How pathological can the boundary of an open, simply connected subset of $mathbb{C}$ be?












4












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















4












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














4












4








4





$begingroup$


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







share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











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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


















  • $begingroup$
    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










1 Answer
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$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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