Mixing
Is there a non-empty open set in $mathbb{R}^3$ whose elements satisfy the following equality?
$begingroup$
Is there a non-empty open set $S$ in $mathbb{R}^3$ such that the equality $x+2xz^2=y+2x^2y$ holds for all $(x,y,z)in S$ ? If not, how to see non-existence of such a non-empty open set? Could someone give a clue?
analysis
asked Jan 18 at 11:21
serenusserenus
24316
$endgroup$
add a comment |
$begingroup$
Is there a non-empty open set $S$ in $mathbb{R}^3$ such that the equality $x+2xz^2=y+2x^2y$ holds for all $(x,y,z)in S$ ? If not, how to see non-existence of such a non-empty open set? Could someone give a clue?
analysis
asked Jan 18 at 11:21
serenusserenus
24316
$endgroup$
$begingroup$
Sorry, I should be more explicit!
$endgroup$
– serenus
Jan 18 at 11:27
$begingroup$
I mean, I asked for a non-empty open subset $S$ of $mathbb{R}^3$.
$endgroup$
– serenus
Jan 18 at 11:54
$begingroup$
Suppose $a,binmathbb R$ and $bgt0$. Is there a non-empty open set $S$ in $mathbb R$ such that the equality $a=by$ holds for all $yin S$?
$endgroup$
– bof
Jan 18 at 12:05
$begingroup$
Of course, there is no such an open set $S$ in $mathbb{R}$.
$endgroup$
– serenus
Feb 16 at 20:57
$begingroup$
Then that answers your question, doesn't it? Choose a point $(x_0,y_0,z_0)in S$. Then $T={y:(x_0,y,z_0)in S}$ is a nonempty open subset of $mathbb R$. Let $a=x_0+2x_0z_0^2$ and $b=1+2x_0^2gt0$; then $a=by$ holds for all $yin T$. I thought that was a pretty good hint.
$endgroup$
– bof
Feb 16 at 22:30
add a comment |
$begingroup$
Is there a non-empty open set $S$ in $mathbb{R}^3$ such that the equality $x+2xz^2=y+2x^2y$ holds for all $(x,y,z)in S$ ? If not, how to see non-existence of such a non-empty open set? Could someone give a clue?
analysis
asked Jan 18 at 11:21
serenusserenus
24316
$endgroup$
Is there a non-empty open set $S$ in $mathbb{R}^3$ such that the equality $x+2xz^2=y+2x^2y$ holds for all $(x,y,z)in S$ ? If not, how to see non-existence of such a non-empty open set? Could someone give a clue?
analysis
analysis
asked Jan 18 at 11:21
serenusserenus
24316
asked Jan 18 at 11:21
serenusserenus
24316
edited Jan 18 at 11:26
serenus
asked Jan 18 at 11:21
serenusserenus
24316
asked Jan 18 at 11:21
serenusserenus
24316
asked Jan 18 at 11:21
serenusserenus
24316
24316
$begingroup$
Sorry, I should be more explicit!
$endgroup$
– serenus
Jan 18 at 11:27
$begingroup$
I mean, I asked for a non-empty open subset $S$ of $mathbb{R}^3$.
$endgroup$
– serenus
Jan 18 at 11:54
$begingroup$
Suppose $a,binmathbb R$ and $bgt0$. Is there a non-empty open set $S$ in $mathbb R$ such that the equality $a=by$ holds for all $yin S$?
$endgroup$
– bof
Jan 18 at 12:05
$begingroup$
Of course, there is no such an open set $S$ in $mathbb{R}$.
$endgroup$
– serenus
Feb 16 at 20:57
$begingroup$
Then that answers your question, doesn't it? Choose a point $(x_0,y_0,z_0)in S$. Then $T={y:(x_0,y,z_0)in S}$ is a nonempty open subset of $mathbb R$. Let $a=x_0+2x_0z_0^2$ and $b=1+2x_0^2gt0$; then $a=by$ holds for all $yin T$. I thought that was a pretty good hint.
$endgroup$
– bof
Feb 16 at 22:30
add a comment |
$begingroup$
Sorry, I should be more explicit!
$endgroup$
– serenus
Jan 18 at 11:27
$begingroup$
I mean, I asked for a non-empty open subset $S$ of $mathbb{R}^3$.
$endgroup$
– serenus
Jan 18 at 11:54
$begingroup$
Suppose $a,binmathbb R$ and $bgt0$. Is there a non-empty open set $S$ in $mathbb R$ such that the equality $a=by$ holds for all $yin S$?
$endgroup$
– bof
Jan 18 at 12:05
$begingroup$
Of course, there is no such an open set $S$ in $mathbb{R}$.
$endgroup$
– serenus
Feb 16 at 20:57
$begingroup$
Then that answers your question, doesn't it? Choose a point $(x_0,y_0,z_0)in S$. Then $T={y:(x_0,y,z_0)in S}$ is a nonempty open subset of $mathbb R$. Let $a=x_0+2x_0z_0^2$ and $b=1+2x_0^2gt0$; then $a=by$ holds for all $yin T$. I thought that was a pretty good hint.
$endgroup$
– bof
Feb 16 at 22:30
$begingroup$
Sorry, I should be more explicit!
$endgroup$
– serenus
Jan 18 at 11:27
$begingroup$
Sorry, I should be more explicit!
$endgroup$
– serenus
Jan 18 at 11:27
$begingroup$
I mean, I asked for a non-empty open subset $S$ of $mathbb{R}^3$.
$endgroup$
– serenus
Jan 18 at 11:54
$begingroup$
I mean, I asked for a non-empty open subset $S$ of $mathbb{R}^3$.
$endgroup$
– serenus
Jan 18 at 11:54
$begingroup$
Suppose $a,binmathbb R$ and $bgt0$. Is there a non-empty open set $S$ in $mathbb R$ such that the equality $a=by$ holds for all $yin S$?
$endgroup$
– bof
Jan 18 at 12:05
$begingroup$
Suppose $a,binmathbb R$ and $bgt0$. Is there a non-empty open set $S$ in $mathbb R$ such that the equality $a=by$ holds for all $yin S$?
$endgroup$
– bof
Jan 18 at 12:05
$begingroup$
Of course, there is no such an open set $S$ in $mathbb{R}$.
$endgroup$
– serenus
Feb 16 at 20:57
$begingroup$
Of course, there is no such an open set $S$ in $mathbb{R}$.
$endgroup$
– serenus
Feb 16 at 20:57
$begingroup$
Then that answers your question, doesn't it? Choose a point $(x_0,y_0,z_0)in S$. Then $T={y:(x_0,y,z_0)in S}$ is a nonempty open subset of $mathbb R$. Let $a=x_0+2x_0z_0^2$ and $b=1+2x_0^2gt0$; then $a=by$ holds for all $yin T$. I thought that was a pretty good hint.
$endgroup$
– bof
Feb 16 at 22:30
$begingroup$
Then that answers your question, doesn't it? Choose a point $(x_0,y_0,z_0)in S$. Then $T={y:(x_0,y,z_0)in S}$ is a nonempty open subset of $mathbb R$. Let $a=x_0+2x_0z_0^2$ and $b=1+2x_0^2gt0$; then $a=by$ holds for all $yin T$. I thought that was a pretty good hint.
$endgroup$
– bof
Feb 16 at 22:30
add a comment |
1 Answer
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$begingroup$
Suppose that there is a non- empty open set $S$ with the above properties. Let $(x_0,y_0,z_0) in S$. Since $S$ is open, there is $t>0$ sich that $(x_0,y_0+t,z_0) in S$.
Therefore $x_0+2x_0z_0^2=y_0+t+2x_0^2(y_0+t)$. Since $x_0+2x_0z_0^2=y_0+2x_0^2y_0$, this gives $t+2x_0^2t=0$. This is a contradiction, since $t+2x_0^2t ge t>0.$
answered Jan 18 at 12:05
FredFred
47k1848
$endgroup$
$begingroup$
Thank you for the proof.
$endgroup$
– serenus
Jan 18 at 13:33
add a comment |
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1 Answer
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$begingroup$
Suppose that there is a non- empty open set $S$ with the above properties. Let $(x_0,y_0,z_0) in S$. Since $S$ is open, there is $t>0$ sich that $(x_0,y_0+t,z_0) in S$.
Therefore $x_0+2x_0z_0^2=y_0+t+2x_0^2(y_0+t)$. Since $x_0+2x_0z_0^2=y_0+2x_0^2y_0$, this gives $t+2x_0^2t=0$. This is a contradiction, since $t+2x_0^2t ge t>0.$
answered Jan 18 at 12:05
FredFred
47k1848
$endgroup$
$begingroup$
Thank you for the proof.
$endgroup$
– serenus
Jan 18 at 13:33
add a comment |
$begingroup$
Suppose that there is a non- empty open set $S$ with the above properties. Let $(x_0,y_0,z_0) in S$. Since $S$ is open, there is $t>0$ sich that $(x_0,y_0+t,z_0) in S$.
Therefore $x_0+2x_0z_0^2=y_0+t+2x_0^2(y_0+t)$. Since $x_0+2x_0z_0^2=y_0+2x_0^2y_0$, this gives $t+2x_0^2t=0$. This is a contradiction, since $t+2x_0^2t ge t>0.$
answered Jan 18 at 12:05
FredFred
47k1848
$endgroup$
$begingroup$
Thank you for the proof.
$endgroup$
– serenus
Jan 18 at 13:33
add a comment |
$begingroup$
Suppose that there is a non- empty open set $S$ with the above properties. Let $(x_0,y_0,z_0) in S$. Since $S$ is open, there is $t>0$ sich that $(x_0,y_0+t,z_0) in S$.
Therefore $x_0+2x_0z_0^2=y_0+t+2x_0^2(y_0+t)$. Since $x_0+2x_0z_0^2=y_0+2x_0^2y_0$, this gives $t+2x_0^2t=0$. This is a contradiction, since $t+2x_0^2t ge t>0.$
answered Jan 18 at 12:05
FredFred
47k1848
$endgroup$
Suppose that there is a non- empty open set $S$ with the above properties. Let $(x_0,y_0,z_0) in S$. Since $S$ is open, there is $t>0$ sich that $(x_0,y_0+t,z_0) in S$.
Therefore $x_0+2x_0z_0^2=y_0+t+2x_0^2(y_0+t)$. Since $x_0+2x_0z_0^2=y_0+2x_0^2y_0$, this gives $t+2x_0^2t=0$. This is a contradiction, since $t+2x_0^2t ge t>0.$
answered Jan 18 at 12:05
FredFred
47k1848
answered Jan 18 at 12:05
FredFred
47k1848
answered Jan 18 at 12:05
FredFred
47k1848
answered Jan 18 at 12:05
FredFred
47k1848
47k1848
$begingroup$
Thank you for the proof.
$endgroup$
– serenus
Jan 18 at 13:33
add a comment |
$begingroup$
Thank you for the proof.
$endgroup$
– serenus
Jan 18 at 13:33
$begingroup$
Thank you for the proof.
$endgroup$
– serenus
Jan 18 at 13:33
$begingroup$
Thank you for the proof.
$endgroup$
– serenus
Jan 18 at 13:33
add a comment |
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$begingroup$
Sorry, I should be more explicit!
$endgroup$
– serenus
Jan 18 at 11:27
$begingroup$
I mean, I asked for a non-empty open subset $S$ of $mathbb{R}^3$.
$endgroup$
– serenus
Jan 18 at 11:54
$begingroup$
Suppose $a,binmathbb R$ and $bgt0$. Is there a non-empty open set $S$ in $mathbb R$ such that the equality $a=by$ holds for all $yin S$?
$endgroup$
– bof
Jan 18 at 12:05
$begingroup$
Of course, there is no such an open set $S$ in $mathbb{R}$.
$endgroup$
– serenus
Feb 16 at 20:57
$begingroup$
Then that answers your question, doesn't it? Choose a point $(x_0,y_0,z_0)in S$. Then $T={y:(x_0,y,z_0)in S}$ is a nonempty open subset of $mathbb R$. Let $a=x_0+2x_0z_0^2$ and $b=1+2x_0^2gt0$; then $a=by$ holds for all $yin T$. I thought that was a pretty good hint.
$endgroup$
– bof
Feb 16 at 22:30