$ in $Gamma$ has no neighbourhood in $U$ that is homeomorphic with an open set $D$ of $Bbb{R}^2$. To prove this takes a little work, but it should be clear visually: any neighbourhood of $e$ in $Gamma$ looks locally like the union along their edges of three half-planes and no open subset of $Bbb{R}^2$ looks like that.
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
$endgroup$
$begingroup$
So must the definition be ammended so that $text{Im}(sigma)$ must contain a neighbourhood of every point of $Gamma$?
$endgroup$
– Elo
Dec 9 '18 at 23:32
$begingroup$
I've changed the notation in my answer to match yours better. The definitions don't need changing. What I am saying is that there can't be a coordinate patch that maps to an open subset of $Gamma$ containing the point $e$.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:38
$begingroup$
I mean the definition in my question. The condition that it maps to an ‘open subset of’ $Gamma$ isn’t implied by the definition in the original question
$endgroup$
– Elo
Dec 9 '18 at 23:53
$begingroup$
Or maybe I’m misunderstanding the implications of being mapped onto by a coordinate patch
$endgroup$
– Elo
Dec 9 '18 at 23:57
$begingroup$
I was assuming your definition of coordinate patch was a homeomorphism between an open subset of $\Bbb{R}^2$ and a neighbourhood in $Gamma$ (and if you have one of those, you can find a slight smaller coordinate patch that gives a homeomorphism with an open neighbourhood in $Gamma$). To go into this in more detail, you need to tell us exactly what definition of "coordinate patch" you are using.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:59
|
show 4 more comments
$begingroup$
Let $S$ be the unit sphere ${(x, y, z) mid x^2+y^2+z^2=1}$ and $T = {(x, y, 0) mid x^2 +y^2<1}$ be the two surfaces you mention and let $Gamma = S cup T$. Then the point $e = (1, 0, 0)$ in $Gamma$ has no neighbourhood in $U$ that is homeomorphic with an open set $D$ of $Bbb{R}^2$. To prove this takes a little work, but it should be clear visually: any neighbourhood of $e$ in $Gamma$ looks locally like the union along their edges of three half-planes and no open subset of $Bbb{R}^2$ looks like that.
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
$endgroup$
Let $S$ be the unit sphere ${(x, y, z) mid x^2+y^2+z^2=1}$ and $T = {(x, y, 0) mid x^2 +y^2<1}$ be the two surfaces you mention and let $Gamma = S cup T$. Then the point $e = (1, 0, 0)$ in $Gamma$ has no neighbourhood in $U$ that is homeomorphic with an open set $D$ of $Bbb{R}^2$. To prove this takes a little work, but it should be clear visually: any neighbourhood of $e$ in $Gamma$ looks locally like the union along their edges of three half-planes and no open subset of $Bbb{R}^2$ looks like that.
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
edited Dec 9 '18 at 23:36
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
answered Dec 9 '18 at 23:24
Rob ArthanRob Arthan
29.2k42966
29.2k42966
$begingroup$
So must the definition be ammended so that $text{Im}(sigma)$ must contain a neighbourhood of every point of $Gamma$?
$endgroup$
– Elo
Dec 9 '18 at 23:32
$begingroup$
I've changed the notation in my answer to match yours better. The definitions don't need changing. What I am saying is that there can't be a coordinate patch that maps to an open subset of $Gamma$ containing the point $e$.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:38
$begingroup$
I mean the definition in my question. The condition that it maps to an ‘open subset of’ $Gamma$ isn’t implied by the definition in the original question
$endgroup$
– Elo
Dec 9 '18 at 23:53
$begingroup$
Or maybe I’m misunderstanding the implications of being mapped onto by a coordinate patch
$endgroup$
– Elo
Dec 9 '18 at 23:57
$begingroup$
I was assuming your definition of coordinate patch was a homeomorphism between an open subset of $\Bbb{R}^2$ and a neighbourhood in $Gamma$ (and if you have one of those, you can find a slight smaller coordinate patch that gives a homeomorphism with an open neighbourhood in $Gamma$). To go into this in more detail, you need to tell us exactly what definition of "coordinate patch" you are using.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:59
|
show 4 more comments
$begingroup$
So must the definition be ammended so that $text{Im}(sigma)$ must contain a neighbourhood of every point of $Gamma$?
$endgroup$
– Elo
Dec 9 '18 at 23:32
$begingroup$
I've changed the notation in my answer to match yours better. The definitions don't need changing. What I am saying is that there can't be a coordinate patch that maps to an open subset of $Gamma$ containing the point $e$.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:38
$begingroup$
I mean the definition in my question. The condition that it maps to an ‘open subset of’ $Gamma$ isn’t implied by the definition in the original question
$endgroup$
– Elo
Dec 9 '18 at 23:53
$begingroup$
Or maybe I’m misunderstanding the implications of being mapped onto by a coordinate patch
$endgroup$
– Elo
Dec 9 '18 at 23:57
$begingroup$
I was assuming your definition of coordinate patch was a homeomorphism between an open subset of $\Bbb{R}^2$ and a neighbourhood in $Gamma$ (and if you have one of those, you can find a slight smaller coordinate patch that gives a homeomorphism with an open neighbourhood in $Gamma$). To go into this in more detail, you need to tell us exactly what definition of "coordinate patch" you are using.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:59
$begingroup$
So must the definition be ammended so that $text{Im}(sigma)$ must contain a neighbourhood of every point of $Gamma$?
$endgroup$
– Elo
Dec 9 '18 at 23:32
$begingroup$
So must the definition be ammended so that $text{Im}(sigma)$ must contain a neighbourhood of every point of $Gamma$?
$endgroup$
– Elo
Dec 9 '18 at 23:32
$begingroup$
I've changed the notation in my answer to match yours better. The definitions don't need changing. What I am saying is that there can't be a coordinate patch that maps to an open subset of $Gamma$ containing the point $e$.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:38
$begingroup$
I've changed the notation in my answer to match yours better. The definitions don't need changing. What I am saying is that there can't be a coordinate patch that maps to an open subset of $Gamma$ containing the point $e$.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:38
$begingroup$
I mean the definition in my question. The condition that it maps to an ‘open subset of’ $Gamma$ isn’t implied by the definition in the original question
$endgroup$
– Elo
Dec 9 '18 at 23:53
$begingroup$
I mean the definition in my question. The condition that it maps to an ‘open subset of’ $Gamma$ isn’t implied by the definition in the original question
$endgroup$
– Elo
Dec 9 '18 at 23:53
$begingroup$
Or maybe I’m misunderstanding the implications of being mapped onto by a coordinate patch
$endgroup$
– Elo
Dec 9 '18 at 23:57
$begingroup$
Or maybe I’m misunderstanding the implications of being mapped onto by a coordinate patch
$endgroup$
– Elo
Dec 9 '18 at 23:57
$begingroup$
I was assuming your definition of coordinate patch was a homeomorphism between an open subset of $\Bbb{R}^2$ and a neighbourhood in $Gamma$ (and if you have one of those, you can find a slight smaller coordinate patch that gives a homeomorphism with an open neighbourhood in $Gamma$). To go into this in more detail, you need to tell us exactly what definition of "coordinate patch" you are using.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:59
$begingroup$
I was assuming your definition of coordinate patch was a homeomorphism between an open subset of $\Bbb{R}^2$ and a neighbourhood in $Gamma$ (and if you have one of those, you can find a slight smaller coordinate patch that gives a homeomorphism with an open neighbourhood in $Gamma$). To go into this in more detail, you need to tell us exactly what definition of "coordinate patch" you are using.
$endgroup$
– Rob Arthan
Dec 9 '18 at 23:59
|
show 4 more comments
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$begingroup$
The union of two surfaces is hardly ever a surface.
$endgroup$
– Saucy O'Path
Dec 9 '18 at 22:40
$begingroup$
If the two surfaces are disjoint then their union may be a surface. Not connected but still a surface. Did your definition mention connectedness?
$endgroup$
– user25959
Dec 9 '18 at 22:47
$begingroup$
Nothing about connectedness, no
$endgroup$
– Elo
Dec 9 '18 at 22:51
$begingroup$
Depending on what you mean by coordinate patch, this is the definition of an embedded surface, or of an immersed surface, or even something else.
$endgroup$
– Michał Miśkiewicz
Dec 9 '18 at 22:58