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How to prove a region remains connected if one point is excluded?
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I'm in trouble to prove this:
Let $G$ be a region such that $a in G$. Prove that $G backslash a$ is a
region.
MY ATTEMPT: $G backslash a$ is open as it is nothing but $G cap a^c$; intersection of two open sets. I have tried to prove connectedness from contradiction. We assume it is not connected, i.e. there exists two open sets $U$ and $V $ such that $G$ $a$ $=$ $U cup V$ and $U cap V = emptyset$.
Now I claim that there exists an open ball, say $B(a,r)$, contained in either $U cup a$ or in $V cup a$. If my claim is true then $G =U_1cup V_1$, with $U_1 = U cup a$ or $U_1 =U$. Similarly I take $V_1 = Vcup a$ or $V$. If $B(a,r)$ is contained in $V cup a$ i rename it $V_1$ and choose $U_1 = U$ and similarly if $B(a,r) subset U cup p$ , I rename it $U_1$ and take $V_1 =V$.
Clearly $U_1 cap V_1 = emptyset$, which draws a contradiction that $G$ is not connected i.e. not a region.
My problem is HOW TO PROVE MY CLAIM that there exists such an open ball $B(a,r) subset U cup a$. I have thought about an intuitive approch; as $G$ is open there exists an open ball $B(a,r') subset G =Ucup V cup a$. If $B(a,r') subset$ one of $U cup a$ or $V cup a$ , I take $r' = r$ and my claim is proved and I'm happy. If that open ball is not contained in any of those two I assume that there is no radius $r$ for which an open ball containing $a$ is a subset of $U cup a$, so my intution says that if I repeatedly decrease the radius after a certain amount of time that open ball will be contained in $G backslash U = V backslash a$. So again my claim is proved.
But I think that I need to be more rigorous, which I'm not doing here. How can I do it? Or is my intution is wrong? Thanks for reading.
general-topology complex-analysis
edited Jul 27 at 12:23
Aloizio Macedo
22.5k23283
asked Jul 27 at 9:14
Subhajit Saha
239112
add a comment |Â
up vote
0
down vote
favorite
I'm in trouble to prove this:
Let $G$ be a region such that $a in G$. Prove that $G backslash a$ is a
region.
MY ATTEMPT: $G backslash a$ is open as it is nothing but $G cap a^c$; intersection of two open sets. I have tried to prove connectedness from contradiction. We assume it is not connected, i.e. there exists two open sets $U$ and $V $ such that $G$ $a$ $=$ $U cup V$ and $U cap V = emptyset$.
Now I claim that there exists an open ball, say $B(a,r)$, contained in either $U cup a$ or in $V cup a$. If my claim is true then $G =U_1cup V_1$, with $U_1 = U cup a$ or $U_1 =U$. Similarly I take $V_1 = Vcup a$ or $V$. If $B(a,r)$ is contained in $V cup a$ i rename it $V_1$ and choose $U_1 = U$ and similarly if $B(a,r) subset U cup p$ , I rename it $U_1$ and take $V_1 =V$.
Clearly $U_1 cap V_1 = emptyset$, which draws a contradiction that $G$ is not connected i.e. not a region.
My problem is HOW TO PROVE MY CLAIM that there exists such an open ball $B(a,r) subset U cup a$. I have thought about an intuitive approch; as $G$ is open there exists an open ball $B(a,r') subset G =Ucup V cup a$. If $B(a,r') subset$ one of $U cup a$ or $V cup a$ , I take $r' = r$ and my claim is proved and I'm happy. If that open ball is not contained in any of those two I assume that there is no radius $r$ for which an open ball containing $a$ is a subset of $U cup a$, so my intution says that if I repeatedly decrease the radius after a certain amount of time that open ball will be contained in $G backslash U = V backslash a$. So again my claim is proved.
But I think that I need to be more rigorous, which I'm not doing here. How can I do it? Or is my intution is wrong? Thanks for reading.
general-topology complex-analysis
edited Jul 27 at 12:23
Aloizio Macedo
22.5k23283
asked Jul 27 at 9:14
Subhajit Saha
239112
2
I assume by region you mean an open connected set in an euclidean space (therefore it is path-connected). Maybe you can argue directly that surrounding $a$ you can take an open ball $B(a,r)$ inside the region, which remains connected after substracting $a$ (this is easy to show). Since this balls lies in the interior of the region, you could build a path from any point inside the ball to the outside.
– Javi
Jul 27 at 9:24
1
@Javi,thanks I have just made my problem too complicated . It was an easy math if you apply path connectedness.
– Subhajit Saha
Jul 27 at 9:40
1
The proposition is false if the region is one dimensional. Otherwise removing countably points should leave the region connected.
– William Elliot
Jul 27 at 10:32
Your post had several Mathjax issues and some issues with grammar, I edited it in order to address them. To be more specific, in the future try to avoid separating equations ($A=B$ is dollarsignA=Bdollarsign, not dollarsignAdollarsign dollarsign = dollarsign dollarsignBdollarsign), for example. For a more thorough reference, you can see this link.
– Aloizio Macedo
Jul 27 at 12:29
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm in trouble to prove this:
Let $G$ be a region such that $a in G$. Prove that $G backslash a$ is a
region.
MY ATTEMPT: $G backslash a$ is open as it is nothing but $G cap a^c$; intersection of two open sets. I have tried to prove connectedness from contradiction. We assume it is not connected, i.e. there exists two open sets $U$ and $V $ such that $G$ $a$ $=$ $U cup V$ and $U cap V = emptyset$.
Now I claim that there exists an open ball, say $B(a,r)$, contained in either $U cup a$ or in $V cup a$. If my claim is true then $G =U_1cup V_1$, with $U_1 = U cup a$ or $U_1 =U$. Similarly I take $V_1 = Vcup a$ or $V$. If $B(a,r)$ is contained in $V cup a$ i rename it $V_1$ and choose $U_1 = U$ and similarly if $B(a,r) subset U cup p$ , I rename it $U_1$ and take $V_1 =V$.
Clearly $U_1 cap V_1 = emptyset$, which draws a contradiction that $G$ is not connected i.e. not a region.
My problem is HOW TO PROVE MY CLAIM that there exists such an open ball $B(a,r) subset U cup a$. I have thought about an intuitive approch; as $G$ is open there exists an open ball $B(a,r') subset G =Ucup V cup a$. If $B(a,r') subset$ one of $U cup a$ or $V cup a$ , I take $r' = r$ and my claim is proved and I'm happy. If that open ball is not contained in any of those two I assume that there is no radius $r$ for which an open ball containing $a$ is a subset of $U cup a$, so my intution says that if I repeatedly decrease the radius after a certain amount of time that open ball will be contained in $G backslash U = V backslash a$. So again my claim is proved.
But I think that I need to be more rigorous, which I'm not doing here. How can I do it? Or is my intution is wrong? Thanks for reading.
general-topology complex-analysis
edited Jul 27 at 12:23
Aloizio Macedo
22.5k23283
asked Jul 27 at 9:14
Subhajit Saha
239112
I'm in trouble to prove this:
Let $G$ be a region such that $a in G$. Prove that $G backslash a$ is a
region.
MY ATTEMPT: $G backslash a$ is open as it is nothing but $G cap a^c$; intersection of two open sets. I have tried to prove connectedness from contradiction. We assume it is not connected, i.e. there exists two open sets $U$ and $V $ such that $G$ $a$ $=$ $U cup V$ and $U cap V = emptyset$.
Now I claim that there exists an open ball, say $B(a,r)$, contained in either $U cup a$ or in $V cup a$. If my claim is true then $G =U_1cup V_1$, with $U_1 = U cup a$ or $U_1 =U$. Similarly I take $V_1 = Vcup a$ or $V$. If $B(a,r)$ is contained in $V cup a$ i rename it $V_1$ and choose $U_1 = U$ and similarly if $B(a,r) subset U cup p$ , I rename it $U_1$ and take $V_1 =V$.
Clearly $U_1 cap V_1 = emptyset$, which draws a contradiction that $G$ is not connected i.e. not a region.
My problem is HOW TO PROVE MY CLAIM that there exists such an open ball $B(a,r) subset U cup a$. I have thought about an intuitive approch; as $G$ is open there exists an open ball $B(a,r') subset G =Ucup V cup a$. If $B(a,r') subset$ one of $U cup a$ or $V cup a$ , I take $r' = r$ and my claim is proved and I'm happy. If that open ball is not contained in any of those two I assume that there is no radius $r$ for which an open ball containing $a$ is a subset of $U cup a$, so my intution says that if I repeatedly decrease the radius after a certain amount of time that open ball will be contained in $G backslash U = V backslash a$. So again my claim is proved.
But I think that I need to be more rigorous, which I'm not doing here. How can I do it? Or is my intution is wrong? Thanks for reading.
general-topology complex-analysis
edited Jul 27 at 12:23
Aloizio Macedo
22.5k23283
asked Jul 27 at 9:14
Subhajit Saha
239112
edited Jul 27 at 12:23
Aloizio Macedo
22.5k23283
edited Jul 27 at 12:23
Aloizio Macedo
22.5k23283
edited Jul 27 at 12:23
Aloizio Macedo
22.5k23283
22.5k23283
asked Jul 27 at 9:14
Subhajit Saha
239112
asked Jul 27 at 9:14
Subhajit Saha
239112
asked Jul 27 at 9:14
Subhajit Saha
239112
239112
2
I assume by region you mean an open connected set in an euclidean space (therefore it is path-connected). Maybe you can argue directly that surrounding $a$ you can take an open ball $B(a,r)$ inside the region, which remains connected after substracting $a$ (this is easy to show). Since this balls lies in the interior of the region, you could build a path from any point inside the ball to the outside.
– Javi
Jul 27 at 9:24
1
@Javi,thanks I have just made my problem too complicated . It was an easy math if you apply path connectedness.
– Subhajit Saha
Jul 27 at 9:40
1
The proposition is false if the region is one dimensional. Otherwise removing countably points should leave the region connected.
– William Elliot
Jul 27 at 10:32
Your post had several Mathjax issues and some issues with grammar, I edited it in order to address them. To be more specific, in the future try to avoid separating equations ($A=B$ is dollarsignA=Bdollarsign, not dollarsignAdollarsign dollarsign = dollarsign dollarsignBdollarsign), for example. For a more thorough reference, you can see this link.
– Aloizio Macedo
Jul 27 at 12:29
add a comment |Â
2
I assume by region you mean an open connected set in an euclidean space (therefore it is path-connected). Maybe you can argue directly that surrounding $a$ you can take an open ball $B(a,r)$ inside the region, which remains connected after substracting $a$ (this is easy to show). Since this balls lies in the interior of the region, you could build a path from any point inside the ball to the outside.
– Javi
Jul 27 at 9:24
1
@Javi,thanks I have just made my problem too complicated . It was an easy math if you apply path connectedness.
– Subhajit Saha
Jul 27 at 9:40
1
The proposition is false if the region is one dimensional. Otherwise removing countably points should leave the region connected.
– William Elliot
Jul 27 at 10:32
Your post had several Mathjax issues and some issues with grammar, I edited it in order to address them. To be more specific, in the future try to avoid separating equations ($A=B$ is dollarsignA=Bdollarsign, not dollarsignAdollarsign dollarsign = dollarsign dollarsignBdollarsign), for example. For a more thorough reference, you can see this link.
– Aloizio Macedo
Jul 27 at 12:29
2
2
I assume by region you mean an open connected set in an euclidean space (therefore it is path-connected). Maybe you can argue directly that surrounding $a$ you can take an open ball $B(a,r)$ inside the region, which remains connected after substracting $a$ (this is easy to show). Since this balls lies in the interior of the region, you could build a path from any point inside the ball to the outside.
– Javi
Jul 27 at 9:24
I assume by region you mean an open connected set in an euclidean space (therefore it is path-connected). Maybe you can argue directly that surrounding $a$ you can take an open ball $B(a,r)$ inside the region, which remains connected after substracting $a$ (this is easy to show). Since this balls lies in the interior of the region, you could build a path from any point inside the ball to the outside.
– Javi
Jul 27 at 9:24
1
1
@Javi,thanks I have just made my problem too complicated . It was an easy math if you apply path connectedness.
– Subhajit Saha
Jul 27 at 9:40
@Javi,thanks I have just made my problem too complicated . It was an easy math if you apply path connectedness.
– Subhajit Saha
Jul 27 at 9:40
1
1
The proposition is false if the region is one dimensional. Otherwise removing countably points should leave the region connected.
– William Elliot
Jul 27 at 10:32
The proposition is false if the region is one dimensional. Otherwise removing countably points should leave the region connected.
– William Elliot
Jul 27 at 10:32
Your post had several Mathjax issues and some issues with grammar, I edited it in order to address them. To be more specific, in the future try to avoid separating equations ($A=B$ is dollarsignA=Bdollarsign, not dollarsignAdollarsign dollarsign = dollarsign dollarsignBdollarsign), for example. For a more thorough reference, you can see this link.
– Aloizio Macedo
Jul 27 at 12:29
Your post had several Mathjax issues and some issues with grammar, I edited it in order to address them. To be more specific, in the future try to avoid separating equations ($A=B$ is dollarsignA=Bdollarsign, not dollarsignAdollarsign dollarsign = dollarsign dollarsignBdollarsign), for example. For a more thorough reference, you can see this link.
– Aloizio Macedo
Jul 27 at 12:29
add a comment |Â
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2
I assume by region you mean an open connected set in an euclidean space (therefore it is path-connected). Maybe you can argue directly that surrounding $a$ you can take an open ball $B(a,r)$ inside the region, which remains connected after substracting $a$ (this is easy to show). Since this balls lies in the interior of the region, you could build a path from any point inside the ball to the outside.
– Javi
Jul 27 at 9:24
1
@Javi,thanks I have just made my problem too complicated . It was an easy math if you apply path connectedness.
– Subhajit Saha
Jul 27 at 9:40
1
The proposition is false if the region is one dimensional. Otherwise removing countably points should leave the region connected.
– William Elliot
Jul 27 at 10:32
Your post had several Mathjax issues and some issues with grammar, I edited it in order to address them. To be more specific, in the future try to avoid separating equations ($A=B$ is dollarsignA=Bdollarsign, not dollarsignAdollarsign dollarsign = dollarsign dollarsignBdollarsign), for example. For a more thorough reference, you can see this link.
– Aloizio Macedo
Jul 27 at 12:29