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How to show $operatorname{Int}(z)$ is not empty?
For any closed smooth curve $z:[a,b]to Bbb C$, we define the interior $operatorname{Int}(z)$ of $z$ as follows:
Because $operatorname{Im}(z)$ is bounded, we can find a circle $C$ such that $operatorname{Im}(z)$ is contained in the interior of $C$, fix a point $v_0$ outside the circle $C$, $forall win Bbb C-operatorname{Im}(z)$, if there exists a path (a continuous map $[0,1]to Bbb C$) from $w$ to $v_0$ which doesn't intersect $operatorname{Im}(z)$, we say $wnotin operatorname{Int}(z)$, otherwise $win operatorname{Int}(z)$.
$1.$ How to show $operatorname{Int}(z)$ is not empty?
$2.$ How to show $operatorname{Int}(z)$ is open?
general-topology complex-analysis differential-topology geometric-topology low-dimensional-topology
asked Nov 20 '18 at 13:32
Born to be proud
778510
add a comment |
For any closed smooth curve $z:[a,b]to Bbb C$, we define the interior $operatorname{Int}(z)$ of $z$ as follows:
Because $operatorname{Im}(z)$ is bounded, we can find a circle $C$ such that $operatorname{Im}(z)$ is contained in the interior of $C$, fix a point $v_0$ outside the circle $C$, $forall win Bbb C-operatorname{Im}(z)$, if there exists a path (a continuous map $[0,1]to Bbb C$) from $w$ to $v_0$ which doesn't intersect $operatorname{Im}(z)$, we say $wnotin operatorname{Int}(z)$, otherwise $win operatorname{Int}(z)$.
$1.$ How to show $operatorname{Int}(z)$ is not empty?
$2.$ How to show $operatorname{Int}(z)$ is open?
general-topology complex-analysis differential-topology geometric-topology low-dimensional-topology
asked Nov 20 '18 at 13:32
Born to be proud
778510
For 2, notice that $mathrm{Im}(z)$ is compact. So given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
– Federico
Nov 20 '18 at 14:11
For 1. this may be related .
– Crostul
Nov 20 '18 at 14:36
add a comment |
For any closed smooth curve $z:[a,b]to Bbb C$, we define the interior $operatorname{Int}(z)$ of $z$ as follows:
Because $operatorname{Im}(z)$ is bounded, we can find a circle $C$ such that $operatorname{Im}(z)$ is contained in the interior of $C$, fix a point $v_0$ outside the circle $C$, $forall win Bbb C-operatorname{Im}(z)$, if there exists a path (a continuous map $[0,1]to Bbb C$) from $w$ to $v_0$ which doesn't intersect $operatorname{Im}(z)$, we say $wnotin operatorname{Int}(z)$, otherwise $win operatorname{Int}(z)$.
$1.$ How to show $operatorname{Int}(z)$ is not empty?
$2.$ How to show $operatorname{Int}(z)$ is open?
general-topology complex-analysis differential-topology geometric-topology low-dimensional-topology
asked Nov 20 '18 at 13:32
Born to be proud
778510
For any closed smooth curve $z:[a,b]to Bbb C$, we define the interior $operatorname{Int}(z)$ of $z$ as follows:
Because $operatorname{Im}(z)$ is bounded, we can find a circle $C$ such that $operatorname{Im}(z)$ is contained in the interior of $C$, fix a point $v_0$ outside the circle $C$, $forall win Bbb C-operatorname{Im}(z)$, if there exists a path (a continuous map $[0,1]to Bbb C$) from $w$ to $v_0$ which doesn't intersect $operatorname{Im}(z)$, we say $wnotin operatorname{Int}(z)$, otherwise $win operatorname{Int}(z)$.
$1.$ How to show $operatorname{Int}(z)$ is not empty?
$2.$ How to show $operatorname{Int}(z)$ is open?
general-topology complex-analysis differential-topology geometric-topology low-dimensional-topology
general-topology complex-analysis differential-topology geometric-topology low-dimensional-topology
asked Nov 20 '18 at 13:32
Born to be proud
778510
asked Nov 20 '18 at 13:32
Born to be proud
778510
edited Nov 20 '18 at 14:30
asked Nov 20 '18 at 13:32
Born to be proud
778510
asked Nov 20 '18 at 13:32
Born to be proud
778510
asked Nov 20 '18 at 13:32
Born to be proud
778510
778510
For 2, notice that $mathrm{Im}(z)$ is compact. So given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
– Federico
Nov 20 '18 at 14:11
For 1. this may be related .
– Crostul
Nov 20 '18 at 14:36
add a comment |
For 2, notice that $mathrm{Im}(z)$ is compact. So given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
– Federico
Nov 20 '18 at 14:11
For 1. this may be related .
– Crostul
Nov 20 '18 at 14:36
For 2, notice that $mathrm{Im}(z)$ is compact. So given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
– Federico
Nov 20 '18 at 14:11
For 2, notice that $mathrm{Im}(z)$ is compact. So given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
– Federico
Nov 20 '18 at 14:11
For 1. this may be related .
– Crostul
Nov 20 '18 at 14:36
For 1. this may be related .
– Crostul
Nov 20 '18 at 14:36
add a comment |
1 Answer
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For 2, notice that $mathrm{Im}(z)$ is compact. So, given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
answered Nov 20 '18 at 14:15
Federico
4,679514
add a comment |
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For 2, notice that $mathrm{Im}(z)$ is compact. So, given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
answered Nov 20 '18 at 14:15
Federico
4,679514
add a comment |
For 2, notice that $mathrm{Im}(z)$ is compact. So, given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
answered Nov 20 '18 at 14:15
Federico
4,679514
add a comment |
For 2, notice that $mathrm{Im}(z)$ is compact. So, given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
answered Nov 20 '18 at 14:15
Federico
4,679514
For 2, notice that $mathrm{Im}(z)$ is compact. So, given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
answered Nov 20 '18 at 14:15
Federico
4,679514
answered Nov 20 '18 at 14:15
Federico
4,679514
answered Nov 20 '18 at 14:15
Federico
4,679514
answered Nov 20 '18 at 14:15
Federico
4,679514
4,679514
add a comment |
add a comment |
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For 2, notice that $mathrm{Im}(z)$ is compact. So given $wnotinmathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $mathrm{Int}(z)$, or none of them does.
– Federico
Nov 20 '18 at 14:11
For 1. this may be related .
– Crostul
Nov 20 '18 at 14:36