Is an open connected set “grid connected?”
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Let $S subseteq mathbb C$ be an open connected set. This implies that it the set is polygon connected. That means, given $a,bin S$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ which are all contained in $S$.
But is it "grid connected?" (I'm inventing this terminology on the spot, so bear with me.) I define a set $Esubseteq mathbb C$ to be "grid connected" if, given points $a,bin E$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ each of which is either horizontal or vertical.
For example, the plane $mathbb C$ is "grid connected" because, given distinct points $a+bi$ and $c+di$, we can connect them with the line segments $[a+bi, c+bi]$ and $[c+bi,c+di]$. (i.e. with a horizontal line segment and a vertical line segment.
general-topology complex-analysis
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
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add a comment |
$begingroup$
Let $S subseteq mathbb C$ be an open connected set. This implies that it the set is polygon connected. That means, given $a,bin S$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ which are all contained in $S$.
But is it "grid connected?" (I'm inventing this terminology on the spot, so bear with me.) I define a set $Esubseteq mathbb C$ to be "grid connected" if, given points $a,bin E$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ each of which is either horizontal or vertical.
For example, the plane $mathbb C$ is "grid connected" because, given distinct points $a+bi$ and $c+di$, we can connect them with the line segments $[a+bi, c+bi]$ and $[c+bi,c+di]$. (i.e. with a horizontal line segment and a vertical line segment.
general-topology complex-analysis
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
$endgroup$
$begingroup$
What proof do you know of the fact that open connected sets on the plane are "polygon connected"? It will probably be adaptable to show that they are "grid connected" as well. (For instance, the proof by defining an equivalence relation given by $x sim y$ if $x$ can reach $y$ applies mutatis mutandis.)
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– Aloizio Macedo♦
Jan 24 at 6:13
add a comment |
$begingroup$
Let $S subseteq mathbb C$ be an open connected set. This implies that it the set is polygon connected. That means, given $a,bin S$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ which are all contained in $S$.
But is it "grid connected?" (I'm inventing this terminology on the spot, so bear with me.) I define a set $Esubseteq mathbb C$ to be "grid connected" if, given points $a,bin E$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ each of which is either horizontal or vertical.
For example, the plane $mathbb C$ is "grid connected" because, given distinct points $a+bi$ and $c+di$, we can connect them with the line segments $[a+bi, c+bi]$ and $[c+bi,c+di]$. (i.e. with a horizontal line segment and a vertical line segment.
general-topology complex-analysis
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
$endgroup$
Let $S subseteq mathbb C$ be an open connected set. This implies that it the set is polygon connected. That means, given $a,bin S$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ which are all contained in $S$.
But is it "grid connected?" (I'm inventing this terminology on the spot, so bear with me.) I define a set $Esubseteq mathbb C$ to be "grid connected" if, given points $a,bin E$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ each of which is either horizontal or vertical.
For example, the plane $mathbb C$ is "grid connected" because, given distinct points $a+bi$ and $c+di$, we can connect them with the line segments $[a+bi, c+bi]$ and $[c+bi,c+di]$. (i.e. with a horizontal line segment and a vertical line segment.
general-topology complex-analysis
general-topology complex-analysis
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
edited Jan 24 at 3:51
Pascal's Wager
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
asked Jan 24 at 3:45
Pascal's WagerPascal's Wager
371315
371315
$begingroup$
What proof do you know of the fact that open connected sets on the plane are "polygon connected"? It will probably be adaptable to show that they are "grid connected" as well. (For instance, the proof by defining an equivalence relation given by $x sim y$ if $x$ can reach $y$ applies mutatis mutandis.)
$endgroup$
– Aloizio Macedo♦
Jan 24 at 6:13
add a comment |
$begingroup$
What proof do you know of the fact that open connected sets on the plane are "polygon connected"? It will probably be adaptable to show that they are "grid connected" as well. (For instance, the proof by defining an equivalence relation given by $x sim y$ if $x$ can reach $y$ applies mutatis mutandis.)
$endgroup$
– Aloizio Macedo♦
Jan 24 at 6:13
$begingroup$
What proof do you know of the fact that open connected sets on the plane are "polygon connected"? It will probably be adaptable to show that they are "grid connected" as well. (For instance, the proof by defining an equivalence relation given by $x sim y$ if $x$ can reach $y$ applies mutatis mutandis.)
$endgroup$
– Aloizio Macedo♦
Jan 24 at 6:13
$begingroup$
What proof do you know of the fact that open connected sets on the plane are "polygon connected"? It will probably be adaptable to show that they are "grid connected" as well. (For instance, the proof by defining an equivalence relation given by $x sim y$ if $x$ can reach $y$ applies mutatis mutandis.)
$endgroup$
– Aloizio Macedo♦
Jan 24 at 6:13
add a comment |
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Since you agree that we can connect using line segments, it suffices to prove that for each segment $[z,w]$ there are a finite number of vertical or horizontal segments $[a_1,a_2]$, $[a_2,a_3]$, ... $[a_{k-1}, a_k]$, contained in the original set $S$ such that $a_1 = z$ and $a_k = w$.
Since $zin S$ and $S$ is open, there is an open neighborhood of $z$ contained in $S$. There we can connect with horizontal and vertical lines as shown in the picture, and, importantly, we're still in $S$.
Finally, since $[z,w]$ is compact, only finitely many of those neighborhoods are needed.
answered Jan 24 at 4:09
DosidisDosidis
41129
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$begingroup$
Since you agree that we can connect using line segments, it suffices to prove that for each segment $[z,w]$ there are a finite number of vertical or horizontal segments $[a_1,a_2]$, $[a_2,a_3]$, ... $[a_{k-1}, a_k]$, contained in the original set $S$ such that $a_1 = z$ and $a_k = w$.
Since $zin S$ and $S$ is open, there is an open neighborhood of $z$ contained in $S$. There we can connect with horizontal and vertical lines as shown in the picture, and, importantly, we're still in $S$.
Finally, since $[z,w]$ is compact, only finitely many of those neighborhoods are needed.
answered Jan 24 at 4:09
DosidisDosidis
41129
$endgroup$
add a comment |
$begingroup$
Since you agree that we can connect using line segments, it suffices to prove that for each segment $[z,w]$ there are a finite number of vertical or horizontal segments $[a_1,a_2]$, $[a_2,a_3]$, ... $[a_{k-1}, a_k]$, contained in the original set $S$ such that $a_1 = z$ and $a_k = w$.
Since $zin S$ and $S$ is open, there is an open neighborhood of $z$ contained in $S$. There we can connect with horizontal and vertical lines as shown in the picture, and, importantly, we're still in $S$.
Finally, since $[z,w]$ is compact, only finitely many of those neighborhoods are needed.
answered Jan 24 at 4:09
DosidisDosidis
41129
$endgroup$
add a comment |
$begingroup$
Since you agree that we can connect using line segments, it suffices to prove that for each segment $[z,w]$ there are a finite number of vertical or horizontal segments $[a_1,a_2]$, $[a_2,a_3]$, ... $[a_{k-1}, a_k]$, contained in the original set $S$ such that $a_1 = z$ and $a_k = w$.
Since $zin S$ and $S$ is open, there is an open neighborhood of $z$ contained in $S$. There we can connect with horizontal and vertical lines as shown in the picture, and, importantly, we're still in $S$.
Finally, since $[z,w]$ is compact, only finitely many of those neighborhoods are needed.
answered Jan 24 at 4:09
DosidisDosidis
41129
$endgroup$
Since you agree that we can connect using line segments, it suffices to prove that for each segment $[z,w]$ there are a finite number of vertical or horizontal segments $[a_1,a_2]$, $[a_2,a_3]$, ... $[a_{k-1}, a_k]$, contained in the original set $S$ such that $a_1 = z$ and $a_k = w$.
Since $zin S$ and $S$ is open, there is an open neighborhood of $z$ contained in $S$. There we can connect with horizontal and vertical lines as shown in the picture, and, importantly, we're still in $S$.
Finally, since $[z,w]$ is compact, only finitely many of those neighborhoods are needed.
answered Jan 24 at 4:09
DosidisDosidis
41129
answered Jan 24 at 4:09
DosidisDosidis
41129
answered Jan 24 at 4:09
DosidisDosidis
41129
answered Jan 24 at 4:09
DosidisDosidis
41129
41129
add a comment |
add a comment |
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$begingroup$
What proof do you know of the fact that open connected sets on the plane are "polygon connected"? It will probably be adaptable to show that they are "grid connected" as well. (For instance, the proof by defining an equivalence relation given by $x sim y$ if $x$ can reach $y$ applies mutatis mutandis.)
$endgroup$
– Aloizio Macedo♦
Jan 24 at 6:13