Is the Irrational Plane Path-Connected? [duplicate]If a topological space is a path-connected, it is also connected. The converse fails. Can this be remedied?Path homotopy in the planeQuestion on separations of path-connected spaces.There is no homeomorphic copy of $[0,1]$ in the plane which contains an open ballShowing that arcs do not separate the plane $mathbbR^2$Are the sets connected or path-connected?If $A subseteq mathbbR^n$ is path-connected, with path-connected complement, is $A$ necessarily simply-connected?Determine whether the following spaces are connected, totally disconnected or neitherPathological Continua which are Path Connected and Locally Path Connected.Is the product of path connected spaces also path connected in a topology other than the product topology?
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Is the Irrational Plane Path-Connected? [duplicate]
If a topological space is a path-connected, it is also connected. The converse fails. Can this be remedied?Path homotopy in the planeQuestion on separations of path-connected spaces.There is no homeomorphic copy of $[0,1]$ in the plane which contains an open ballShowing that arcs do not separate the plane $mathbbR^2$Are the sets connected or path-connected?If $A subseteq mathbbR^n$ is path-connected, with path-connected complement, is $A$ necessarily simply-connected?Determine whether the following spaces are connected, totally disconnected or neitherPathological Continua which are Path Connected and Locally Path Connected.Is the product of path connected spaces also path connected in a topology other than the product topology?
$begingroup$
This question already has an answer here:
Is the Cartesian square of the set of irrational numbers path connected?
1 answer
Is the set $I times I subset mathbbR^2$, where $I = mathbbRsetminus mathbbQ$ path connected?
It seems like it could be. Consider the path $gamma : (0,1) to mathbbR times mathbbR $ given by $gamma(t) = left(t, (1 - t^3)^frac13right)$. Fermat's last theorem guarantees $gamma((0,1)) subset I times I$. Thus, $gamma$ is a path of non-zero length contained in $I times I$. Perhaps paths like $gamma$ could be glued together to connect any two points of $I times I$?
Edit
As has been pointed out below, it is not the case that $gamma((0,1)) subset I times I$. Clearly $gamma(q) in mathbbQ times mathbbR$ when $q in mathbbQ$. Fermat's theorem only guarantees that $gamma((0,1)) subset mathbbR times I$. I had in mind the arc given implicitly by the equation $x^3 + y^3 = 1$ for $x,y in (0,1)$. While this arc will be a subset of $I times I$, it is in fact totally disconnected.
general-topology analysis
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
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marked as duplicate by Asaf Karagila♦ general-topology
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2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is the Cartesian square of the set of irrational numbers path connected?
1 answer
Is the set $I times I subset mathbbR^2$, where $I = mathbbRsetminus mathbbQ$ path connected?
It seems like it could be. Consider the path $gamma : (0,1) to mathbbR times mathbbR $ given by $gamma(t) = left(t, (1 - t^3)^frac13right)$. Fermat's last theorem guarantees $gamma((0,1)) subset I times I$. Thus, $gamma$ is a path of non-zero length contained in $I times I$. Perhaps paths like $gamma$ could be glued together to connect any two points of $I times I$?
Edit
As has been pointed out below, it is not the case that $gamma((0,1)) subset I times I$. Clearly $gamma(q) in mathbbQ times mathbbR$ when $q in mathbbQ$. Fermat's theorem only guarantees that $gamma((0,1)) subset mathbbR times I$. I had in mind the arc given implicitly by the equation $x^3 + y^3 = 1$ for $x,y in (0,1)$. While this arc will be a subset of $I times I$, it is in fact totally disconnected.
general-topology analysis
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
$endgroup$
marked as duplicate by Asaf Karagila♦ general-topology
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2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Fermat's last theorem is a nice thought, but unfortunately $gamma((0,1)) nsubseteq I times I$, because $I times I$ is missing all points with rational $x$-coordinate, for instance.
$endgroup$
– Jane Doé
3 hours ago
$begingroup$
If $pi_1,pi_2$ are the projections onto the first and second coordinates, then $alpha_1=pi_1circ gamma$ and $alpha_2=pi_2circgamma$ are continuous. Since $(0,1)$ is connected then its image by $alpha_1$ and $alpha_2$ are connected. This mean that they contain only one point.
$endgroup$
– logarithm
3 hours ago
$begingroup$
I see now that the use of Fermat's last theorem was misguided. The arc composed of points $x, y in (0,1)$ such that $x^3 + y^3 = 1$ certainly lives in $I times I$, but is not connected. Meanwhile the arc I described contains points with rational coordinates.
$endgroup$
– Charles Hudgins
3 hours ago
add a comment |
$begingroup$
This question already has an answer here:
Is the Cartesian square of the set of irrational numbers path connected?
1 answer
Is the set $I times I subset mathbbR^2$, where $I = mathbbRsetminus mathbbQ$ path connected?
It seems like it could be. Consider the path $gamma : (0,1) to mathbbR times mathbbR $ given by $gamma(t) = left(t, (1 - t^3)^frac13right)$. Fermat's last theorem guarantees $gamma((0,1)) subset I times I$. Thus, $gamma$ is a path of non-zero length contained in $I times I$. Perhaps paths like $gamma$ could be glued together to connect any two points of $I times I$?
Edit
As has been pointed out below, it is not the case that $gamma((0,1)) subset I times I$. Clearly $gamma(q) in mathbbQ times mathbbR$ when $q in mathbbQ$. Fermat's theorem only guarantees that $gamma((0,1)) subset mathbbR times I$. I had in mind the arc given implicitly by the equation $x^3 + y^3 = 1$ for $x,y in (0,1)$. While this arc will be a subset of $I times I$, it is in fact totally disconnected.
general-topology analysis
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
$endgroup$
This question already has an answer here:
Is the Cartesian square of the set of irrational numbers path connected?
1 answer
Is the set $I times I subset mathbbR^2$, where $I = mathbbRsetminus mathbbQ$ path connected?
It seems like it could be. Consider the path $gamma : (0,1) to mathbbR times mathbbR $ given by $gamma(t) = left(t, (1 - t^3)^frac13right)$. Fermat's last theorem guarantees $gamma((0,1)) subset I times I$. Thus, $gamma$ is a path of non-zero length contained in $I times I$. Perhaps paths like $gamma$ could be glued together to connect any two points of $I times I$?
Edit
As has been pointed out below, it is not the case that $gamma((0,1)) subset I times I$. Clearly $gamma(q) in mathbbQ times mathbbR$ when $q in mathbbQ$. Fermat's theorem only guarantees that $gamma((0,1)) subset mathbbR times I$. I had in mind the arc given implicitly by the equation $x^3 + y^3 = 1$ for $x,y in (0,1)$. While this arc will be a subset of $I times I$, it is in fact totally disconnected.
This question already has an answer here:
Is the Cartesian square of the set of irrational numbers path connected?
1 answer
general-topology analysis
general-topology analysis
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
edited 3 hours ago
Charles Hudgins
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
asked 3 hours ago
Charles HudginsCharles Hudgins
4357
4357
marked as duplicate by Asaf Karagila♦ general-topology
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marked as duplicate by Asaf Karagila♦ general-topology
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2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Fermat's last theorem is a nice thought, but unfortunately $gamma((0,1)) nsubseteq I times I$, because $I times I$ is missing all points with rational $x$-coordinate, for instance.
$endgroup$
– Jane Doé
3 hours ago
$begingroup$
If $pi_1,pi_2$ are the projections onto the first and second coordinates, then $alpha_1=pi_1circ gamma$ and $alpha_2=pi_2circgamma$ are continuous. Since $(0,1)$ is connected then its image by $alpha_1$ and $alpha_2$ are connected. This mean that they contain only one point.
$endgroup$
– logarithm
3 hours ago
$begingroup$
I see now that the use of Fermat's last theorem was misguided. The arc composed of points $x, y in (0,1)$ such that $x^3 + y^3 = 1$ certainly lives in $I times I$, but is not connected. Meanwhile the arc I described contains points with rational coordinates.
$endgroup$
– Charles Hudgins
3 hours ago
add a comment |
1
$begingroup$
Fermat's last theorem is a nice thought, but unfortunately $gamma((0,1)) nsubseteq I times I$, because $I times I$ is missing all points with rational $x$-coordinate, for instance.
$endgroup$
– Jane Doé
3 hours ago
$begingroup$
If $pi_1,pi_2$ are the projections onto the first and second coordinates, then $alpha_1=pi_1circ gamma$ and $alpha_2=pi_2circgamma$ are continuous. Since $(0,1)$ is connected then its image by $alpha_1$ and $alpha_2$ are connected. This mean that they contain only one point.
$endgroup$
– logarithm
3 hours ago
$begingroup$
I see now that the use of Fermat's last theorem was misguided. The arc composed of points $x, y in (0,1)$ such that $x^3 + y^3 = 1$ certainly lives in $I times I$, but is not connected. Meanwhile the arc I described contains points with rational coordinates.
$endgroup$
– Charles Hudgins
3 hours ago
1
1
$begingroup$
Fermat's last theorem is a nice thought, but unfortunately $gamma((0,1)) nsubseteq I times I$, because $I times I$ is missing all points with rational $x$-coordinate, for instance.
$endgroup$
– Jane Doé
3 hours ago
$begingroup$
Fermat's last theorem is a nice thought, but unfortunately $gamma((0,1)) nsubseteq I times I$, because $I times I$ is missing all points with rational $x$-coordinate, for instance.
$endgroup$
– Jane Doé
3 hours ago
$begingroup$
If $pi_1,pi_2$ are the projections onto the first and second coordinates, then $alpha_1=pi_1circ gamma$ and $alpha_2=pi_2circgamma$ are continuous. Since $(0,1)$ is connected then its image by $alpha_1$ and $alpha_2$ are connected. This mean that they contain only one point.
$endgroup$
– logarithm
3 hours ago
$begingroup$
If $pi_1,pi_2$ are the projections onto the first and second coordinates, then $alpha_1=pi_1circ gamma$ and $alpha_2=pi_2circgamma$ are continuous. Since $(0,1)$ is connected then its image by $alpha_1$ and $alpha_2$ are connected. This mean that they contain only one point.
$endgroup$
– logarithm
3 hours ago
$begingroup$
I see now that the use of Fermat's last theorem was misguided. The arc composed of points $x, y in (0,1)$ such that $x^3 + y^3 = 1$ certainly lives in $I times I$, but is not connected. Meanwhile the arc I described contains points with rational coordinates.
$endgroup$
– Charles Hudgins
3 hours ago
$begingroup$
I see now that the use of Fermat's last theorem was misguided. The arc composed of points $x, y in (0,1)$ such that $x^3 + y^3 = 1$ certainly lives in $I times I$, but is not connected. Meanwhile the arc I described contains points with rational coordinates.
$endgroup$
– Charles Hudgins
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$Itimes I$ is not path connected (or connected, for that matter), because the projection $pi(x,y)=x$ is a continuous map and $pi[Itimes I]$ is the disconnected set $I$.
In fact, in your candidate path, $gammaleft(frac12right)=left(frac12,fracsqrt[3]72right)notin Itimes I$.
On the other hand, $(Bbb RtimesBbb R)setminus(Bbb Qtimes Bbb Q)$ is path-connected, just like all complements of countable subsets of $Bbb R^2$ are.
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
$endgroup$
add a comment |
$begingroup$
No. It's not even connected. Consider the map$$beginarrayrcccfcolon&Itimes I&longrightarrow&0,1\&(x,y)&mapsto&begincases1&text if y>frac12\0&text if y<frac12.endcasesendarray$$Then $f$ is continuous and $f(Itimes I)=0,1$, which is disconnected.
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$Itimes I$ is not path connected (or connected, for that matter), because the projection $pi(x,y)=x$ is a continuous map and $pi[Itimes I]$ is the disconnected set $I$.
In fact, in your candidate path, $gammaleft(frac12right)=left(frac12,fracsqrt[3]72right)notin Itimes I$.
On the other hand, $(Bbb RtimesBbb R)setminus(Bbb Qtimes Bbb Q)$ is path-connected, just like all complements of countable subsets of $Bbb R^2$ are.
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
$endgroup$
add a comment |
$begingroup$
$Itimes I$ is not path connected (or connected, for that matter), because the projection $pi(x,y)=x$ is a continuous map and $pi[Itimes I]$ is the disconnected set $I$.
In fact, in your candidate path, $gammaleft(frac12right)=left(frac12,fracsqrt[3]72right)notin Itimes I$.
On the other hand, $(Bbb RtimesBbb R)setminus(Bbb Qtimes Bbb Q)$ is path-connected, just like all complements of countable subsets of $Bbb R^2$ are.
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
$endgroup$
add a comment |
$begingroup$
$Itimes I$ is not path connected (or connected, for that matter), because the projection $pi(x,y)=x$ is a continuous map and $pi[Itimes I]$ is the disconnected set $I$.
In fact, in your candidate path, $gammaleft(frac12right)=left(frac12,fracsqrt[3]72right)notin Itimes I$.
On the other hand, $(Bbb RtimesBbb R)setminus(Bbb Qtimes Bbb Q)$ is path-connected, just like all complements of countable subsets of $Bbb R^2$ are.
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
$endgroup$
$Itimes I$ is not path connected (or connected, for that matter), because the projection $pi(x,y)=x$ is a continuous map and $pi[Itimes I]$ is the disconnected set $I$.
In fact, in your candidate path, $gammaleft(frac12right)=left(frac12,fracsqrt[3]72right)notin Itimes I$.
On the other hand, $(Bbb RtimesBbb R)setminus(Bbb Qtimes Bbb Q)$ is path-connected, just like all complements of countable subsets of $Bbb R^2$ are.
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
edited 2 hours ago
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
answered 3 hours ago
Saucy O'PathSaucy O'Path
7,3891827
7,3891827
add a comment |
add a comment |
$begingroup$
No. It's not even connected. Consider the map$$beginarrayrcccfcolon&Itimes I&longrightarrow&0,1\&(x,y)&mapsto&begincases1&text if y>frac12\0&text if y<frac12.endcasesendarray$$Then $f$ is continuous and $f(Itimes I)=0,1$, which is disconnected.
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
$endgroup$
add a comment |
$begingroup$
No. It's not even connected. Consider the map$$beginarrayrcccfcolon&Itimes I&longrightarrow&0,1\&(x,y)&mapsto&begincases1&text if y>frac12\0&text if y<frac12.endcasesendarray$$Then $f$ is continuous and $f(Itimes I)=0,1$, which is disconnected.
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
$endgroup$
add a comment |
$begingroup$
No. It's not even connected. Consider the map$$beginarrayrcccfcolon&Itimes I&longrightarrow&0,1\&(x,y)&mapsto&begincases1&text if y>frac12\0&text if y<frac12.endcasesendarray$$Then $f$ is continuous and $f(Itimes I)=0,1$, which is disconnected.
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
$endgroup$
No. It's not even connected. Consider the map$$beginarrayrcccfcolon&Itimes I&longrightarrow&0,1\&(x,y)&mapsto&begincases1&text if y>frac12\0&text if y<frac12.endcasesendarray$$Then $f$ is continuous and $f(Itimes I)=0,1$, which is disconnected.
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
answered 3 hours ago
José Carlos SantosJosé Carlos Santos
181k24142256
181k24142256
add a comment |
add a comment |
CB8axUYtdRyY52oZWt
1
$begingroup$
Fermat's last theorem is a nice thought, but unfortunately $gamma((0,1)) nsubseteq I times I$, because $I times I$ is missing all points with rational $x$-coordinate, for instance.
$endgroup$
– Jane Doé
3 hours ago
$begingroup$
If $pi_1,pi_2$ are the projections onto the first and second coordinates, then $alpha_1=pi_1circ gamma$ and $alpha_2=pi_2circgamma$ are continuous. Since $(0,1)$ is connected then its image by $alpha_1$ and $alpha_2$ are connected. This mean that they contain only one point.
$endgroup$
– logarithm
3 hours ago
$begingroup$
I see now that the use of Fermat's last theorem was misguided. The arc composed of points $x, y in (0,1)$ such that $x^3 + y^3 = 1$ certainly lives in $I times I$, but is not connected. Meanwhile the arc I described contains points with rational coordinates.
$endgroup$
– Charles Hudgins
3 hours ago