How can I prove that $mathbb Q times mathbb Q$ is connected?












0












$begingroup$


How can I prove that $mathbb Q times mathbb Q$ is connected? I was trying to prove that it isn't because $mathbb Q$ is not connected but I can't find two open $A,B$ to prove it.






general-topology connectedness







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry, I´m new in it and my english level isn´t the best. $Q$ are the rational numbers. And they are not connected. But I have to demostrate that $Q$X$Q$ are or not connected. I was trying to prove that it isn´t connected but I couldn´t find any open partition for prove it. That´s why I think it´s connected but I don´t know who can I prove it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 17:11










  • $begingroup$
    @Meliodas Use detexify.kirelabs.org/classify.html to help yourself with tex commands.
    $endgroup$
    – Wolfgang
    Dec 24 '18 at 17:19












  • $begingroup$
    thank you very much, I will use it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:13










  • $begingroup$
    This is definitely not the best way to prove that $Bbb Q^2$ is not connected, but it might be a useful fact to know anyway: $Bbb Q^2$ and $Bbb Q$ are homeomorphic, so in particular the former is disconnected since the latter is
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:44
















0












$begingroup$


How can I prove that $mathbb Q times mathbb Q$ is connected? I was trying to prove that it isn't because $mathbb Q$ is not connected but I can't find two open $A,B$ to prove it.






general-topology connectedness







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry, I´m new in it and my english level isn´t the best. $Q$ are the rational numbers. And they are not connected. But I have to demostrate that $Q$X$Q$ are or not connected. I was trying to prove that it isn´t connected but I couldn´t find any open partition for prove it. That´s why I think it´s connected but I don´t know who can I prove it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 17:11










  • $begingroup$
    @Meliodas Use detexify.kirelabs.org/classify.html to help yourself with tex commands.
    $endgroup$
    – Wolfgang
    Dec 24 '18 at 17:19












  • $begingroup$
    thank you very much, I will use it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:13










  • $begingroup$
    This is definitely not the best way to prove that $Bbb Q^2$ is not connected, but it might be a useful fact to know anyway: $Bbb Q^2$ and $Bbb Q$ are homeomorphic, so in particular the former is disconnected since the latter is
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:44














0












0








0





$begingroup$


How can I prove that $mathbb Q times mathbb Q$ is connected? I was trying to prove that it isn't because $mathbb Q$ is not connected but I can't find two open $A,B$ to prove it.






general-topology connectedness







share|cite|improve this question











$endgroup$




How can I prove that $mathbb Q times mathbb Q$ is connected? I was trying to prove that it isn't because $mathbb Q$ is not connected but I can't find two open $A,B$ to prove it.






general-topology connectedness




general-topology connectedness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 17:46









José Carlos Santos

167k22132235




167k22132235










asked Dec 24 '18 at 17:03









MeliodasMeliodas

215




215












  • $begingroup$
    Sorry, I´m new in it and my english level isn´t the best. $Q$ are the rational numbers. And they are not connected. But I have to demostrate that $Q$X$Q$ are or not connected. I was trying to prove that it isn´t connected but I couldn´t find any open partition for prove it. That´s why I think it´s connected but I don´t know who can I prove it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 17:11










  • $begingroup$
    @Meliodas Use detexify.kirelabs.org/classify.html to help yourself with tex commands.
    $endgroup$
    – Wolfgang
    Dec 24 '18 at 17:19












  • $begingroup$
    thank you very much, I will use it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:13










  • $begingroup$
    This is definitely not the best way to prove that $Bbb Q^2$ is not connected, but it might be a useful fact to know anyway: $Bbb Q^2$ and $Bbb Q$ are homeomorphic, so in particular the former is disconnected since the latter is
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:44


















  • $begingroup$
    Sorry, I´m new in it and my english level isn´t the best. $Q$ are the rational numbers. And they are not connected. But I have to demostrate that $Q$X$Q$ are or not connected. I was trying to prove that it isn´t connected but I couldn´t find any open partition for prove it. That´s why I think it´s connected but I don´t know who can I prove it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 17:11










  • $begingroup$
    @Meliodas Use detexify.kirelabs.org/classify.html to help yourself with tex commands.
    $endgroup$
    – Wolfgang
    Dec 24 '18 at 17:19












  • $begingroup$
    thank you very much, I will use it.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:13










  • $begingroup$
    This is definitely not the best way to prove that $Bbb Q^2$ is not connected, but it might be a useful fact to know anyway: $Bbb Q^2$ and $Bbb Q$ are homeomorphic, so in particular the former is disconnected since the latter is
    $endgroup$
    – Alessandro Codenotti
    Dec 25 '18 at 16:44
















$begingroup$
Sorry, I´m new in it and my english level isn´t the best. $Q$ are the rational numbers. And they are not connected. But I have to demostrate that $Q$X$Q$ are or not connected. I was trying to prove that it isn´t connected but I couldn´t find any open partition for prove it. That´s why I think it´s connected but I don´t know who can I prove it.
$endgroup$
– Meliodas
Dec 24 '18 at 17:11




$begingroup$
Sorry, I´m new in it and my english level isn´t the best. $Q$ are the rational numbers. And they are not connected. But I have to demostrate that $Q$X$Q$ are or not connected. I was trying to prove that it isn´t connected but I couldn´t find any open partition for prove it. That´s why I think it´s connected but I don´t know who can I prove it.
$endgroup$
– Meliodas
Dec 24 '18 at 17:11












$begingroup$
@Meliodas Use detexify.kirelabs.org/classify.html to help yourself with tex commands.
$endgroup$
– Wolfgang
Dec 24 '18 at 17:19






$begingroup$
@Meliodas Use detexify.kirelabs.org/classify.html to help yourself with tex commands.
$endgroup$
– Wolfgang
Dec 24 '18 at 17:19














$begingroup$
thank you very much, I will use it.
$endgroup$
– Meliodas
Dec 24 '18 at 18:13




$begingroup$
thank you very much, I will use it.
$endgroup$
– Meliodas
Dec 24 '18 at 18:13












$begingroup$
This is definitely not the best way to prove that $Bbb Q^2$ is not connected, but it might be a useful fact to know anyway: $Bbb Q^2$ and $Bbb Q$ are homeomorphic, so in particular the former is disconnected since the latter is
$endgroup$
– Alessandro Codenotti
Dec 25 '18 at 16:44




$begingroup$
This is definitely not the best way to prove that $Bbb Q^2$ is not connected, but it might be a useful fact to know anyway: $Bbb Q^2$ and $Bbb Q$ are homeomorphic, so in particular the former is disconnected since the latter is
$endgroup$
– Alessandro Codenotti
Dec 25 '18 at 16:44










3 Answers
3






active

oldest

votes


















4












$begingroup$

You know that $mathbb Q$ is not connected. That means that there is a set $Asubsetmathbb Q$ such that $A$ is open and closed in $mathbb Q$ and that furthermore $Aneqemptyset,mathbb Q$. So, consider the set $Atimes A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, is very clear if you use this way.Thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:17



















1












$begingroup$

The map $f(x,y)=x$ is continuous on $mathbb R^2.$ If $mathbb Qtimes mathbb Q $ were connected, then $f(mathbb Qtimes mathbb Q)$ would be connected. But $f(mathbb Qtimes mathbb Q)=mathbb Q,$ contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If f(x,y) continuos then you go of once connected space to another one?
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:21










  • $begingroup$
    If $E$ is connected and $f$ is continuous, then $f(E)$ is connected.
    $endgroup$
    – zhw.
    Dec 24 '18 at 18:30










  • $begingroup$
    Perfect,thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:32



















0












$begingroup$

Your initial intuition was correct - $mathbb{Q}^2$ is indeed not connected.



Think about it intuitively. $mathbb{Q}^2$ is basically a bunch of points sprinkled in the plane $mathbb{R}^2$. The easiest way to show that it's disconnected would be to "cut it in half with a line" - that is, find a line $l$ such that $(1)$ there are rational points (= elements of $mathbb{Q}^2$) on both sides of $l$ and $(2)$ there are no rational points actually on $l$ itself.




  • Do you see why this would in fact disconnect $mathbb{Q}^2$?


Now $(1)$ is basically trivial - $mathbb{Q}^2$ is scattered throughout $mathbb{R}^2$ so that there are rational points on both sides of any line. So we're just down to $(2)$: can you think of a line in $mathbb{R}^2$ not containing any rational points? HINT: there's even a horizontal (or vertical, if you prefer) example - think about $pi$ ...)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand. I was thinking only in the horizontal of x=0 but if I find one with only irrationals points I have the solution. Thanks for the explanation.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:25










  • $begingroup$
    @Meliodas Yes, and it's not hard to find such a line - do you see how?
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:01










  • $begingroup$
    I thought in x=irrational number. In this way (x,y) don´t belong to rational numbers for all the y of the vertical, Is this enough?
    $endgroup$
    – Meliodas
    Dec 25 '18 at 16:29











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

You know that $mathbb Q$ is not connected. That means that there is a set $Asubsetmathbb Q$ such that $A$ is open and closed in $mathbb Q$ and that furthermore $Aneqemptyset,mathbb Q$. So, consider the set $Atimes A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, is very clear if you use this way.Thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:17
















4












$begingroup$

You know that $mathbb Q$ is not connected. That means that there is a set $Asubsetmathbb Q$ such that $A$ is open and closed in $mathbb Q$ and that furthermore $Aneqemptyset,mathbb Q$. So, consider the set $Atimes A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, is very clear if you use this way.Thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:17














4












4








4





$begingroup$

You know that $mathbb Q$ is not connected. That means that there is a set $Asubsetmathbb Q$ such that $A$ is open and closed in $mathbb Q$ and that furthermore $Aneqemptyset,mathbb Q$. So, consider the set $Atimes A$.






share|cite|improve this answer









$endgroup$



You know that $mathbb Q$ is not connected. That means that there is a set $Asubsetmathbb Q$ such that $A$ is open and closed in $mathbb Q$ and that furthermore $Aneqemptyset,mathbb Q$. So, consider the set $Atimes A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 17:13









José Carlos SantosJosé Carlos Santos

167k22132235




167k22132235












  • $begingroup$
    Yes, is very clear if you use this way.Thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:17


















  • $begingroup$
    Yes, is very clear if you use this way.Thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:17
















$begingroup$
Yes, is very clear if you use this way.Thanks!
$endgroup$
– Meliodas
Dec 24 '18 at 18:17




$begingroup$
Yes, is very clear if you use this way.Thanks!
$endgroup$
– Meliodas
Dec 24 '18 at 18:17











1












$begingroup$

The map $f(x,y)=x$ is continuous on $mathbb R^2.$ If $mathbb Qtimes mathbb Q $ were connected, then $f(mathbb Qtimes mathbb Q)$ would be connected. But $f(mathbb Qtimes mathbb Q)=mathbb Q,$ contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If f(x,y) continuos then you go of once connected space to another one?
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:21










  • $begingroup$
    If $E$ is connected and $f$ is continuous, then $f(E)$ is connected.
    $endgroup$
    – zhw.
    Dec 24 '18 at 18:30










  • $begingroup$
    Perfect,thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:32
















1












$begingroup$

The map $f(x,y)=x$ is continuous on $mathbb R^2.$ If $mathbb Qtimes mathbb Q $ were connected, then $f(mathbb Qtimes mathbb Q)$ would be connected. But $f(mathbb Qtimes mathbb Q)=mathbb Q,$ contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If f(x,y) continuos then you go of once connected space to another one?
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:21










  • $begingroup$
    If $E$ is connected and $f$ is continuous, then $f(E)$ is connected.
    $endgroup$
    – zhw.
    Dec 24 '18 at 18:30










  • $begingroup$
    Perfect,thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:32














1












1








1





$begingroup$

The map $f(x,y)=x$ is continuous on $mathbb R^2.$ If $mathbb Qtimes mathbb Q $ were connected, then $f(mathbb Qtimes mathbb Q)$ would be connected. But $f(mathbb Qtimes mathbb Q)=mathbb Q,$ contradiction.






share|cite|improve this answer









$endgroup$



The map $f(x,y)=x$ is continuous on $mathbb R^2.$ If $mathbb Qtimes mathbb Q $ were connected, then $f(mathbb Qtimes mathbb Q)$ would be connected. But $f(mathbb Qtimes mathbb Q)=mathbb Q,$ contradiction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 17:34









zhw.zhw.

74.1k43175




74.1k43175












  • $begingroup$
    If f(x,y) continuos then you go of once connected space to another one?
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:21










  • $begingroup$
    If $E$ is connected and $f$ is continuous, then $f(E)$ is connected.
    $endgroup$
    – zhw.
    Dec 24 '18 at 18:30










  • $begingroup$
    Perfect,thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:32


















  • $begingroup$
    If f(x,y) continuos then you go of once connected space to another one?
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:21










  • $begingroup$
    If $E$ is connected and $f$ is continuous, then $f(E)$ is connected.
    $endgroup$
    – zhw.
    Dec 24 '18 at 18:30










  • $begingroup$
    Perfect,thanks!
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:32
















$begingroup$
If f(x,y) continuos then you go of once connected space to another one?
$endgroup$
– Meliodas
Dec 24 '18 at 18:21




$begingroup$
If f(x,y) continuos then you go of once connected space to another one?
$endgroup$
– Meliodas
Dec 24 '18 at 18:21












$begingroup$
If $E$ is connected and $f$ is continuous, then $f(E)$ is connected.
$endgroup$
– zhw.
Dec 24 '18 at 18:30




$begingroup$
If $E$ is connected and $f$ is continuous, then $f(E)$ is connected.
$endgroup$
– zhw.
Dec 24 '18 at 18:30












$begingroup$
Perfect,thanks!
$endgroup$
– Meliodas
Dec 24 '18 at 18:32




$begingroup$
Perfect,thanks!
$endgroup$
– Meliodas
Dec 24 '18 at 18:32











0












$begingroup$

Your initial intuition was correct - $mathbb{Q}^2$ is indeed not connected.



Think about it intuitively. $mathbb{Q}^2$ is basically a bunch of points sprinkled in the plane $mathbb{R}^2$. The easiest way to show that it's disconnected would be to "cut it in half with a line" - that is, find a line $l$ such that $(1)$ there are rational points (= elements of $mathbb{Q}^2$) on both sides of $l$ and $(2)$ there are no rational points actually on $l$ itself.




  • Do you see why this would in fact disconnect $mathbb{Q}^2$?


Now $(1)$ is basically trivial - $mathbb{Q}^2$ is scattered throughout $mathbb{R}^2$ so that there are rational points on both sides of any line. So we're just down to $(2)$: can you think of a line in $mathbb{R}^2$ not containing any rational points? HINT: there's even a horizontal (or vertical, if you prefer) example - think about $pi$ ...)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand. I was thinking only in the horizontal of x=0 but if I find one with only irrationals points I have the solution. Thanks for the explanation.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:25










  • $begingroup$
    @Meliodas Yes, and it's not hard to find such a line - do you see how?
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:01










  • $begingroup$
    I thought in x=irrational number. In this way (x,y) don´t belong to rational numbers for all the y of the vertical, Is this enough?
    $endgroup$
    – Meliodas
    Dec 25 '18 at 16:29
















0












$begingroup$

Your initial intuition was correct - $mathbb{Q}^2$ is indeed not connected.



Think about it intuitively. $mathbb{Q}^2$ is basically a bunch of points sprinkled in the plane $mathbb{R}^2$. The easiest way to show that it's disconnected would be to "cut it in half with a line" - that is, find a line $l$ such that $(1)$ there are rational points (= elements of $mathbb{Q}^2$) on both sides of $l$ and $(2)$ there are no rational points actually on $l$ itself.




  • Do you see why this would in fact disconnect $mathbb{Q}^2$?


Now $(1)$ is basically trivial - $mathbb{Q}^2$ is scattered throughout $mathbb{R}^2$ so that there are rational points on both sides of any line. So we're just down to $(2)$: can you think of a line in $mathbb{R}^2$ not containing any rational points? HINT: there's even a horizontal (or vertical, if you prefer) example - think about $pi$ ...)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand. I was thinking only in the horizontal of x=0 but if I find one with only irrationals points I have the solution. Thanks for the explanation.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:25










  • $begingroup$
    @Meliodas Yes, and it's not hard to find such a line - do you see how?
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:01










  • $begingroup$
    I thought in x=irrational number. In this way (x,y) don´t belong to rational numbers for all the y of the vertical, Is this enough?
    $endgroup$
    – Meliodas
    Dec 25 '18 at 16:29














0












0








0





$begingroup$

Your initial intuition was correct - $mathbb{Q}^2$ is indeed not connected.



Think about it intuitively. $mathbb{Q}^2$ is basically a bunch of points sprinkled in the plane $mathbb{R}^2$. The easiest way to show that it's disconnected would be to "cut it in half with a line" - that is, find a line $l$ such that $(1)$ there are rational points (= elements of $mathbb{Q}^2$) on both sides of $l$ and $(2)$ there are no rational points actually on $l$ itself.




  • Do you see why this would in fact disconnect $mathbb{Q}^2$?


Now $(1)$ is basically trivial - $mathbb{Q}^2$ is scattered throughout $mathbb{R}^2$ so that there are rational points on both sides of any line. So we're just down to $(2)$: can you think of a line in $mathbb{R}^2$ not containing any rational points? HINT: there's even a horizontal (or vertical, if you prefer) example - think about $pi$ ...)






share|cite|improve this answer









$endgroup$



Your initial intuition was correct - $mathbb{Q}^2$ is indeed not connected.



Think about it intuitively. $mathbb{Q}^2$ is basically a bunch of points sprinkled in the plane $mathbb{R}^2$. The easiest way to show that it's disconnected would be to "cut it in half with a line" - that is, find a line $l$ such that $(1)$ there are rational points (= elements of $mathbb{Q}^2$) on both sides of $l$ and $(2)$ there are no rational points actually on $l$ itself.




  • Do you see why this would in fact disconnect $mathbb{Q}^2$?


Now $(1)$ is basically trivial - $mathbb{Q}^2$ is scattered throughout $mathbb{R}^2$ so that there are rational points on both sides of any line. So we're just down to $(2)$: can you think of a line in $mathbb{R}^2$ not containing any rational points? HINT: there's even a horizontal (or vertical, if you prefer) example - think about $pi$ ...)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 17:52









Noah SchweberNoah Schweber

127k10151290




127k10151290












  • $begingroup$
    I understand. I was thinking only in the horizontal of x=0 but if I find one with only irrationals points I have the solution. Thanks for the explanation.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:25










  • $begingroup$
    @Meliodas Yes, and it's not hard to find such a line - do you see how?
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:01










  • $begingroup$
    I thought in x=irrational number. In this way (x,y) don´t belong to rational numbers for all the y of the vertical, Is this enough?
    $endgroup$
    – Meliodas
    Dec 25 '18 at 16:29


















  • $begingroup$
    I understand. I was thinking only in the horizontal of x=0 but if I find one with only irrationals points I have the solution. Thanks for the explanation.
    $endgroup$
    – Meliodas
    Dec 24 '18 at 18:25










  • $begingroup$
    @Meliodas Yes, and it's not hard to find such a line - do you see how?
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:01










  • $begingroup$
    I thought in x=irrational number. In this way (x,y) don´t belong to rational numbers for all the y of the vertical, Is this enough?
    $endgroup$
    – Meliodas
    Dec 25 '18 at 16:29
















$begingroup$
I understand. I was thinking only in the horizontal of x=0 but if I find one with only irrationals points I have the solution. Thanks for the explanation.
$endgroup$
– Meliodas
Dec 24 '18 at 18:25




$begingroup$
I understand. I was thinking only in the horizontal of x=0 but if I find one with only irrationals points I have the solution. Thanks for the explanation.
$endgroup$
– Meliodas
Dec 24 '18 at 18:25












$begingroup$
@Meliodas Yes, and it's not hard to find such a line - do you see how?
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:01




$begingroup$
@Meliodas Yes, and it's not hard to find such a line - do you see how?
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:01












$begingroup$
I thought in x=irrational number. In this way (x,y) don´t belong to rational numbers for all the y of the vertical, Is this enough?
$endgroup$
– Meliodas
Dec 25 '18 at 16:29




$begingroup$
I thought in x=irrational number. In this way (x,y) don´t belong to rational numbers for all the y of the vertical, Is this enough?
$endgroup$
– Meliodas
Dec 25 '18 at 16:29


















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