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How can I prove $(0,2]$ is a smooth manifold? And something questions about the choice of the open set of a...
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I am studying manifold, when I see the definition about the topological manifold, I am confused about the choice about the open set. For example, when we study the real line $mathbb{R}$ we usually choose $(a,b)$ to be the open set we consider, but when it comes to the form $(a,b]$, we have to choose $(c,b] subset (a,b]$ to be the open set to cover the $(a,b]$ (am I right?). But if we want to show that $(a,b]$ is a smooth manifold, we have to find the chart. So how can I find the proper homeomorphism which makes the open set of $(a,b]$ is homeomorphic to the open subset of $mathbb{R}$?
differential-geometry manifolds differential-topology
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
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|
show 6 more comments
$begingroup$
I am studying manifold, when I see the definition about the topological manifold, I am confused about the choice about the open set. For example, when we study the real line $mathbb{R}$ we usually choose $(a,b)$ to be the open set we consider, but when it comes to the form $(a,b]$, we have to choose $(c,b] subset (a,b]$ to be the open set to cover the $(a,b]$ (am I right?). But if we want to show that $(a,b]$ is a smooth manifold, we have to find the chart. So how can I find the proper homeomorphism which makes the open set of $(a,b]$ is homeomorphic to the open subset of $mathbb{R}$?
differential-geometry manifolds differential-topology
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
$endgroup$
$begingroup$
A half-open interval is not a manifold but a manifold with boundary.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 15:51
$begingroup$
@MoisheCohen I met a problem today. This problem wants me to show $mathbb{R}/2pi mathbb{Z}$ is a manifold, and $mathbb{R}/2pi mathbb{Z}$ is just $(0,2pi]$ (am I right?)
$endgroup$
– R.Sherlock
Jan 20 '18 at 15:54
$begingroup$
As a set you could say yes but an interval can be disconnected by removing one point and the circle cannot, so there is a topological difference
$endgroup$
– Calvin Khor
Jan 20 '18 at 16:03
1
$begingroup$
@R.Sherlock: No, you are wrong.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 16:03
1
$begingroup$
As a set, $Bbb R/2piBbb Z ne (0,2pi]$, it is instead the set $left.Big{{x + 2kpi mid k in Bbb Z}right| xin Bbb RBig}$. Topologically, $Bbb R/2piBbb Z$ is also not homeomorphic to $(0,2pi]$. Instead it is homeomorphic to the circle.
$endgroup$
– Paul Sinclair
Jan 20 '18 at 21:06
|
show 6 more comments
$begingroup$
I am studying manifold, when I see the definition about the topological manifold, I am confused about the choice about the open set. For example, when we study the real line $mathbb{R}$ we usually choose $(a,b)$ to be the open set we consider, but when it comes to the form $(a,b]$, we have to choose $(c,b] subset (a,b]$ to be the open set to cover the $(a,b]$ (am I right?). But if we want to show that $(a,b]$ is a smooth manifold, we have to find the chart. So how can I find the proper homeomorphism which makes the open set of $(a,b]$ is homeomorphic to the open subset of $mathbb{R}$?
differential-geometry manifolds differential-topology
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
$endgroup$
I am studying manifold, when I see the definition about the topological manifold, I am confused about the choice about the open set. For example, when we study the real line $mathbb{R}$ we usually choose $(a,b)$ to be the open set we consider, but when it comes to the form $(a,b]$, we have to choose $(c,b] subset (a,b]$ to be the open set to cover the $(a,b]$ (am I right?). But if we want to show that $(a,b]$ is a smooth manifold, we have to find the chart. So how can I find the proper homeomorphism which makes the open set of $(a,b]$ is homeomorphic to the open subset of $mathbb{R}$?
differential-geometry manifolds differential-topology
differential-geometry manifolds differential-topology
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
asked Jan 20 '18 at 15:40
R.SherlockR.Sherlock
383112
383112
$begingroup$
A half-open interval is not a manifold but a manifold with boundary.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 15:51
$begingroup$
@MoisheCohen I met a problem today. This problem wants me to show $mathbb{R}/2pi mathbb{Z}$ is a manifold, and $mathbb{R}/2pi mathbb{Z}$ is just $(0,2pi]$ (am I right?)
$endgroup$
– R.Sherlock
Jan 20 '18 at 15:54
$begingroup$
As a set you could say yes but an interval can be disconnected by removing one point and the circle cannot, so there is a topological difference
$endgroup$
– Calvin Khor
Jan 20 '18 at 16:03
1
$begingroup$
@R.Sherlock: No, you are wrong.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 16:03
1
$begingroup$
As a set, $Bbb R/2piBbb Z ne (0,2pi]$, it is instead the set $left.Big{{x + 2kpi mid k in Bbb Z}right| xin Bbb RBig}$. Topologically, $Bbb R/2piBbb Z$ is also not homeomorphic to $(0,2pi]$. Instead it is homeomorphic to the circle.
$endgroup$
– Paul Sinclair
Jan 20 '18 at 21:06
|
show 6 more comments
$begingroup$
A half-open interval is not a manifold but a manifold with boundary.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 15:51
$begingroup$
@MoisheCohen I met a problem today. This problem wants me to show $mathbb{R}/2pi mathbb{Z}$ is a manifold, and $mathbb{R}/2pi mathbb{Z}$ is just $(0,2pi]$ (am I right?)
$endgroup$
– R.Sherlock
Jan 20 '18 at 15:54
$begingroup$
As a set you could say yes but an interval can be disconnected by removing one point and the circle cannot, so there is a topological difference
$endgroup$
– Calvin Khor
Jan 20 '18 at 16:03
1
$begingroup$
@R.Sherlock: No, you are wrong.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 16:03
1
$begingroup$
As a set, $Bbb R/2piBbb Z ne (0,2pi]$, it is instead the set $left.Big{{x + 2kpi mid k in Bbb Z}right| xin Bbb RBig}$. Topologically, $Bbb R/2piBbb Z$ is also not homeomorphic to $(0,2pi]$. Instead it is homeomorphic to the circle.
$endgroup$
– Paul Sinclair
Jan 20 '18 at 21:06
$begingroup$
A half-open interval is not a manifold but a manifold with boundary.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 15:51
$begingroup$
A half-open interval is not a manifold but a manifold with boundary.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 15:51
$begingroup$
@MoisheCohen I met a problem today. This problem wants me to show $mathbb{R}/2pi mathbb{Z}$ is a manifold, and $mathbb{R}/2pi mathbb{Z}$ is just $(0,2pi]$ (am I right?)
$endgroup$
– R.Sherlock
Jan 20 '18 at 15:54
$begingroup$
@MoisheCohen I met a problem today. This problem wants me to show $mathbb{R}/2pi mathbb{Z}$ is a manifold, and $mathbb{R}/2pi mathbb{Z}$ is just $(0,2pi]$ (am I right?)
$endgroup$
– R.Sherlock
Jan 20 '18 at 15:54
$begingroup$
As a set you could say yes but an interval can be disconnected by removing one point and the circle cannot, so there is a topological difference
$endgroup$
– Calvin Khor
Jan 20 '18 at 16:03
$begingroup$
As a set you could say yes but an interval can be disconnected by removing one point and the circle cannot, so there is a topological difference
$endgroup$
– Calvin Khor
Jan 20 '18 at 16:03
1
1
$begingroup$
@R.Sherlock: No, you are wrong.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 16:03
$begingroup$
@R.Sherlock: No, you are wrong.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 16:03
1
1
$begingroup$
As a set, $Bbb R/2piBbb Z ne (0,2pi]$, it is instead the set $left.Big{{x + 2kpi mid k in Bbb Z}right| xin Bbb RBig}$. Topologically, $Bbb R/2piBbb Z$ is also not homeomorphic to $(0,2pi]$. Instead it is homeomorphic to the circle.
$endgroup$
– Paul Sinclair
Jan 20 '18 at 21:06
$begingroup$
As a set, $Bbb R/2piBbb Z ne (0,2pi]$, it is instead the set $left.Big{{x + 2kpi mid k in Bbb Z}right| xin Bbb RBig}$. Topologically, $Bbb R/2piBbb Z$ is also not homeomorphic to $(0,2pi]$. Instead it is homeomorphic to the circle.
$endgroup$
– Paul Sinclair
Jan 20 '18 at 21:06
|
show 6 more comments
1 Answer
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It turns out that $M= (0, 2]$ is a topological manifold of dimension $1$ with boundary.
Here is an interior chart. Define $phi : (0, 2) subseteq M to mathbb{R}^1$ by $phi(x) = x$, then $((0, 2), phi)$ is an interior chart since $phi$ is clearly a homeomorphism. So every point $p in (0, 2) subseteq M$ is contained in a neighborhood of $M$ homeomorphic to an open susbet of $mathbb{R}^1$ and so all such $p$ are (manifold) interior points of $M$.
Now here is a boundary chart, define $psi : (1, 2] subseteq M to mathbb{H}^1 = [0, infty)$ by $psi(x)= -x+2$, then $operatorname{Im}(psi) = [0, 1)$ which is open in $mathbb{H}^1$ and $psi$ is clearly a homeomorphism. Thus ${2}$ is contained in a neighborhood of $M$ homeomorphic to an open subset of $mathbb{H}^1$ and so $2$ is a (manifold) boundary point of $M$.
Thus $M$ is a topological manifold of dimension $1$ with boundary.
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
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1 Answer
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$begingroup$
It turns out that $M= (0, 2]$ is a topological manifold of dimension $1$ with boundary.
Here is an interior chart. Define $phi : (0, 2) subseteq M to mathbb{R}^1$ by $phi(x) = x$, then $((0, 2), phi)$ is an interior chart since $phi$ is clearly a homeomorphism. So every point $p in (0, 2) subseteq M$ is contained in a neighborhood of $M$ homeomorphic to an open susbet of $mathbb{R}^1$ and so all such $p$ are (manifold) interior points of $M$.
Now here is a boundary chart, define $psi : (1, 2] subseteq M to mathbb{H}^1 = [0, infty)$ by $psi(x)= -x+2$, then $operatorname{Im}(psi) = [0, 1)$ which is open in $mathbb{H}^1$ and $psi$ is clearly a homeomorphism. Thus ${2}$ is contained in a neighborhood of $M$ homeomorphic to an open subset of $mathbb{H}^1$ and so $2$ is a (manifold) boundary point of $M$.
Thus $M$ is a topological manifold of dimension $1$ with boundary.
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
$endgroup$
add a comment |
$begingroup$
It turns out that $M= (0, 2]$ is a topological manifold of dimension $1$ with boundary.
Here is an interior chart. Define $phi : (0, 2) subseteq M to mathbb{R}^1$ by $phi(x) = x$, then $((0, 2), phi)$ is an interior chart since $phi$ is clearly a homeomorphism. So every point $p in (0, 2) subseteq M$ is contained in a neighborhood of $M$ homeomorphic to an open susbet of $mathbb{R}^1$ and so all such $p$ are (manifold) interior points of $M$.
Now here is a boundary chart, define $psi : (1, 2] subseteq M to mathbb{H}^1 = [0, infty)$ by $psi(x)= -x+2$, then $operatorname{Im}(psi) = [0, 1)$ which is open in $mathbb{H}^1$ and $psi$ is clearly a homeomorphism. Thus ${2}$ is contained in a neighborhood of $M$ homeomorphic to an open subset of $mathbb{H}^1$ and so $2$ is a (manifold) boundary point of $M$.
Thus $M$ is a topological manifold of dimension $1$ with boundary.
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
$endgroup$
add a comment |
$begingroup$
It turns out that $M= (0, 2]$ is a topological manifold of dimension $1$ with boundary.
Here is an interior chart. Define $phi : (0, 2) subseteq M to mathbb{R}^1$ by $phi(x) = x$, then $((0, 2), phi)$ is an interior chart since $phi$ is clearly a homeomorphism. So every point $p in (0, 2) subseteq M$ is contained in a neighborhood of $M$ homeomorphic to an open susbet of $mathbb{R}^1$ and so all such $p$ are (manifold) interior points of $M$.
Now here is a boundary chart, define $psi : (1, 2] subseteq M to mathbb{H}^1 = [0, infty)$ by $psi(x)= -x+2$, then $operatorname{Im}(psi) = [0, 1)$ which is open in $mathbb{H}^1$ and $psi$ is clearly a homeomorphism. Thus ${2}$ is contained in a neighborhood of $M$ homeomorphic to an open subset of $mathbb{H}^1$ and so $2$ is a (manifold) boundary point of $M$.
Thus $M$ is a topological manifold of dimension $1$ with boundary.
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
$endgroup$
It turns out that $M= (0, 2]$ is a topological manifold of dimension $1$ with boundary.
Here is an interior chart. Define $phi : (0, 2) subseteq M to mathbb{R}^1$ by $phi(x) = x$, then $((0, 2), phi)$ is an interior chart since $phi$ is clearly a homeomorphism. So every point $p in (0, 2) subseteq M$ is contained in a neighborhood of $M$ homeomorphic to an open susbet of $mathbb{R}^1$ and so all such $p$ are (manifold) interior points of $M$.
Now here is a boundary chart, define $psi : (1, 2] subseteq M to mathbb{H}^1 = [0, infty)$ by $psi(x)= -x+2$, then $operatorname{Im}(psi) = [0, 1)$ which is open in $mathbb{H}^1$ and $psi$ is clearly a homeomorphism. Thus ${2}$ is contained in a neighborhood of $M$ homeomorphic to an open subset of $mathbb{H}^1$ and so $2$ is a (manifold) boundary point of $M$.
Thus $M$ is a topological manifold of dimension $1$ with boundary.
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
answered Jan 9 at 0:14
PerturbativePerturbative
4,29811551
4,29811551
add a comment |
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$begingroup$
A half-open interval is not a manifold but a manifold with boundary.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 15:51
$begingroup$
@MoisheCohen I met a problem today. This problem wants me to show $mathbb{R}/2pi mathbb{Z}$ is a manifold, and $mathbb{R}/2pi mathbb{Z}$ is just $(0,2pi]$ (am I right?)
$endgroup$
– R.Sherlock
Jan 20 '18 at 15:54
$begingroup$
As a set you could say yes but an interval can be disconnected by removing one point and the circle cannot, so there is a topological difference
$endgroup$
– Calvin Khor
Jan 20 '18 at 16:03
1
$begingroup$
@R.Sherlock: No, you are wrong.
$endgroup$
– Moishe Cohen
Jan 20 '18 at 16:03
1
$begingroup$
As a set, $Bbb R/2piBbb Z ne (0,2pi]$, it is instead the set $left.Big{{x + 2kpi mid k in Bbb Z}right| xin Bbb RBig}$. Topologically, $Bbb R/2piBbb Z$ is also not homeomorphic to $(0,2pi]$. Instead it is homeomorphic to the circle.
$endgroup$
– Paul Sinclair
Jan 20 '18 at 21:06