How pathological can the boundary of an open, simply connected subset of $mathbb{C}$ be?
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In my complex analysis class we’re currently covering the Riemann Mapping Theorem, and as is well-known there are no conditions imposed on the actual shape of the region $Omega$, except that it must be open, simply connected, and not all of $mathbb{C}$. So I was wondering how pathological such a region can really be, and particularly its boundary $partial Omega$. For example, can the boundary be a non-nullset? No-where differentiable? Can it be that the interior of the closure of $Omega$ is not the same as $Omega$ itself? And so on.
This question may also be extended to $mathbb{R}^n$. However, I feel like “simply connected” is no longer the right analogue here, because $mathbb{R}^3$ with a disjoint set of points removed is simply connected, which is not true of $mathbb{R}^2$. I think to generalize, the question would have to be “How pathological can the boundary of an open, connected subset $Omega subset mathbb{R}^n$ be, given that $mathbb{R}^n cup {infty} setminus Omega$ is also connected?”.
complex-analysis analysis examples-counterexamples
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
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add a comment |
$begingroup$
In my complex analysis class we’re currently covering the Riemann Mapping Theorem, and as is well-known there are no conditions imposed on the actual shape of the region $Omega$, except that it must be open, simply connected, and not all of $mathbb{C}$. So I was wondering how pathological such a region can really be, and particularly its boundary $partial Omega$. For example, can the boundary be a non-nullset? No-where differentiable? Can it be that the interior of the closure of $Omega$ is not the same as $Omega$ itself? And so on.
This question may also be extended to $mathbb{R}^n$. However, I feel like “simply connected” is no longer the right analogue here, because $mathbb{R}^3$ with a disjoint set of points removed is simply connected, which is not true of $mathbb{R}^2$. I think to generalize, the question would have to be “How pathological can the boundary of an open, connected subset $Omega subset mathbb{R}^n$ be, given that $mathbb{R}^n cup {infty} setminus Omega$ is also connected?”.
complex-analysis analysis examples-counterexamples
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
$endgroup$
$begingroup$
Your question whether the boundary can be nowhere differentiable only makes sense if the boundary is a simple closed curve.
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– Paul Frost
Dec 12 '18 at 14:23
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Concerning the Riemann mapping theorem see also math.stackexchange.com/q/3020805.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:29
add a comment |
$begingroup$
In my complex analysis class we’re currently covering the Riemann Mapping Theorem, and as is well-known there are no conditions imposed on the actual shape of the region $Omega$, except that it must be open, simply connected, and not all of $mathbb{C}$. So I was wondering how pathological such a region can really be, and particularly its boundary $partial Omega$. For example, can the boundary be a non-nullset? No-where differentiable? Can it be that the interior of the closure of $Omega$ is not the same as $Omega$ itself? And so on.
This question may also be extended to $mathbb{R}^n$. However, I feel like “simply connected” is no longer the right analogue here, because $mathbb{R}^3$ with a disjoint set of points removed is simply connected, which is not true of $mathbb{R}^2$. I think to generalize, the question would have to be “How pathological can the boundary of an open, connected subset $Omega subset mathbb{R}^n$ be, given that $mathbb{R}^n cup {infty} setminus Omega$ is also connected?”.
complex-analysis analysis examples-counterexamples
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
$endgroup$
In my complex analysis class we’re currently covering the Riemann Mapping Theorem, and as is well-known there are no conditions imposed on the actual shape of the region $Omega$, except that it must be open, simply connected, and not all of $mathbb{C}$. So I was wondering how pathological such a region can really be, and particularly its boundary $partial Omega$. For example, can the boundary be a non-nullset? No-where differentiable? Can it be that the interior of the closure of $Omega$ is not the same as $Omega$ itself? And so on.
This question may also be extended to $mathbb{R}^n$. However, I feel like “simply connected” is no longer the right analogue here, because $mathbb{R}^3$ with a disjoint set of points removed is simply connected, which is not true of $mathbb{R}^2$. I think to generalize, the question would have to be “How pathological can the boundary of an open, connected subset $Omega subset mathbb{R}^n$ be, given that $mathbb{R}^n cup {infty} setminus Omega$ is also connected?”.
complex-analysis analysis examples-counterexamples
complex-analysis analysis examples-counterexamples
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
asked Dec 12 '18 at 12:13
AdarainAdarain
1907
1907
$begingroup$
Your question whether the boundary can be nowhere differentiable only makes sense if the boundary is a simple closed curve.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:23
$begingroup$
Concerning the Riemann mapping theorem see also math.stackexchange.com/q/3020805.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:29
add a comment |
$begingroup$
Your question whether the boundary can be nowhere differentiable only makes sense if the boundary is a simple closed curve.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:23
$begingroup$
Concerning the Riemann mapping theorem see also math.stackexchange.com/q/3020805.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:29
$begingroup$
Your question whether the boundary can be nowhere differentiable only makes sense if the boundary is a simple closed curve.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:23
$begingroup$
Your question whether the boundary can be nowhere differentiable only makes sense if the boundary is a simple closed curve.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:23
$begingroup$
Concerning the Riemann mapping theorem see also math.stackexchange.com/q/3020805.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:29
$begingroup$
Concerning the Riemann mapping theorem see also math.stackexchange.com/q/3020805.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:29
add a comment |
1 Answer
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active
oldest
votes
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Let $E subset [0,1]$ be a "fat Cantor set": closed, nowhere dense, with positive Lebesgue measure. Then $((-1,1) times (-1,2)) backslash ([0,1] times E) $ is open in $mathbb R^2$ and simply connected, and its boundary contains
$[0,1] times E$ which has positive two-dimensional Lebesgue measure.
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
Robert Israel's answer also provides an example that the interior of the closure of $Omega$ is not the same as $Omega$. A more elementary example is $((-1,1) times (-1,1)) setminus ([0,1] times { 0 })$. See also math.stackexchange.com/q/3025281.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $E subset [0,1]$ be a "fat Cantor set": closed, nowhere dense, with positive Lebesgue measure. Then $((-1,1) times (-1,2)) backslash ([0,1] times E) $ is open in $mathbb R^2$ and simply connected, and its boundary contains
$[0,1] times E$ which has positive two-dimensional Lebesgue measure.
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
Robert Israel's answer also provides an example that the interior of the closure of $Omega$ is not the same as $Omega$. A more elementary example is $((-1,1) times (-1,1)) setminus ([0,1] times { 0 })$. See also math.stackexchange.com/q/3025281.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:26
add a comment |
$begingroup$
Let $E subset [0,1]$ be a "fat Cantor set": closed, nowhere dense, with positive Lebesgue measure. Then $((-1,1) times (-1,2)) backslash ([0,1] times E) $ is open in $mathbb R^2$ and simply connected, and its boundary contains
$[0,1] times E$ which has positive two-dimensional Lebesgue measure.
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
Robert Israel's answer also provides an example that the interior of the closure of $Omega$ is not the same as $Omega$. A more elementary example is $((-1,1) times (-1,1)) setminus ([0,1] times { 0 })$. See also math.stackexchange.com/q/3025281.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:26
add a comment |
$begingroup$
Let $E subset [0,1]$ be a "fat Cantor set": closed, nowhere dense, with positive Lebesgue measure. Then $((-1,1) times (-1,2)) backslash ([0,1] times E) $ is open in $mathbb R^2$ and simply connected, and its boundary contains
$[0,1] times E$ which has positive two-dimensional Lebesgue measure.
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
$endgroup$
Let $E subset [0,1]$ be a "fat Cantor set": closed, nowhere dense, with positive Lebesgue measure. Then $((-1,1) times (-1,2)) backslash ([0,1] times E) $ is open in $mathbb R^2$ and simply connected, and its boundary contains
$[0,1] times E$ which has positive two-dimensional Lebesgue measure.
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
answered Dec 12 '18 at 12:57
Robert IsraelRobert Israel
324k23214468
324k23214468
$begingroup$
Robert Israel's answer also provides an example that the interior of the closure of $Omega$ is not the same as $Omega$. A more elementary example is $((-1,1) times (-1,1)) setminus ([0,1] times { 0 })$. See also math.stackexchange.com/q/3025281.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:26
add a comment |
$begingroup$
Robert Israel's answer also provides an example that the interior of the closure of $Omega$ is not the same as $Omega$. A more elementary example is $((-1,1) times (-1,1)) setminus ([0,1] times { 0 })$. See also math.stackexchange.com/q/3025281.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:26
$begingroup$
Robert Israel's answer also provides an example that the interior of the closure of $Omega$ is not the same as $Omega$. A more elementary example is $((-1,1) times (-1,1)) setminus ([0,1] times { 0 })$. See also math.stackexchange.com/q/3025281.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:26
$begingroup$
Robert Israel's answer also provides an example that the interior of the closure of $Omega$ is not the same as $Omega$. A more elementary example is $((-1,1) times (-1,1)) setminus ([0,1] times { 0 })$. See also math.stackexchange.com/q/3025281.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:26
add a comment |
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$begingroup$
Your question whether the boundary can be nowhere differentiable only makes sense if the boundary is a simple closed curve.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:23
$begingroup$
Concerning the Riemann mapping theorem see also math.stackexchange.com/q/3020805.
$endgroup$
– Paul Frost
Dec 12 '18 at 14:29