Connected component of ${(x_1,…,x_n) ; | ; x_1^2+…+{x_{n-1}}^2-x_n^2 neq 0}$












1












$begingroup$


The question is in the title, actually for $n=2$, it's okay, for $n=3$, I thought that :



${(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 > 0 }$ are the connected component.



Actually, for $n geq 4$, I have the impression that ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 > 0 }$ are the connected component, cause each are open and closed, but I need to prove that it's connected.



Anyone could help me, please ?



Thank you !






general-topology analysis connectedness path-connected







share|cite|improve this question









$endgroup$












  • $begingroup$
    "I thought that" - what makes you think so?
    $endgroup$
    – lisyarus
    Jan 16 at 17:08










  • $begingroup$
    The two sets are closed and open. So, for sure there is at least two connex components. And each set seems connected, and even path connected, but I didn't prove it in an explicit way, that's why I've written "I thought".
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:10










  • $begingroup$
    @ChocoSavour ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }$ is open but not close.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:19






  • 1




    $begingroup$
    @mathcounterexamples.net I meant closed in ${(x_1,..., x_n) ; | ; x_1^2+x_2^2+...-x_n^2 neq 0 }$, for the topology induced by $mathbb{R}^n$, isn't it ?
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:21












  • $begingroup$
    Then you should precise your question saying that you're looking for the connected components for the induced topology. As stated I understood connected components in $mathbb R^n$.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:25


















1












$begingroup$


The question is in the title, actually for $n=2$, it's okay, for $n=3$, I thought that :



${(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 > 0 }$ are the connected component.



Actually, for $n geq 4$, I have the impression that ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 > 0 }$ are the connected component, cause each are open and closed, but I need to prove that it's connected.



Anyone could help me, please ?



Thank you !






general-topology analysis connectedness path-connected







share|cite|improve this question









$endgroup$












  • $begingroup$
    "I thought that" - what makes you think so?
    $endgroup$
    – lisyarus
    Jan 16 at 17:08










  • $begingroup$
    The two sets are closed and open. So, for sure there is at least two connex components. And each set seems connected, and even path connected, but I didn't prove it in an explicit way, that's why I've written "I thought".
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:10










  • $begingroup$
    @ChocoSavour ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }$ is open but not close.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:19






  • 1




    $begingroup$
    @mathcounterexamples.net I meant closed in ${(x_1,..., x_n) ; | ; x_1^2+x_2^2+...-x_n^2 neq 0 }$, for the topology induced by $mathbb{R}^n$, isn't it ?
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:21












  • $begingroup$
    Then you should precise your question saying that you're looking for the connected components for the induced topology. As stated I understood connected components in $mathbb R^n$.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:25
















1












1








1


0



$begingroup$


The question is in the title, actually for $n=2$, it's okay, for $n=3$, I thought that :



${(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 > 0 }$ are the connected component.



Actually, for $n geq 4$, I have the impression that ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 > 0 }$ are the connected component, cause each are open and closed, but I need to prove that it's connected.



Anyone could help me, please ?



Thank you !






general-topology analysis connectedness path-connected







share|cite|improve this question









$endgroup$




The question is in the title, actually for $n=2$, it's okay, for $n=3$, I thought that :



${(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+x_2^2-{x_{3}}^2 > 0 }$ are the connected component.



Actually, for $n geq 4$, I have the impression that ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }, {(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 > 0 }$ are the connected component, cause each are open and closed, but I need to prove that it's connected.



Anyone could help me, please ?



Thank you !






general-topology analysis connectedness path-connected




general-topology analysis connectedness path-connected






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 17:00









ChocoSavourChocoSavour

34818




34818












  • $begingroup$
    "I thought that" - what makes you think so?
    $endgroup$
    – lisyarus
    Jan 16 at 17:08










  • $begingroup$
    The two sets are closed and open. So, for sure there is at least two connex components. And each set seems connected, and even path connected, but I didn't prove it in an explicit way, that's why I've written "I thought".
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:10










  • $begingroup$
    @ChocoSavour ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }$ is open but not close.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:19






  • 1




    $begingroup$
    @mathcounterexamples.net I meant closed in ${(x_1,..., x_n) ; | ; x_1^2+x_2^2+...-x_n^2 neq 0 }$, for the topology induced by $mathbb{R}^n$, isn't it ?
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:21












  • $begingroup$
    Then you should precise your question saying that you're looking for the connected components for the induced topology. As stated I understood connected components in $mathbb R^n$.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:25




















  • $begingroup$
    "I thought that" - what makes you think so?
    $endgroup$
    – lisyarus
    Jan 16 at 17:08










  • $begingroup$
    The two sets are closed and open. So, for sure there is at least two connex components. And each set seems connected, and even path connected, but I didn't prove it in an explicit way, that's why I've written "I thought".
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:10










  • $begingroup$
    @ChocoSavour ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }$ is open but not close.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:19






  • 1




    $begingroup$
    @mathcounterexamples.net I meant closed in ${(x_1,..., x_n) ; | ; x_1^2+x_2^2+...-x_n^2 neq 0 }$, for the topology induced by $mathbb{R}^n$, isn't it ?
    $endgroup$
    – ChocoSavour
    Jan 16 at 17:21












  • $begingroup$
    Then you should precise your question saying that you're looking for the connected components for the induced topology. As stated I understood connected components in $mathbb R^n$.
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 17:25


















$begingroup$
"I thought that" - what makes you think so?
$endgroup$
– lisyarus
Jan 16 at 17:08




$begingroup$
"I thought that" - what makes you think so?
$endgroup$
– lisyarus
Jan 16 at 17:08












$begingroup$
The two sets are closed and open. So, for sure there is at least two connex components. And each set seems connected, and even path connected, but I didn't prove it in an explicit way, that's why I've written "I thought".
$endgroup$
– ChocoSavour
Jan 16 at 17:10




$begingroup$
The two sets are closed and open. So, for sure there is at least two connex components. And each set seems connected, and even path connected, but I didn't prove it in an explicit way, that's why I've written "I thought".
$endgroup$
– ChocoSavour
Jan 16 at 17:10












$begingroup$
@ChocoSavour ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }$ is open but not close.
$endgroup$
– mathcounterexamples.net
Jan 16 at 17:19




$begingroup$
@ChocoSavour ${(x_1,...,x_n) ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 < 0 }$ is open but not close.
$endgroup$
– mathcounterexamples.net
Jan 16 at 17:19




1




1




$begingroup$
@mathcounterexamples.net I meant closed in ${(x_1,..., x_n) ; | ; x_1^2+x_2^2+...-x_n^2 neq 0 }$, for the topology induced by $mathbb{R}^n$, isn't it ?
$endgroup$
– ChocoSavour
Jan 16 at 17:21






$begingroup$
@mathcounterexamples.net I meant closed in ${(x_1,..., x_n) ; | ; x_1^2+x_2^2+...-x_n^2 neq 0 }$, for the topology induced by $mathbb{R}^n$, isn't it ?
$endgroup$
– ChocoSavour
Jan 16 at 17:21














$begingroup$
Then you should precise your question saying that you're looking for the connected components for the induced topology. As stated I understood connected components in $mathbb R^n$.
$endgroup$
– mathcounterexamples.net
Jan 16 at 17:25






$begingroup$
Then you should precise your question saying that you're looking for the connected components for the induced topology. As stated I understood connected components in $mathbb R^n$.
$endgroup$
– mathcounterexamples.net
Jan 16 at 17:25












1 Answer
1






active

oldest

votes


















1












$begingroup$

Let’s note $S_n= {(x_1,...,x_n) in mathbb R^n ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 neq 0}$



Case $n=2$



$S_2={(x,y) in mathbb R^2 mid x^2 neq y^2} ={(x,y) in mathbb R^2 mid x neq y wedge x neq -y}$ has 4 connected components.



Case $n ge 3$



Let’s prove that in that case, $S_n$ has 3 connected components, namely:



$$begin{aligned}
S_n^+ &= left{(x_1,...,x_n) in mathbb R^n mid sqrt{x_1^2+...+{x_{n-1}}^2}< x_nright}\
S_n^0 &= left{(x_1,...,x_n) in mathbb R^n mid -sqrt{x_1^2+...+{x_{n-1}}^2}< x_n < sqrt{x_1^2+...+{x_{n-1}}^2}right}\
S_n^- &= left{(x_1,...,x_n) in mathbb R^n mid x_n < -sqrt{x_1^2+...+{x_{n-1}}^2}right}
end{aligned}$$



Those 3 opens are in fact path connected. $S_n^+$ and $S_n^-$ are convex. Following question: when is the epigraph a convex cone? is a good basis for the proof.



$S_n^0$ is path connected and therefore connected. Use «  vertical » lines going to the hyperplane $x_n=0$ and one or two segment lines in the hyperplane $x_n=0$, to avoid passing through the origin.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much !
    $endgroup$
    – ChocoSavour
    Jan 22 at 16:31












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let’s note $S_n= {(x_1,...,x_n) in mathbb R^n ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 neq 0}$



Case $n=2$



$S_2={(x,y) in mathbb R^2 mid x^2 neq y^2} ={(x,y) in mathbb R^2 mid x neq y wedge x neq -y}$ has 4 connected components.



Case $n ge 3$



Let’s prove that in that case, $S_n$ has 3 connected components, namely:



$$begin{aligned}
S_n^+ &= left{(x_1,...,x_n) in mathbb R^n mid sqrt{x_1^2+...+{x_{n-1}}^2}< x_nright}\
S_n^0 &= left{(x_1,...,x_n) in mathbb R^n mid -sqrt{x_1^2+...+{x_{n-1}}^2}< x_n < sqrt{x_1^2+...+{x_{n-1}}^2}right}\
S_n^- &= left{(x_1,...,x_n) in mathbb R^n mid x_n < -sqrt{x_1^2+...+{x_{n-1}}^2}right}
end{aligned}$$



Those 3 opens are in fact path connected. $S_n^+$ and $S_n^-$ are convex. Following question: when is the epigraph a convex cone? is a good basis for the proof.



$S_n^0$ is path connected and therefore connected. Use «  vertical » lines going to the hyperplane $x_n=0$ and one or two segment lines in the hyperplane $x_n=0$, to avoid passing through the origin.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much !
    $endgroup$
    – ChocoSavour
    Jan 22 at 16:31
















1












$begingroup$

Let’s note $S_n= {(x_1,...,x_n) in mathbb R^n ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 neq 0}$



Case $n=2$



$S_2={(x,y) in mathbb R^2 mid x^2 neq y^2} ={(x,y) in mathbb R^2 mid x neq y wedge x neq -y}$ has 4 connected components.



Case $n ge 3$



Let’s prove that in that case, $S_n$ has 3 connected components, namely:



$$begin{aligned}
S_n^+ &= left{(x_1,...,x_n) in mathbb R^n mid sqrt{x_1^2+...+{x_{n-1}}^2}< x_nright}\
S_n^0 &= left{(x_1,...,x_n) in mathbb R^n mid -sqrt{x_1^2+...+{x_{n-1}}^2}< x_n < sqrt{x_1^2+...+{x_{n-1}}^2}right}\
S_n^- &= left{(x_1,...,x_n) in mathbb R^n mid x_n < -sqrt{x_1^2+...+{x_{n-1}}^2}right}
end{aligned}$$



Those 3 opens are in fact path connected. $S_n^+$ and $S_n^-$ are convex. Following question: when is the epigraph a convex cone? is a good basis for the proof.



$S_n^0$ is path connected and therefore connected. Use «  vertical » lines going to the hyperplane $x_n=0$ and one or two segment lines in the hyperplane $x_n=0$, to avoid passing through the origin.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much !
    $endgroup$
    – ChocoSavour
    Jan 22 at 16:31














1












1








1





$begingroup$

Let’s note $S_n= {(x_1,...,x_n) in mathbb R^n ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 neq 0}$



Case $n=2$



$S_2={(x,y) in mathbb R^2 mid x^2 neq y^2} ={(x,y) in mathbb R^2 mid x neq y wedge x neq -y}$ has 4 connected components.



Case $n ge 3$



Let’s prove that in that case, $S_n$ has 3 connected components, namely:



$$begin{aligned}
S_n^+ &= left{(x_1,...,x_n) in mathbb R^n mid sqrt{x_1^2+...+{x_{n-1}}^2}< x_nright}\
S_n^0 &= left{(x_1,...,x_n) in mathbb R^n mid -sqrt{x_1^2+...+{x_{n-1}}^2}< x_n < sqrt{x_1^2+...+{x_{n-1}}^2}right}\
S_n^- &= left{(x_1,...,x_n) in mathbb R^n mid x_n < -sqrt{x_1^2+...+{x_{n-1}}^2}right}
end{aligned}$$



Those 3 opens are in fact path connected. $S_n^+$ and $S_n^-$ are convex. Following question: when is the epigraph a convex cone? is a good basis for the proof.



$S_n^0$ is path connected and therefore connected. Use «  vertical » lines going to the hyperplane $x_n=0$ and one or two segment lines in the hyperplane $x_n=0$, to avoid passing through the origin.






share|cite|improve this answer











$endgroup$



Let’s note $S_n= {(x_1,...,x_n) in mathbb R^n ; | ; x_1^2+...+{x_{n-1}}^2-x_n^2 neq 0}$



Case $n=2$



$S_2={(x,y) in mathbb R^2 mid x^2 neq y^2} ={(x,y) in mathbb R^2 mid x neq y wedge x neq -y}$ has 4 connected components.



Case $n ge 3$



Let’s prove that in that case, $S_n$ has 3 connected components, namely:



$$begin{aligned}
S_n^+ &= left{(x_1,...,x_n) in mathbb R^n mid sqrt{x_1^2+...+{x_{n-1}}^2}< x_nright}\
S_n^0 &= left{(x_1,...,x_n) in mathbb R^n mid -sqrt{x_1^2+...+{x_{n-1}}^2}< x_n < sqrt{x_1^2+...+{x_{n-1}}^2}right}\
S_n^- &= left{(x_1,...,x_n) in mathbb R^n mid x_n < -sqrt{x_1^2+...+{x_{n-1}}^2}right}
end{aligned}$$



Those 3 opens are in fact path connected. $S_n^+$ and $S_n^-$ are convex. Following question: when is the epigraph a convex cone? is a good basis for the proof.



$S_n^0$ is path connected and therefore connected. Use «  vertical » lines going to the hyperplane $x_n=0$ and one or two segment lines in the hyperplane $x_n=0$, to avoid passing through the origin.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 at 7:05

























answered Jan 16 at 20:49









mathcounterexamples.netmathcounterexamples.net

26.9k22158




26.9k22158












  • $begingroup$
    Thank you very much !
    $endgroup$
    – ChocoSavour
    Jan 22 at 16:31


















  • $begingroup$
    Thank you very much !
    $endgroup$
    – ChocoSavour
    Jan 22 at 16:31
















$begingroup$
Thank you very much !
$endgroup$
– ChocoSavour
Jan 22 at 16:31




$begingroup$
Thank you very much !
$endgroup$
– ChocoSavour
Jan 22 at 16:31


















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