Connectedness of sets in the plane with at least one coordinate rational [closed]
$begingroup$
Let $A$ be the space :
$A = {(x,y)in R^2:x in Qtext{ or }y in Q}$
equipped with the subspace topology for the standard metric topology on $R^2$.
$(a)$ Show that $A$ is connected.
$(b)$ Show that the intersection of $A$ and $S^1$ is not connected.
general-topology
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
$endgroup$
closed as off-topic by mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039 Dec 19 '18 at 5:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
Let $A$ be the space :
$A = {(x,y)in R^2:x in Qtext{ or }y in Q}$
equipped with the subspace topology for the standard metric topology on $R^2$.
$(a)$ Show that $A$ is connected.
$(b)$ Show that the intersection of $A$ and $S^1$ is not connected.
general-topology
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
$endgroup$
closed as off-topic by mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039 Dec 19 '18 at 5:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $A$ be the space :
$A = {(x,y)in R^2:x in Qtext{ or }y in Q}$
equipped with the subspace topology for the standard metric topology on $R^2$.
$(a)$ Show that $A$ is connected.
$(b)$ Show that the intersection of $A$ and $S^1$ is not connected.
general-topology
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
$endgroup$
Let $A$ be the space :
$A = {(x,y)in R^2:x in Qtext{ or }y in Q}$
equipped with the subspace topology for the standard metric topology on $R^2$.
$(a)$ Show that $A$ is connected.
$(b)$ Show that the intersection of $A$ and $S^1$ is not connected.
general-topology
general-topology
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
edited Dec 18 '18 at 21:08
mrtaurho
5,51551439
5,51551439
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
asked Dec 18 '18 at 20:52
John TipotasJohn Tipotas
11
11
closed as off-topic by mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039 Dec 19 '18 at 5:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039 Dec 19 '18 at 5:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, metamorphy, Tianlalu, Brahadeesh, user 170039
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $f:Arightarrow {0,1}$ be a continuous function such that $f(0,0)=0$, since $rtimes mathbb{R}$ is connected, for $rinmathbb{Q}$, $f(rtimes mathbb{R})=f(r,0)$. You also have $f(mathbb{R}times {0})=f(0,0)$. This implies that $f(rtimesmathbb{R})=f(r,0)=f(0,0)$. A similar argument shows that $f(mathbb{R}times{r})=f(0,0)$. This implies that $f$ is constant and $A$ is connected.
To show that $Acap S^1$ is connected, consider the points $p=(cos(pi/4),(sin(pi/4)); q=(-cos(pi/4),sin(pi/4))in mathbb{R}^2-Acap S^1$ and the line $l$ throught $p$ and $q$, $mathbb{R}^2-l$ has two connected components $U,V$ and $Ucap Acap S^1$ and $Vcap Acap S^1$ are not empty. Since $(0,-1)$ and $(0,1)$ are elements of a distinct connected component.
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f:Arightarrow {0,1}$ be a continuous function such that $f(0,0)=0$, since $rtimes mathbb{R}$ is connected, for $rinmathbb{Q}$, $f(rtimes mathbb{R})=f(r,0)$. You also have $f(mathbb{R}times {0})=f(0,0)$. This implies that $f(rtimesmathbb{R})=f(r,0)=f(0,0)$. A similar argument shows that $f(mathbb{R}times{r})=f(0,0)$. This implies that $f$ is constant and $A$ is connected.
To show that $Acap S^1$ is connected, consider the points $p=(cos(pi/4),(sin(pi/4)); q=(-cos(pi/4),sin(pi/4))in mathbb{R}^2-Acap S^1$ and the line $l$ throught $p$ and $q$, $mathbb{R}^2-l$ has two connected components $U,V$ and $Ucap Acap S^1$ and $Vcap Acap S^1$ are not empty. Since $(0,-1)$ and $(0,1)$ are elements of a distinct connected component.
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
$endgroup$
add a comment |
$begingroup$
Let $f:Arightarrow {0,1}$ be a continuous function such that $f(0,0)=0$, since $rtimes mathbb{R}$ is connected, for $rinmathbb{Q}$, $f(rtimes mathbb{R})=f(r,0)$. You also have $f(mathbb{R}times {0})=f(0,0)$. This implies that $f(rtimesmathbb{R})=f(r,0)=f(0,0)$. A similar argument shows that $f(mathbb{R}times{r})=f(0,0)$. This implies that $f$ is constant and $A$ is connected.
To show that $Acap S^1$ is connected, consider the points $p=(cos(pi/4),(sin(pi/4)); q=(-cos(pi/4),sin(pi/4))in mathbb{R}^2-Acap S^1$ and the line $l$ throught $p$ and $q$, $mathbb{R}^2-l$ has two connected components $U,V$ and $Ucap Acap S^1$ and $Vcap Acap S^1$ are not empty. Since $(0,-1)$ and $(0,1)$ are elements of a distinct connected component.
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
$endgroup$
add a comment |
$begingroup$
Let $f:Arightarrow {0,1}$ be a continuous function such that $f(0,0)=0$, since $rtimes mathbb{R}$ is connected, for $rinmathbb{Q}$, $f(rtimes mathbb{R})=f(r,0)$. You also have $f(mathbb{R}times {0})=f(0,0)$. This implies that $f(rtimesmathbb{R})=f(r,0)=f(0,0)$. A similar argument shows that $f(mathbb{R}times{r})=f(0,0)$. This implies that $f$ is constant and $A$ is connected.
To show that $Acap S^1$ is connected, consider the points $p=(cos(pi/4),(sin(pi/4)); q=(-cos(pi/4),sin(pi/4))in mathbb{R}^2-Acap S^1$ and the line $l$ throught $p$ and $q$, $mathbb{R}^2-l$ has two connected components $U,V$ and $Ucap Acap S^1$ and $Vcap Acap S^1$ are not empty. Since $(0,-1)$ and $(0,1)$ are elements of a distinct connected component.
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
$endgroup$
Let $f:Arightarrow {0,1}$ be a continuous function such that $f(0,0)=0$, since $rtimes mathbb{R}$ is connected, for $rinmathbb{Q}$, $f(rtimes mathbb{R})=f(r,0)$. You also have $f(mathbb{R}times {0})=f(0,0)$. This implies that $f(rtimesmathbb{R})=f(r,0)=f(0,0)$. A similar argument shows that $f(mathbb{R}times{r})=f(0,0)$. This implies that $f$ is constant and $A$ is connected.
To show that $Acap S^1$ is connected, consider the points $p=(cos(pi/4),(sin(pi/4)); q=(-cos(pi/4),sin(pi/4))in mathbb{R}^2-Acap S^1$ and the line $l$ throught $p$ and $q$, $mathbb{R}^2-l$ has two connected components $U,V$ and $Ucap Acap S^1$ and $Vcap Acap S^1$ are not empty. Since $(0,-1)$ and $(0,1)$ are elements of a distinct connected component.
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
edited Dec 18 '18 at 21:09
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
answered Dec 18 '18 at 21:02
Tsemo AristideTsemo Aristide
58.5k11445
58.5k11445
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