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Is this open subset $E subset mathbb R^n$ connected?
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Suppose $E subset mathbb R^n$ is an open subset with $0 notin E$. $E$ has following properties:(1) if $x = (x_1, dots, x_n) in E$, then $x(alpha) = (alpha x_1, alpha^2 x_2, dots, alpha^n x_n) in E$ for every positive real number $alpha in mathbb R_+$ (2) if $y=(y_1, dots, y_n) in E$, then $y_i < 0$ for every $i =1, dots, n$. I am wondering whether $E$ is connected.
My idea: Let $x, y in E$. We first connect $x to x(alpha)$ and $y to y(beta)$ for $alpha, beta$ sufficiently small and then leverage the openness of $E$. But this does not work since even if $x(alpha), y(beta)$ are sufficiently close to $0$, we cannot guarantee they belong to the same norm ball.
Edit: I am trying to abstract an application problem. The second property was missing before. There was an answer below to show the set is not connected with only property $1$. Sorry to cause confusion.
general-topology connectedness path-connected
asked yesterday
user9527
923524
add a comment |Â
up vote
0
down vote
favorite
Suppose $E subset mathbb R^n$ is an open subset with $0 notin E$. $E$ has following properties:(1) if $x = (x_1, dots, x_n) in E$, then $x(alpha) = (alpha x_1, alpha^2 x_2, dots, alpha^n x_n) in E$ for every positive real number $alpha in mathbb R_+$ (2) if $y=(y_1, dots, y_n) in E$, then $y_i < 0$ for every $i =1, dots, n$. I am wondering whether $E$ is connected.
My idea: Let $x, y in E$. We first connect $x to x(alpha)$ and $y to y(beta)$ for $alpha, beta$ sufficiently small and then leverage the openness of $E$. But this does not work since even if $x(alpha), y(beta)$ are sufficiently close to $0$, we cannot guarantee they belong to the same norm ball.
Edit: I am trying to abstract an application problem. The second property was missing before. There was an answer below to show the set is not connected with only property $1$. Sorry to cause confusion.
general-topology connectedness path-connected
asked yesterday
user9527
923524
I suspect you are missing an extra hypothesis.
– Lord Shark the Unknown
yesterday
@user9527 see my updated answer
– mathworker21
6 hours ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $E subset mathbb R^n$ is an open subset with $0 notin E$. $E$ has following properties:(1) if $x = (x_1, dots, x_n) in E$, then $x(alpha) = (alpha x_1, alpha^2 x_2, dots, alpha^n x_n) in E$ for every positive real number $alpha in mathbb R_+$ (2) if $y=(y_1, dots, y_n) in E$, then $y_i < 0$ for every $i =1, dots, n$. I am wondering whether $E$ is connected.
My idea: Let $x, y in E$. We first connect $x to x(alpha)$ and $y to y(beta)$ for $alpha, beta$ sufficiently small and then leverage the openness of $E$. But this does not work since even if $x(alpha), y(beta)$ are sufficiently close to $0$, we cannot guarantee they belong to the same norm ball.
Edit: I am trying to abstract an application problem. The second property was missing before. There was an answer below to show the set is not connected with only property $1$. Sorry to cause confusion.
general-topology connectedness path-connected
asked yesterday
user9527
923524
Suppose $E subset mathbb R^n$ is an open subset with $0 notin E$. $E$ has following properties:(1) if $x = (x_1, dots, x_n) in E$, then $x(alpha) = (alpha x_1, alpha^2 x_2, dots, alpha^n x_n) in E$ for every positive real number $alpha in mathbb R_+$ (2) if $y=(y_1, dots, y_n) in E$, then $y_i < 0$ for every $i =1, dots, n$. I am wondering whether $E$ is connected.
My idea: Let $x, y in E$. We first connect $x to x(alpha)$ and $y to y(beta)$ for $alpha, beta$ sufficiently small and then leverage the openness of $E$. But this does not work since even if $x(alpha), y(beta)$ are sufficiently close to $0$, we cannot guarantee they belong to the same norm ball.
Edit: I am trying to abstract an application problem. The second property was missing before. There was an answer below to show the set is not connected with only property $1$. Sorry to cause confusion.
general-topology connectedness path-connected
asked yesterday
user9527
923524
edited 7 hours ago
asked yesterday
user9527
923524
asked yesterday
user9527
923524
asked yesterday
user9527
923524
923524
I suspect you are missing an extra hypothesis.
– Lord Shark the Unknown
yesterday
@user9527 see my updated answer
– mathworker21
6 hours ago
add a comment |Â
I suspect you are missing an extra hypothesis.
– Lord Shark the Unknown
yesterday
@user9527 see my updated answer
– mathworker21
6 hours ago
I suspect you are missing an extra hypothesis.
– Lord Shark the Unknown
yesterday
I suspect you are missing an extra hypothesis.
– Lord Shark the Unknown
yesterday
@user9527 see my updated answer
– mathworker21
6 hours ago
@user9527 see my updated answer
– mathworker21
6 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
For ease, I'm replacing your property (2) with $y_i > 0$ for each $i$ (which is the same property for all intents and purposes).
Take $n=2$ and let $E = E_1 sqcup E_2$ for $E_1 = (alpha x_1,alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 > 1$ and $E_2 = (alpha x_1, alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 < 1$. The point is that the sets $E_1,E_2$ are well-defined: indeed, if $P := (alpha x,alpha^2 y)$ satisfies $fracyx^2 > 1$, then if we represented $P$ also as $(beta fracalpha xbeta, beta^2 fracalpha^2 ybeta^2)$, then $fracfracalpha^2 ybeta^2(fracalpha xbeta)^2 = fracyx^2$. And clearly $E_1,E_2$ are disjoint open sets not containing $0$ that satisfy properties (1) and (2).
answered yesterday
mathworker21
6,4081727
Condition 2) is not satisfied.
– Kavi Rama Murthy
7 hours ago
@KaviRamaMurthy That property wasn't there when I first posted my answer. But now I've updated my answer accordingly.
– mathworker21
6 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For ease, I'm replacing your property (2) with $y_i > 0$ for each $i$ (which is the same property for all intents and purposes).
Take $n=2$ and let $E = E_1 sqcup E_2$ for $E_1 = (alpha x_1,alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 > 1$ and $E_2 = (alpha x_1, alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 < 1$. The point is that the sets $E_1,E_2$ are well-defined: indeed, if $P := (alpha x,alpha^2 y)$ satisfies $fracyx^2 > 1$, then if we represented $P$ also as $(beta fracalpha xbeta, beta^2 fracalpha^2 ybeta^2)$, then $fracfracalpha^2 ybeta^2(fracalpha xbeta)^2 = fracyx^2$. And clearly $E_1,E_2$ are disjoint open sets not containing $0$ that satisfy properties (1) and (2).
answered yesterday
mathworker21
6,4081727
Condition 2) is not satisfied.
– Kavi Rama Murthy
7 hours ago
@KaviRamaMurthy That property wasn't there when I first posted my answer. But now I've updated my answer accordingly.
– mathworker21
6 hours ago
add a comment |Â
up vote
1
down vote
accepted
For ease, I'm replacing your property (2) with $y_i > 0$ for each $i$ (which is the same property for all intents and purposes).
Take $n=2$ and let $E = E_1 sqcup E_2$ for $E_1 = (alpha x_1,alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 > 1$ and $E_2 = (alpha x_1, alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 < 1$. The point is that the sets $E_1,E_2$ are well-defined: indeed, if $P := (alpha x,alpha^2 y)$ satisfies $fracyx^2 > 1$, then if we represented $P$ also as $(beta fracalpha xbeta, beta^2 fracalpha^2 ybeta^2)$, then $fracfracalpha^2 ybeta^2(fracalpha xbeta)^2 = fracyx^2$. And clearly $E_1,E_2$ are disjoint open sets not containing $0$ that satisfy properties (1) and (2).
answered yesterday
mathworker21
6,4081727
Condition 2) is not satisfied.
– Kavi Rama Murthy
7 hours ago
@KaviRamaMurthy That property wasn't there when I first posted my answer. But now I've updated my answer accordingly.
– mathworker21
6 hours ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For ease, I'm replacing your property (2) with $y_i > 0$ for each $i$ (which is the same property for all intents and purposes).
Take $n=2$ and let $E = E_1 sqcup E_2$ for $E_1 = (alpha x_1,alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 > 1$ and $E_2 = (alpha x_1, alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 < 1$. The point is that the sets $E_1,E_2$ are well-defined: indeed, if $P := (alpha x,alpha^2 y)$ satisfies $fracyx^2 > 1$, then if we represented $P$ also as $(beta fracalpha xbeta, beta^2 fracalpha^2 ybeta^2)$, then $fracfracalpha^2 ybeta^2(fracalpha xbeta)^2 = fracyx^2$. And clearly $E_1,E_2$ are disjoint open sets not containing $0$ that satisfy properties (1) and (2).
answered yesterday
mathworker21
6,4081727
For ease, I'm replacing your property (2) with $y_i > 0$ for each $i$ (which is the same property for all intents and purposes).
Take $n=2$ and let $E = E_1 sqcup E_2$ for $E_1 = (alpha x_1,alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 > 1$ and $E_2 = (alpha x_1, alpha^2 x_2) : x_1,x_2,alpha > 0, fracx_2x_1^2 < 1$. The point is that the sets $E_1,E_2$ are well-defined: indeed, if $P := (alpha x,alpha^2 y)$ satisfies $fracyx^2 > 1$, then if we represented $P$ also as $(beta fracalpha xbeta, beta^2 fracalpha^2 ybeta^2)$, then $fracfracalpha^2 ybeta^2(fracalpha xbeta)^2 = fracyx^2$. And clearly $E_1,E_2$ are disjoint open sets not containing $0$ that satisfy properties (1) and (2).
answered yesterday
mathworker21
6,4081727
edited 6 hours ago
answered yesterday
mathworker21
6,4081727
answered yesterday
mathworker21
6,4081727
answered yesterday
mathworker21
6,4081727
6,4081727
Condition 2) is not satisfied.
– Kavi Rama Murthy
7 hours ago
@KaviRamaMurthy That property wasn't there when I first posted my answer. But now I've updated my answer accordingly.
– mathworker21
6 hours ago
add a comment |Â
Condition 2) is not satisfied.
– Kavi Rama Murthy
7 hours ago
@KaviRamaMurthy That property wasn't there when I first posted my answer. But now I've updated my answer accordingly.
– mathworker21
6 hours ago
Condition 2) is not satisfied.
– Kavi Rama Murthy
7 hours ago
Condition 2) is not satisfied.
– Kavi Rama Murthy
7 hours ago
@KaviRamaMurthy That property wasn't there when I first posted my answer. But now I've updated my answer accordingly.
– mathworker21
6 hours ago
@KaviRamaMurthy That property wasn't there when I first posted my answer. But now I've updated my answer accordingly.
– mathworker21
6 hours ago
add a comment |Â
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I suspect you are missing an extra hypothesis.
– Lord Shark the Unknown
yesterday
@user9527 see my updated answer
– mathworker21
6 hours ago