Boundary of the set of points away from manifold is a hypersurface
$begingroup$
(This is part of Problem 6-6 from Lee's Introduction to Smooth Manifolds textbook.)
Suppose $Msubset mathbb{R}^n$ is a compact embedded submanifold. For
any $epsilon>0$, let $M_epsilon$ be the set of all points in
$mathbb{R}^n$ whose distance from $M$ is less than $epsilon$. Show
that for sufficiently small $epsilon$, $partial M_epsilon$ is a
compact embedded hypersurface in $mathbb{R}^n$.
I thought to use the tubular neighbourhood theorem (since we have an embedded submanifold) and showing for some sufficiently small tubular neighbourhood, maybe we can prove that the boundary of the tubular neighbourhood is an embedded $(n-1)$-dimensional embedded manifold (and relate $epsilon$ to the distance function in the definition of a tubular neighbourhood). However, I have no idea how to prove that (or if it's even the right approach).
For reference, I'm using Lee's definition of a tubular neighbourhood. That is, $E(V)$ where $E(x,v)=x+v$ and $V={(x,v)in NM : |v|<delta(x)}$ for positive continuous $delta: M rightarrow mathbb{R}$.
differential-geometry manifolds smooth-manifolds
$endgroup$
add a comment |
$begingroup$
(This is part of Problem 6-6 from Lee's Introduction to Smooth Manifolds textbook.)
Suppose $Msubset mathbb{R}^n$ is a compact embedded submanifold. For
any $epsilon>0$, let $M_epsilon$ be the set of all points in
$mathbb{R}^n$ whose distance from $M$ is less than $epsilon$. Show
that for sufficiently small $epsilon$, $partial M_epsilon$ is a
compact embedded hypersurface in $mathbb{R}^n$.
I thought to use the tubular neighbourhood theorem (since we have an embedded submanifold) and showing for some sufficiently small tubular neighbourhood, maybe we can prove that the boundary of the tubular neighbourhood is an embedded $(n-1)$-dimensional embedded manifold (and relate $epsilon$ to the distance function in the definition of a tubular neighbourhood). However, I have no idea how to prove that (or if it's even the right approach).
For reference, I'm using Lee's definition of a tubular neighbourhood. That is, $E(V)$ where $E(x,v)=x+v$ and $V={(x,v)in NM : |v|<delta(x)}$ for positive continuous $delta: M rightarrow mathbb{R}$.
differential-geometry manifolds smooth-manifolds
$endgroup$
add a comment |
$begingroup$
(This is part of Problem 6-6 from Lee's Introduction to Smooth Manifolds textbook.)
Suppose $Msubset mathbb{R}^n$ is a compact embedded submanifold. For
any $epsilon>0$, let $M_epsilon$ be the set of all points in
$mathbb{R}^n$ whose distance from $M$ is less than $epsilon$. Show
that for sufficiently small $epsilon$, $partial M_epsilon$ is a
compact embedded hypersurface in $mathbb{R}^n$.
I thought to use the tubular neighbourhood theorem (since we have an embedded submanifold) and showing for some sufficiently small tubular neighbourhood, maybe we can prove that the boundary of the tubular neighbourhood is an embedded $(n-1)$-dimensional embedded manifold (and relate $epsilon$ to the distance function in the definition of a tubular neighbourhood). However, I have no idea how to prove that (or if it's even the right approach).
For reference, I'm using Lee's definition of a tubular neighbourhood. That is, $E(V)$ where $E(x,v)=x+v$ and $V={(x,v)in NM : |v|<delta(x)}$ for positive continuous $delta: M rightarrow mathbb{R}$.
differential-geometry manifolds smooth-manifolds
$endgroup$
(This is part of Problem 6-6 from Lee's Introduction to Smooth Manifolds textbook.)
Suppose $Msubset mathbb{R}^n$ is a compact embedded submanifold. For
any $epsilon>0$, let $M_epsilon$ be the set of all points in
$mathbb{R}^n$ whose distance from $M$ is less than $epsilon$. Show
that for sufficiently small $epsilon$, $partial M_epsilon$ is a
compact embedded hypersurface in $mathbb{R}^n$.
I thought to use the tubular neighbourhood theorem (since we have an embedded submanifold) and showing for some sufficiently small tubular neighbourhood, maybe we can prove that the boundary of the tubular neighbourhood is an embedded $(n-1)$-dimensional embedded manifold (and relate $epsilon$ to the distance function in the definition of a tubular neighbourhood). However, I have no idea how to prove that (or if it's even the right approach).
For reference, I'm using Lee's definition of a tubular neighbourhood. That is, $E(V)$ where $E(x,v)=x+v$ and $V={(x,v)in NM : |v|<delta(x)}$ for positive continuous $delta: M rightarrow mathbb{R}$.
differential-geometry manifolds smooth-manifolds
differential-geometry manifolds smooth-manifolds
edited Dec 9 '18 at 7:06
Frederic Chopin
321111
321111
asked Dec 7 '18 at 0:20
FunctionalDefectFunctionalDefect
255
255
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you suggest, using the tubular neighborhood theorem is the right idea.
Summary. The core point is as follows: Let $y in mathbb{R}^n$ have $d(y,M) = ell$. Then since $M$ is compact there exists a point $x in M$ with $d(y,x) = ell$, and the line segment from $x$ to $y$ in $mathbb{R}^n$ intersects $M$ orthogonally at $x$. (This last fact is a multivariable calculus exercise: The gradient of $d(cdot, y)$ is orthogonal to $M$ at any minimum.) This lets you relate "distance from $M$" to a condition on the normal bundle of $M$, which lets you use tubular neighborhoods.
I'll flesh out some details without being too careful, but I think everything below is fairly rote given the above:
Review of stuff in Lee. Some reminders so that we're on the same page: Let $M subseteq mathbb{R}^n$ be an embedded submanifold. The points of the normal bundle $NM$ of $M$ are pairs $(x, v)$ with $x in M$ and $v in T_{x}mathbb{R}^n$ normal to $M$. There's a map $E : NM to mathbb{R}^n$ given by $E(x,v) = x+v$.
Suppose $f: M to mathbb{R}$ is a positive continuous function on $M$ and let $V$ be the subset of $NM$ given by
$$
V = lbrace (x, v) in NM : |v| < f(x) rbrace.
$$
If $U = E(V)$ is open and $E$ restricts to a diffeomorphism $V cong U$, then $U$ is called a tubular neighborhood of $M$ in $mathbb{R}^n$. Lee proves (it is Theorem 10.19 in my copy) that every embedded $M subseteq mathbb{R}^n$ has a tubular neighborhood.
Notation. For any $delta > 0$ we define $V_delta subset NM$ by
$$
V_delta = lbrace (x, v) in NM : |v| < delta rbrace.
$$
When $E$ gives a diffeomorphism of $V_delta$ onto an open subset of $mathbb{R}^n$, we denote by $U_delta$ the image of $V_delta$.
When $M$ is compact, any $f$ giving a tubular neighborhood has a positive minimum, and so there exists a constant $delta > 0$ such that $M$ has a tubular neighborhood of the form $U_delta = E(V_delta)$.
Let $epsilon < delta$.
The Main Point. $M_epsilon$ is contained within $U_delta = E(V_delta)$. This follows from the discussion in our summary at the top; if $x$ is a point in $M$ closest to $y$, then $y$ is the image under $E$ of $(x, y-x)$, which is in $V_delta$.
In fact, this argument shows that $M_epsilon = U_epsilon$.
It follows that $partial M_epsilon$ is the image under $E$ of the set
$$
partial V_epsilon = lbrace (x, v) in NM : |v| = epsilon rbrace.
$$
Since $partial V_epsilon$ is clearly an embedded submanifold of $V_delta$, and $E$ is a diffeomorphism from $V_delta$ onto $U_delta$, it follows that $partial M_epsilon$ is an embedded submanifold of $U_delta$ (and hence of $mathbb{R}^n$).
Similarly, since $partial V_epsilon$ has codimension $1$, we see that $partial M_epsilon$ does as well.
(In fact this shows that $partial M_epsilon$ is the diffeomorpic to the bundle of unit $(n-m)$-spheres in $NM$; we can picture $partial M_epsilon$ as a "hollow tube" around $M$ whose "cross-section" at any $x in M$ is an $(n-m)$-sphere. It follows easily from this that $partial M_epsilon$ is compact.)
$endgroup$
$begingroup$
Thank you so much for your help!
$endgroup$
– Frederic Chopin
Dec 10 '18 at 22:14
$begingroup$
Very thorough answer, thanks
$endgroup$
– FunctionalDefect
Dec 10 '18 at 22:16
add a comment |
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$begingroup$
As you suggest, using the tubular neighborhood theorem is the right idea.
Summary. The core point is as follows: Let $y in mathbb{R}^n$ have $d(y,M) = ell$. Then since $M$ is compact there exists a point $x in M$ with $d(y,x) = ell$, and the line segment from $x$ to $y$ in $mathbb{R}^n$ intersects $M$ orthogonally at $x$. (This last fact is a multivariable calculus exercise: The gradient of $d(cdot, y)$ is orthogonal to $M$ at any minimum.) This lets you relate "distance from $M$" to a condition on the normal bundle of $M$, which lets you use tubular neighborhoods.
I'll flesh out some details without being too careful, but I think everything below is fairly rote given the above:
Review of stuff in Lee. Some reminders so that we're on the same page: Let $M subseteq mathbb{R}^n$ be an embedded submanifold. The points of the normal bundle $NM$ of $M$ are pairs $(x, v)$ with $x in M$ and $v in T_{x}mathbb{R}^n$ normal to $M$. There's a map $E : NM to mathbb{R}^n$ given by $E(x,v) = x+v$.
Suppose $f: M to mathbb{R}$ is a positive continuous function on $M$ and let $V$ be the subset of $NM$ given by
$$
V = lbrace (x, v) in NM : |v| < f(x) rbrace.
$$
If $U = E(V)$ is open and $E$ restricts to a diffeomorphism $V cong U$, then $U$ is called a tubular neighborhood of $M$ in $mathbb{R}^n$. Lee proves (it is Theorem 10.19 in my copy) that every embedded $M subseteq mathbb{R}^n$ has a tubular neighborhood.
Notation. For any $delta > 0$ we define $V_delta subset NM$ by
$$
V_delta = lbrace (x, v) in NM : |v| < delta rbrace.
$$
When $E$ gives a diffeomorphism of $V_delta$ onto an open subset of $mathbb{R}^n$, we denote by $U_delta$ the image of $V_delta$.
When $M$ is compact, any $f$ giving a tubular neighborhood has a positive minimum, and so there exists a constant $delta > 0$ such that $M$ has a tubular neighborhood of the form $U_delta = E(V_delta)$.
Let $epsilon < delta$.
The Main Point. $M_epsilon$ is contained within $U_delta = E(V_delta)$. This follows from the discussion in our summary at the top; if $x$ is a point in $M$ closest to $y$, then $y$ is the image under $E$ of $(x, y-x)$, which is in $V_delta$.
In fact, this argument shows that $M_epsilon = U_epsilon$.
It follows that $partial M_epsilon$ is the image under $E$ of the set
$$
partial V_epsilon = lbrace (x, v) in NM : |v| = epsilon rbrace.
$$
Since $partial V_epsilon$ is clearly an embedded submanifold of $V_delta$, and $E$ is a diffeomorphism from $V_delta$ onto $U_delta$, it follows that $partial M_epsilon$ is an embedded submanifold of $U_delta$ (and hence of $mathbb{R}^n$).
Similarly, since $partial V_epsilon$ has codimension $1$, we see that $partial M_epsilon$ does as well.
(In fact this shows that $partial M_epsilon$ is the diffeomorpic to the bundle of unit $(n-m)$-spheres in $NM$; we can picture $partial M_epsilon$ as a "hollow tube" around $M$ whose "cross-section" at any $x in M$ is an $(n-m)$-sphere. It follows easily from this that $partial M_epsilon$ is compact.)
$endgroup$
$begingroup$
Thank you so much for your help!
$endgroup$
– Frederic Chopin
Dec 10 '18 at 22:14
$begingroup$
Very thorough answer, thanks
$endgroup$
– FunctionalDefect
Dec 10 '18 at 22:16
add a comment |
$begingroup$
As you suggest, using the tubular neighborhood theorem is the right idea.
Summary. The core point is as follows: Let $y in mathbb{R}^n$ have $d(y,M) = ell$. Then since $M$ is compact there exists a point $x in M$ with $d(y,x) = ell$, and the line segment from $x$ to $y$ in $mathbb{R}^n$ intersects $M$ orthogonally at $x$. (This last fact is a multivariable calculus exercise: The gradient of $d(cdot, y)$ is orthogonal to $M$ at any minimum.) This lets you relate "distance from $M$" to a condition on the normal bundle of $M$, which lets you use tubular neighborhoods.
I'll flesh out some details without being too careful, but I think everything below is fairly rote given the above:
Review of stuff in Lee. Some reminders so that we're on the same page: Let $M subseteq mathbb{R}^n$ be an embedded submanifold. The points of the normal bundle $NM$ of $M$ are pairs $(x, v)$ with $x in M$ and $v in T_{x}mathbb{R}^n$ normal to $M$. There's a map $E : NM to mathbb{R}^n$ given by $E(x,v) = x+v$.
Suppose $f: M to mathbb{R}$ is a positive continuous function on $M$ and let $V$ be the subset of $NM$ given by
$$
V = lbrace (x, v) in NM : |v| < f(x) rbrace.
$$
If $U = E(V)$ is open and $E$ restricts to a diffeomorphism $V cong U$, then $U$ is called a tubular neighborhood of $M$ in $mathbb{R}^n$. Lee proves (it is Theorem 10.19 in my copy) that every embedded $M subseteq mathbb{R}^n$ has a tubular neighborhood.
Notation. For any $delta > 0$ we define $V_delta subset NM$ by
$$
V_delta = lbrace (x, v) in NM : |v| < delta rbrace.
$$
When $E$ gives a diffeomorphism of $V_delta$ onto an open subset of $mathbb{R}^n$, we denote by $U_delta$ the image of $V_delta$.
When $M$ is compact, any $f$ giving a tubular neighborhood has a positive minimum, and so there exists a constant $delta > 0$ such that $M$ has a tubular neighborhood of the form $U_delta = E(V_delta)$.
Let $epsilon < delta$.
The Main Point. $M_epsilon$ is contained within $U_delta = E(V_delta)$. This follows from the discussion in our summary at the top; if $x$ is a point in $M$ closest to $y$, then $y$ is the image under $E$ of $(x, y-x)$, which is in $V_delta$.
In fact, this argument shows that $M_epsilon = U_epsilon$.
It follows that $partial M_epsilon$ is the image under $E$ of the set
$$
partial V_epsilon = lbrace (x, v) in NM : |v| = epsilon rbrace.
$$
Since $partial V_epsilon$ is clearly an embedded submanifold of $V_delta$, and $E$ is a diffeomorphism from $V_delta$ onto $U_delta$, it follows that $partial M_epsilon$ is an embedded submanifold of $U_delta$ (and hence of $mathbb{R}^n$).
Similarly, since $partial V_epsilon$ has codimension $1$, we see that $partial M_epsilon$ does as well.
(In fact this shows that $partial M_epsilon$ is the diffeomorpic to the bundle of unit $(n-m)$-spheres in $NM$; we can picture $partial M_epsilon$ as a "hollow tube" around $M$ whose "cross-section" at any $x in M$ is an $(n-m)$-sphere. It follows easily from this that $partial M_epsilon$ is compact.)
$endgroup$
$begingroup$
Thank you so much for your help!
$endgroup$
– Frederic Chopin
Dec 10 '18 at 22:14
$begingroup$
Very thorough answer, thanks
$endgroup$
– FunctionalDefect
Dec 10 '18 at 22:16
add a comment |
$begingroup$
As you suggest, using the tubular neighborhood theorem is the right idea.
Summary. The core point is as follows: Let $y in mathbb{R}^n$ have $d(y,M) = ell$. Then since $M$ is compact there exists a point $x in M$ with $d(y,x) = ell$, and the line segment from $x$ to $y$ in $mathbb{R}^n$ intersects $M$ orthogonally at $x$. (This last fact is a multivariable calculus exercise: The gradient of $d(cdot, y)$ is orthogonal to $M$ at any minimum.) This lets you relate "distance from $M$" to a condition on the normal bundle of $M$, which lets you use tubular neighborhoods.
I'll flesh out some details without being too careful, but I think everything below is fairly rote given the above:
Review of stuff in Lee. Some reminders so that we're on the same page: Let $M subseteq mathbb{R}^n$ be an embedded submanifold. The points of the normal bundle $NM$ of $M$ are pairs $(x, v)$ with $x in M$ and $v in T_{x}mathbb{R}^n$ normal to $M$. There's a map $E : NM to mathbb{R}^n$ given by $E(x,v) = x+v$.
Suppose $f: M to mathbb{R}$ is a positive continuous function on $M$ and let $V$ be the subset of $NM$ given by
$$
V = lbrace (x, v) in NM : |v| < f(x) rbrace.
$$
If $U = E(V)$ is open and $E$ restricts to a diffeomorphism $V cong U$, then $U$ is called a tubular neighborhood of $M$ in $mathbb{R}^n$. Lee proves (it is Theorem 10.19 in my copy) that every embedded $M subseteq mathbb{R}^n$ has a tubular neighborhood.
Notation. For any $delta > 0$ we define $V_delta subset NM$ by
$$
V_delta = lbrace (x, v) in NM : |v| < delta rbrace.
$$
When $E$ gives a diffeomorphism of $V_delta$ onto an open subset of $mathbb{R}^n$, we denote by $U_delta$ the image of $V_delta$.
When $M$ is compact, any $f$ giving a tubular neighborhood has a positive minimum, and so there exists a constant $delta > 0$ such that $M$ has a tubular neighborhood of the form $U_delta = E(V_delta)$.
Let $epsilon < delta$.
The Main Point. $M_epsilon$ is contained within $U_delta = E(V_delta)$. This follows from the discussion in our summary at the top; if $x$ is a point in $M$ closest to $y$, then $y$ is the image under $E$ of $(x, y-x)$, which is in $V_delta$.
In fact, this argument shows that $M_epsilon = U_epsilon$.
It follows that $partial M_epsilon$ is the image under $E$ of the set
$$
partial V_epsilon = lbrace (x, v) in NM : |v| = epsilon rbrace.
$$
Since $partial V_epsilon$ is clearly an embedded submanifold of $V_delta$, and $E$ is a diffeomorphism from $V_delta$ onto $U_delta$, it follows that $partial M_epsilon$ is an embedded submanifold of $U_delta$ (and hence of $mathbb{R}^n$).
Similarly, since $partial V_epsilon$ has codimension $1$, we see that $partial M_epsilon$ does as well.
(In fact this shows that $partial M_epsilon$ is the diffeomorpic to the bundle of unit $(n-m)$-spheres in $NM$; we can picture $partial M_epsilon$ as a "hollow tube" around $M$ whose "cross-section" at any $x in M$ is an $(n-m)$-sphere. It follows easily from this that $partial M_epsilon$ is compact.)
$endgroup$
As you suggest, using the tubular neighborhood theorem is the right idea.
Summary. The core point is as follows: Let $y in mathbb{R}^n$ have $d(y,M) = ell$. Then since $M$ is compact there exists a point $x in M$ with $d(y,x) = ell$, and the line segment from $x$ to $y$ in $mathbb{R}^n$ intersects $M$ orthogonally at $x$. (This last fact is a multivariable calculus exercise: The gradient of $d(cdot, y)$ is orthogonal to $M$ at any minimum.) This lets you relate "distance from $M$" to a condition on the normal bundle of $M$, which lets you use tubular neighborhoods.
I'll flesh out some details without being too careful, but I think everything below is fairly rote given the above:
Review of stuff in Lee. Some reminders so that we're on the same page: Let $M subseteq mathbb{R}^n$ be an embedded submanifold. The points of the normal bundle $NM$ of $M$ are pairs $(x, v)$ with $x in M$ and $v in T_{x}mathbb{R}^n$ normal to $M$. There's a map $E : NM to mathbb{R}^n$ given by $E(x,v) = x+v$.
Suppose $f: M to mathbb{R}$ is a positive continuous function on $M$ and let $V$ be the subset of $NM$ given by
$$
V = lbrace (x, v) in NM : |v| < f(x) rbrace.
$$
If $U = E(V)$ is open and $E$ restricts to a diffeomorphism $V cong U$, then $U$ is called a tubular neighborhood of $M$ in $mathbb{R}^n$. Lee proves (it is Theorem 10.19 in my copy) that every embedded $M subseteq mathbb{R}^n$ has a tubular neighborhood.
Notation. For any $delta > 0$ we define $V_delta subset NM$ by
$$
V_delta = lbrace (x, v) in NM : |v| < delta rbrace.
$$
When $E$ gives a diffeomorphism of $V_delta$ onto an open subset of $mathbb{R}^n$, we denote by $U_delta$ the image of $V_delta$.
When $M$ is compact, any $f$ giving a tubular neighborhood has a positive minimum, and so there exists a constant $delta > 0$ such that $M$ has a tubular neighborhood of the form $U_delta = E(V_delta)$.
Let $epsilon < delta$.
The Main Point. $M_epsilon$ is contained within $U_delta = E(V_delta)$. This follows from the discussion in our summary at the top; if $x$ is a point in $M$ closest to $y$, then $y$ is the image under $E$ of $(x, y-x)$, which is in $V_delta$.
In fact, this argument shows that $M_epsilon = U_epsilon$.
It follows that $partial M_epsilon$ is the image under $E$ of the set
$$
partial V_epsilon = lbrace (x, v) in NM : |v| = epsilon rbrace.
$$
Since $partial V_epsilon$ is clearly an embedded submanifold of $V_delta$, and $E$ is a diffeomorphism from $V_delta$ onto $U_delta$, it follows that $partial M_epsilon$ is an embedded submanifold of $U_delta$ (and hence of $mathbb{R}^n$).
Similarly, since $partial V_epsilon$ has codimension $1$, we see that $partial M_epsilon$ does as well.
(In fact this shows that $partial M_epsilon$ is the diffeomorpic to the bundle of unit $(n-m)$-spheres in $NM$; we can picture $partial M_epsilon$ as a "hollow tube" around $M$ whose "cross-section" at any $x in M$ is an $(n-m)$-sphere. It follows easily from this that $partial M_epsilon$ is compact.)
answered Dec 10 '18 at 13:28
mollyerinmollyerin
2,68576
2,68576
$begingroup$
Thank you so much for your help!
$endgroup$
– Frederic Chopin
Dec 10 '18 at 22:14
$begingroup$
Very thorough answer, thanks
$endgroup$
– FunctionalDefect
Dec 10 '18 at 22:16
add a comment |
$begingroup$
Thank you so much for your help!
$endgroup$
– Frederic Chopin
Dec 10 '18 at 22:14
$begingroup$
Very thorough answer, thanks
$endgroup$
– FunctionalDefect
Dec 10 '18 at 22:16
$begingroup$
Thank you so much for your help!
$endgroup$
– Frederic Chopin
Dec 10 '18 at 22:14
$begingroup$
Thank you so much for your help!
$endgroup$
– Frederic Chopin
Dec 10 '18 at 22:14
$begingroup$
Very thorough answer, thanks
$endgroup$
– FunctionalDefect
Dec 10 '18 at 22:16
$begingroup$
Very thorough answer, thanks
$endgroup$
– FunctionalDefect
Dec 10 '18 at 22:16
add a comment |
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