Mixing
Question of whether two given spaces are homeomorphic.
Clash Royale CLAN TAG#URR8PPP
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Let $D^2$ be the closed disk on the plane.
First we pick an arbitrary point $xin bd(D^2)$ on $D^2$, and define $X = D^2-x$.
Then define another space $Y$ by removing a homeomorphic image of the closed interval $I$ from the boundary, that is, $Y = D^2-h(I)$.(The original question defines $Y$ by removing the upper closed semi-circle of $D^2$ while I think this can be generalized.)
I tried to construct the homeomorphism between two spaces but failed. The two spaces are both connected, non-compact and convex so I also failed to prove they are not homeomorphic.
general-topology
asked Jul 17 at 20:54
user413924
1516
add a comment |Â
up vote
6
down vote
favorite
Let $D^2$ be the closed disk on the plane.
First we pick an arbitrary point $xin bd(D^2)$ on $D^2$, and define $X = D^2-x$.
Then define another space $Y$ by removing a homeomorphic image of the closed interval $I$ from the boundary, that is, $Y = D^2-h(I)$.(The original question defines $Y$ by removing the upper closed semi-circle of $D^2$ while I think this can be generalized.)
I tried to construct the homeomorphism between two spaces but failed. The two spaces are both connected, non-compact and convex so I also failed to prove they are not homeomorphic.
general-topology
asked Jul 17 at 20:54
user413924
1516
The boundary-punctured disc is homeomorphic to the closed upper half plane, and if you delete an arc, you instead get the upper half plane with $(-infty, t_0) cup (t_1, infty)$ where $t_0 < t_1$. By scaling and translation, you see that the result of deleting any arc from the boundary of $D^2$ is homeomphic to any other. This still leaves the question of point vs arc open, but that is true (I just need to come up with the argument...)
– Mike Miller
Jul 18 at 2:13
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $D^2$ be the closed disk on the plane.
First we pick an arbitrary point $xin bd(D^2)$ on $D^2$, and define $X = D^2-x$.
Then define another space $Y$ by removing a homeomorphic image of the closed interval $I$ from the boundary, that is, $Y = D^2-h(I)$.(The original question defines $Y$ by removing the upper closed semi-circle of $D^2$ while I think this can be generalized.)
I tried to construct the homeomorphism between two spaces but failed. The two spaces are both connected, non-compact and convex so I also failed to prove they are not homeomorphic.
general-topology
asked Jul 17 at 20:54
user413924
1516
Let $D^2$ be the closed disk on the plane.
First we pick an arbitrary point $xin bd(D^2)$ on $D^2$, and define $X = D^2-x$.
Then define another space $Y$ by removing a homeomorphic image of the closed interval $I$ from the boundary, that is, $Y = D^2-h(I)$.(The original question defines $Y$ by removing the upper closed semi-circle of $D^2$ while I think this can be generalized.)
I tried to construct the homeomorphism between two spaces but failed. The two spaces are both connected, non-compact and convex so I also failed to prove they are not homeomorphic.
general-topology
asked Jul 17 at 20:54
user413924
1516
asked Jul 17 at 20:54
user413924
1516
asked Jul 17 at 20:54
user413924
1516
asked Jul 17 at 20:54
user413924
1516
1516
The boundary-punctured disc is homeomorphic to the closed upper half plane, and if you delete an arc, you instead get the upper half plane with $(-infty, t_0) cup (t_1, infty)$ where $t_0 < t_1$. By scaling and translation, you see that the result of deleting any arc from the boundary of $D^2$ is homeomphic to any other. This still leaves the question of point vs arc open, but that is true (I just need to come up with the argument...)
– Mike Miller
Jul 18 at 2:13
add a comment |Â
The boundary-punctured disc is homeomorphic to the closed upper half plane, and if you delete an arc, you instead get the upper half plane with $(-infty, t_0) cup (t_1, infty)$ where $t_0 < t_1$. By scaling and translation, you see that the result of deleting any arc from the boundary of $D^2$ is homeomphic to any other. This still leaves the question of point vs arc open, but that is true (I just need to come up with the argument...)
– Mike Miller
Jul 18 at 2:13
The boundary-punctured disc is homeomorphic to the closed upper half plane, and if you delete an arc, you instead get the upper half plane with $(-infty, t_0) cup (t_1, infty)$ where $t_0 < t_1$. By scaling and translation, you see that the result of deleting any arc from the boundary of $D^2$ is homeomphic to any other. This still leaves the question of point vs arc open, but that is true (I just need to come up with the argument...)
– Mike Miller
Jul 18 at 2:13
The boundary-punctured disc is homeomorphic to the closed upper half plane, and if you delete an arc, you instead get the upper half plane with $(-infty, t_0) cup (t_1, infty)$ where $t_0 < t_1$. By scaling and translation, you see that the result of deleting any arc from the boundary of $D^2$ is homeomphic to any other. This still leaves the question of point vs arc open, but that is true (I just need to come up with the argument...)
– Mike Miller
Jul 18 at 2:13
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let $e : mathbbR to S^1, e(t) = e^it$. Then a homeomorphic image of a closed interval in $S^1$ is any set having the form $J = e([a,b])$ with $a < b$ and $b - a < 2pi$.
We have $X = D^2 backslash e(c) $ for some $c in mathbbR$ and $Y = D^2 backslash e([a,b])$ for suitable $a,b$ as above. Define $r : [a,a + 2 pi] to [c,c+2pi]$, $r(x) = c$ for $t in [a,b]$, $r(x) = c + frac2pi (x - b)a+2pi-b$ for $t in [b,a+2pi]$. This is a continuous map. Define
$$H : [a,a + 2 pi] times [0,1] to [c,c+2pi], H(x,t) = (1-t)(x -a + c) + t r(x) .$$
This is a continuous map such that (with $H_t(x) = H(x,t)$)
(1) $H_1 = r$
(2) $H_t(a) = c, H_t(a+2pi) = c + 2pi$ for all $t$
(3) $H_t$ is strictly increasing, in particular bijective, for all $t < 1$.
Define
$$G : [a,a + 2 pi] times [0,1] to D^2, G(x,t) = te(H(x,t))$$
which is again continuous. Consider the continuous map
$$p : [a,a + 2 pi] times [0,1] to D^2, p(x,t) = te(x) .$$
Since domain and range are compact, it is an identification map. It is easy to check that if $p(xi) = p(xi')$, then $G(xi) = G(xi')$. Therefore we obtain a unique continuous $g : D^2 to D^2$ such that $g circ p = G$. By construction $g$ maps $Y$ bijectively onto $X$. The restriction $g' : Y to X$ is an open map which shows that it is a homeomorphism: Let $U subset Y$ be open. Then $D^2 backslash U$ is compact, thus $g(D^2 backslash U) ) = D^2 backslash g(U)$ is compact so that $g'(U) = g(U)$ is open in $X$.
answered Jul 24 at 10:52
Paul Frost
3,703420
add a comment |Â
up vote
2
down vote
The answer of @PaulFrost is awesome, and should be the accepted answer. I thought I'd just give an explicit description of his approach.
We'll write the unit disk as
$$ mathbbD = re^pi itmid 0le rle1, -1le tle1$$
And define the lower hemisphere as $H=e^pi itmid -1le tle0$, and the point $q=-1$.
We have a continuous map $h:mathbbDrightarrowmathbbD$, given by
$$ h(re^pi it) = re^t$$
It should be checked this is well-defined (plugging in $t=1$ and $t=-1$). It's also straightforward that $h^-1(q)=H$. Thus we get a map $h:mathbbDsetminus Hrightarrow mathbbDsetminusq$.
An explicit inverse can be given too:
$$ h^-1(rho e^pi itheta) = rho e^(rho+theta)^2/(rho+theta+rho$$
One has to worry about that denominator being zero. The quadratic formula shows that happens when $rho+theta=0$. If $rho<1$, then $-1<theta<0$, and it is easy to check the fraction in the exponent goes to zero. So $h^-1(rho e^-pi irho)=rho$.
When $rho=1$, that's a prpblem, because then we have $theta=pm1$. Luckily we are ignoring that point ($q$), so we get a well-defined map
$$ h^-1:mathbbDsetminus qrightarrowmathbbDsetminus H$$
and so $h$ is the desired homeomorphism.
answered Jul 25 at 0:36
Steve D
2,042419
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $e : mathbbR to S^1, e(t) = e^it$. Then a homeomorphic image of a closed interval in $S^1$ is any set having the form $J = e([a,b])$ with $a < b$ and $b - a < 2pi$.
We have $X = D^2 backslash e(c) $ for some $c in mathbbR$ and $Y = D^2 backslash e([a,b])$ for suitable $a,b$ as above. Define $r : [a,a + 2 pi] to [c,c+2pi]$, $r(x) = c$ for $t in [a,b]$, $r(x) = c + frac2pi (x - b)a+2pi-b$ for $t in [b,a+2pi]$. This is a continuous map. Define
$$H : [a,a + 2 pi] times [0,1] to [c,c+2pi], H(x,t) = (1-t)(x -a + c) + t r(x) .$$
This is a continuous map such that (with $H_t(x) = H(x,t)$)
(1) $H_1 = r$
(2) $H_t(a) = c, H_t(a+2pi) = c + 2pi$ for all $t$
(3) $H_t$ is strictly increasing, in particular bijective, for all $t < 1$.
Define
$$G : [a,a + 2 pi] times [0,1] to D^2, G(x,t) = te(H(x,t))$$
which is again continuous. Consider the continuous map
$$p : [a,a + 2 pi] times [0,1] to D^2, p(x,t) = te(x) .$$
Since domain and range are compact, it is an identification map. It is easy to check that if $p(xi) = p(xi')$, then $G(xi) = G(xi')$. Therefore we obtain a unique continuous $g : D^2 to D^2$ such that $g circ p = G$. By construction $g$ maps $Y$ bijectively onto $X$. The restriction $g' : Y to X$ is an open map which shows that it is a homeomorphism: Let $U subset Y$ be open. Then $D^2 backslash U$ is compact, thus $g(D^2 backslash U) ) = D^2 backslash g(U)$ is compact so that $g'(U) = g(U)$ is open in $X$.
answered Jul 24 at 10:52
Paul Frost
3,703420
add a comment |Â
up vote
3
down vote
accepted
Let $e : mathbbR to S^1, e(t) = e^it$. Then a homeomorphic image of a closed interval in $S^1$ is any set having the form $J = e([a,b])$ with $a < b$ and $b - a < 2pi$.
We have $X = D^2 backslash e(c) $ for some $c in mathbbR$ and $Y = D^2 backslash e([a,b])$ for suitable $a,b$ as above. Define $r : [a,a + 2 pi] to [c,c+2pi]$, $r(x) = c$ for $t in [a,b]$, $r(x) = c + frac2pi (x - b)a+2pi-b$ for $t in [b,a+2pi]$. This is a continuous map. Define
$$H : [a,a + 2 pi] times [0,1] to [c,c+2pi], H(x,t) = (1-t)(x -a + c) + t r(x) .$$
This is a continuous map such that (with $H_t(x) = H(x,t)$)
(1) $H_1 = r$
(2) $H_t(a) = c, H_t(a+2pi) = c + 2pi$ for all $t$
(3) $H_t$ is strictly increasing, in particular bijective, for all $t < 1$.
Define
$$G : [a,a + 2 pi] times [0,1] to D^2, G(x,t) = te(H(x,t))$$
which is again continuous. Consider the continuous map
$$p : [a,a + 2 pi] times [0,1] to D^2, p(x,t) = te(x) .$$
Since domain and range are compact, it is an identification map. It is easy to check that if $p(xi) = p(xi')$, then $G(xi) = G(xi')$. Therefore we obtain a unique continuous $g : D^2 to D^2$ such that $g circ p = G$. By construction $g$ maps $Y$ bijectively onto $X$. The restriction $g' : Y to X$ is an open map which shows that it is a homeomorphism: Let $U subset Y$ be open. Then $D^2 backslash U$ is compact, thus $g(D^2 backslash U) ) = D^2 backslash g(U)$ is compact so that $g'(U) = g(U)$ is open in $X$.
answered Jul 24 at 10:52
Paul Frost
3,703420
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $e : mathbbR to S^1, e(t) = e^it$. Then a homeomorphic image of a closed interval in $S^1$ is any set having the form $J = e([a,b])$ with $a < b$ and $b - a < 2pi$.
We have $X = D^2 backslash e(c) $ for some $c in mathbbR$ and $Y = D^2 backslash e([a,b])$ for suitable $a,b$ as above. Define $r : [a,a + 2 pi] to [c,c+2pi]$, $r(x) = c$ for $t in [a,b]$, $r(x) = c + frac2pi (x - b)a+2pi-b$ for $t in [b,a+2pi]$. This is a continuous map. Define
$$H : [a,a + 2 pi] times [0,1] to [c,c+2pi], H(x,t) = (1-t)(x -a + c) + t r(x) .$$
This is a continuous map such that (with $H_t(x) = H(x,t)$)
(1) $H_1 = r$
(2) $H_t(a) = c, H_t(a+2pi) = c + 2pi$ for all $t$
(3) $H_t$ is strictly increasing, in particular bijective, for all $t < 1$.
Define
$$G : [a,a + 2 pi] times [0,1] to D^2, G(x,t) = te(H(x,t))$$
which is again continuous. Consider the continuous map
$$p : [a,a + 2 pi] times [0,1] to D^2, p(x,t) = te(x) .$$
Since domain and range are compact, it is an identification map. It is easy to check that if $p(xi) = p(xi')$, then $G(xi) = G(xi')$. Therefore we obtain a unique continuous $g : D^2 to D^2$ such that $g circ p = G$. By construction $g$ maps $Y$ bijectively onto $X$. The restriction $g' : Y to X$ is an open map which shows that it is a homeomorphism: Let $U subset Y$ be open. Then $D^2 backslash U$ is compact, thus $g(D^2 backslash U) ) = D^2 backslash g(U)$ is compact so that $g'(U) = g(U)$ is open in $X$.
answered Jul 24 at 10:52
Paul Frost
3,703420
Let $e : mathbbR to S^1, e(t) = e^it$. Then a homeomorphic image of a closed interval in $S^1$ is any set having the form $J = e([a,b])$ with $a < b$ and $b - a < 2pi$.
We have $X = D^2 backslash e(c) $ for some $c in mathbbR$ and $Y = D^2 backslash e([a,b])$ for suitable $a,b$ as above. Define $r : [a,a + 2 pi] to [c,c+2pi]$, $r(x) = c$ for $t in [a,b]$, $r(x) = c + frac2pi (x - b)a+2pi-b$ for $t in [b,a+2pi]$. This is a continuous map. Define
$$H : [a,a + 2 pi] times [0,1] to [c,c+2pi], H(x,t) = (1-t)(x -a + c) + t r(x) .$$
This is a continuous map such that (with $H_t(x) = H(x,t)$)
(1) $H_1 = r$
(2) $H_t(a) = c, H_t(a+2pi) = c + 2pi$ for all $t$
(3) $H_t$ is strictly increasing, in particular bijective, for all $t < 1$.
Define
$$G : [a,a + 2 pi] times [0,1] to D^2, G(x,t) = te(H(x,t))$$
which is again continuous. Consider the continuous map
$$p : [a,a + 2 pi] times [0,1] to D^2, p(x,t) = te(x) .$$
Since domain and range are compact, it is an identification map. It is easy to check that if $p(xi) = p(xi')$, then $G(xi) = G(xi')$. Therefore we obtain a unique continuous $g : D^2 to D^2$ such that $g circ p = G$. By construction $g$ maps $Y$ bijectively onto $X$. The restriction $g' : Y to X$ is an open map which shows that it is a homeomorphism: Let $U subset Y$ be open. Then $D^2 backslash U$ is compact, thus $g(D^2 backslash U) ) = D^2 backslash g(U)$ is compact so that $g'(U) = g(U)$ is open in $X$.
answered Jul 24 at 10:52
Paul Frost
3,703420
answered Jul 24 at 10:52
Paul Frost
3,703420
answered Jul 24 at 10:52
Paul Frost
3,703420
answered Jul 24 at 10:52
Paul Frost
3,703420
3,703420
add a comment |Â
add a comment |Â
up vote
2
down vote
The answer of @PaulFrost is awesome, and should be the accepted answer. I thought I'd just give an explicit description of his approach.
We'll write the unit disk as
$$ mathbbD = re^pi itmid 0le rle1, -1le tle1$$
And define the lower hemisphere as $H=e^pi itmid -1le tle0$, and the point $q=-1$.
We have a continuous map $h:mathbbDrightarrowmathbbD$, given by
$$ h(re^pi it) = re^t$$
It should be checked this is well-defined (plugging in $t=1$ and $t=-1$). It's also straightforward that $h^-1(q)=H$. Thus we get a map $h:mathbbDsetminus Hrightarrow mathbbDsetminusq$.
An explicit inverse can be given too:
$$ h^-1(rho e^pi itheta) = rho e^(rho+theta)^2/(rho+theta+rho$$
One has to worry about that denominator being zero. The quadratic formula shows that happens when $rho+theta=0$. If $rho<1$, then $-1<theta<0$, and it is easy to check the fraction in the exponent goes to zero. So $h^-1(rho e^-pi irho)=rho$.
When $rho=1$, that's a prpblem, because then we have $theta=pm1$. Luckily we are ignoring that point ($q$), so we get a well-defined map
$$ h^-1:mathbbDsetminus qrightarrowmathbbDsetminus H$$
and so $h$ is the desired homeomorphism.
answered Jul 25 at 0:36
Steve D
2,042419
add a comment |Â
up vote
2
down vote
The answer of @PaulFrost is awesome, and should be the accepted answer. I thought I'd just give an explicit description of his approach.
We'll write the unit disk as
$$ mathbbD = re^pi itmid 0le rle1, -1le tle1$$
And define the lower hemisphere as $H=e^pi itmid -1le tle0$, and the point $q=-1$.
We have a continuous map $h:mathbbDrightarrowmathbbD$, given by
$$ h(re^pi it) = re^t$$
It should be checked this is well-defined (plugging in $t=1$ and $t=-1$). It's also straightforward that $h^-1(q)=H$. Thus we get a map $h:mathbbDsetminus Hrightarrow mathbbDsetminusq$.
An explicit inverse can be given too:
$$ h^-1(rho e^pi itheta) = rho e^(rho+theta)^2/(rho+theta+rho$$
One has to worry about that denominator being zero. The quadratic formula shows that happens when $rho+theta=0$. If $rho<1$, then $-1<theta<0$, and it is easy to check the fraction in the exponent goes to zero. So $h^-1(rho e^-pi irho)=rho$.
When $rho=1$, that's a prpblem, because then we have $theta=pm1$. Luckily we are ignoring that point ($q$), so we get a well-defined map
$$ h^-1:mathbbDsetminus qrightarrowmathbbDsetminus H$$
and so $h$ is the desired homeomorphism.
answered Jul 25 at 0:36
Steve D
2,042419
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The answer of @PaulFrost is awesome, and should be the accepted answer. I thought I'd just give an explicit description of his approach.
We'll write the unit disk as
$$ mathbbD = re^pi itmid 0le rle1, -1le tle1$$
And define the lower hemisphere as $H=e^pi itmid -1le tle0$, and the point $q=-1$.
We have a continuous map $h:mathbbDrightarrowmathbbD$, given by
$$ h(re^pi it) = re^t$$
It should be checked this is well-defined (plugging in $t=1$ and $t=-1$). It's also straightforward that $h^-1(q)=H$. Thus we get a map $h:mathbbDsetminus Hrightarrow mathbbDsetminusq$.
An explicit inverse can be given too:
$$ h^-1(rho e^pi itheta) = rho e^(rho+theta)^2/(rho+theta+rho$$
One has to worry about that denominator being zero. The quadratic formula shows that happens when $rho+theta=0$. If $rho<1$, then $-1<theta<0$, and it is easy to check the fraction in the exponent goes to zero. So $h^-1(rho e^-pi irho)=rho$.
When $rho=1$, that's a prpblem, because then we have $theta=pm1$. Luckily we are ignoring that point ($q$), so we get a well-defined map
$$ h^-1:mathbbDsetminus qrightarrowmathbbDsetminus H$$
and so $h$ is the desired homeomorphism.
answered Jul 25 at 0:36
Steve D
2,042419
The answer of @PaulFrost is awesome, and should be the accepted answer. I thought I'd just give an explicit description of his approach.
We'll write the unit disk as
$$ mathbbD = re^pi itmid 0le rle1, -1le tle1$$
And define the lower hemisphere as $H=e^pi itmid -1le tle0$, and the point $q=-1$.
We have a continuous map $h:mathbbDrightarrowmathbbD$, given by
$$ h(re^pi it) = re^t$$
It should be checked this is well-defined (plugging in $t=1$ and $t=-1$). It's also straightforward that $h^-1(q)=H$. Thus we get a map $h:mathbbDsetminus Hrightarrow mathbbDsetminusq$.
An explicit inverse can be given too:
$$ h^-1(rho e^pi itheta) = rho e^(rho+theta)^2/(rho+theta+rho$$
One has to worry about that denominator being zero. The quadratic formula shows that happens when $rho+theta=0$. If $rho<1$, then $-1<theta<0$, and it is easy to check the fraction in the exponent goes to zero. So $h^-1(rho e^-pi irho)=rho$.
When $rho=1$, that's a prpblem, because then we have $theta=pm1$. Luckily we are ignoring that point ($q$), so we get a well-defined map
$$ h^-1:mathbbDsetminus qrightarrowmathbbDsetminus H$$
and so $h$ is the desired homeomorphism.
answered Jul 25 at 0:36
Steve D
2,042419
answered Jul 25 at 0:36
Steve D
2,042419
answered Jul 25 at 0:36
Steve D
2,042419
answered Jul 25 at 0:36
Steve D
2,042419
2,042419
add a comment |Â
add a comment |Â
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The boundary-punctured disc is homeomorphic to the closed upper half plane, and if you delete an arc, you instead get the upper half plane with $(-infty, t_0) cup (t_1, infty)$ where $t_0 < t_1$. By scaling and translation, you see that the result of deleting any arc from the boundary of $D^2$ is homeomphic to any other. This still leaves the question of point vs arc open, but that is true (I just need to come up with the argument...)
– Mike Miller
Jul 18 at 2:13