Mixing
Limit points of $A={(x,y)mid (x^2+y^2)(y^2-x^2+1)le0}$
$begingroup$
I’m having a hard time trying to figure out accumulation points in this set. I have an easy enough time with accumulation points in 1 dimension but this is really twisting my brain. Any tips?
complex-analysis
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
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add a comment |
$begingroup$
I’m having a hard time trying to figure out accumulation points in this set. I have an easy enough time with accumulation points in 1 dimension but this is really twisting my brain. Any tips?
complex-analysis
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
$endgroup$
add a comment |
$begingroup$
I’m having a hard time trying to figure out accumulation points in this set. I have an easy enough time with accumulation points in 1 dimension but this is really twisting my brain. Any tips?
complex-analysis
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
$endgroup$
I’m having a hard time trying to figure out accumulation points in this set. I have an easy enough time with accumulation points in 1 dimension but this is really twisting my brain. Any tips?
complex-analysis
complex-analysis
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
edited Feb 1 at 19:51
mathcounterexamples.net
26.9k22158
26.9k22158
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
asked Feb 1 at 19:44
Thomas BeesonThomas Beeson
41
41
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
You have
$$A= {(0,0)} cup A^prime$$ where
$$A^prime={(x,y)mid y^2-x^2+1le0}$$
$A^prime$ is a closed subset of $mathbb R^2$ as it is the inverse image of the closed subset $(-infty, 0]$ under the continuous map $(x,y) mapsto y^2-x^2+1$.
As $(0,0) notin A^prime$, $(0,0)$ is not a limit point of $A$.
And as $A^prime$ is closed, the set of limit points of $A$ is included in $A^prime$.
Now, it is fairly easy to prove that all $x in A^prime$ is a limit point of $A$.
Finally the set of limit points of $A$ is $A^prime$.
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
If we break apart the factors.
$(x^2+y^2) ge 0$
The origin in the set.
Or the other factor is less than $0$
$y^2 - x^2 + 1 le 0\
x^2 - y^2 ge 1$
That describes region defined by a hyperbola.
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You have
$$A= {(0,0)} cup A^prime$$ where
$$A^prime={(x,y)mid y^2-x^2+1le0}$$
$A^prime$ is a closed subset of $mathbb R^2$ as it is the inverse image of the closed subset $(-infty, 0]$ under the continuous map $(x,y) mapsto y^2-x^2+1$.
As $(0,0) notin A^prime$, $(0,0)$ is not a limit point of $A$.
And as $A^prime$ is closed, the set of limit points of $A$ is included in $A^prime$.
Now, it is fairly easy to prove that all $x in A^prime$ is a limit point of $A$.
Finally the set of limit points of $A$ is $A^prime$.
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
You have
$$A= {(0,0)} cup A^prime$$ where
$$A^prime={(x,y)mid y^2-x^2+1le0}$$
$A^prime$ is a closed subset of $mathbb R^2$ as it is the inverse image of the closed subset $(-infty, 0]$ under the continuous map $(x,y) mapsto y^2-x^2+1$.
As $(0,0) notin A^prime$, $(0,0)$ is not a limit point of $A$.
And as $A^prime$ is closed, the set of limit points of $A$ is included in $A^prime$.
Now, it is fairly easy to prove that all $x in A^prime$ is a limit point of $A$.
Finally the set of limit points of $A$ is $A^prime$.
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
You have
$$A= {(0,0)} cup A^prime$$ where
$$A^prime={(x,y)mid y^2-x^2+1le0}$$
$A^prime$ is a closed subset of $mathbb R^2$ as it is the inverse image of the closed subset $(-infty, 0]$ under the continuous map $(x,y) mapsto y^2-x^2+1$.
As $(0,0) notin A^prime$, $(0,0)$ is not a limit point of $A$.
And as $A^prime$ is closed, the set of limit points of $A$ is included in $A^prime$.
Now, it is fairly easy to prove that all $x in A^prime$ is a limit point of $A$.
Finally the set of limit points of $A$ is $A^prime$.
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
You have
$$A= {(0,0)} cup A^prime$$ where
$$A^prime={(x,y)mid y^2-x^2+1le0}$$
$A^prime$ is a closed subset of $mathbb R^2$ as it is the inverse image of the closed subset $(-infty, 0]$ under the continuous map $(x,y) mapsto y^2-x^2+1$.
As $(0,0) notin A^prime$, $(0,0)$ is not a limit point of $A$.
And as $A^prime$ is closed, the set of limit points of $A$ is included in $A^prime$.
Now, it is fairly easy to prove that all $x in A^prime$ is a limit point of $A$.
Finally the set of limit points of $A$ is $A^prime$.
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Feb 1 at 20:11
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 1 at 20:01
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
add a comment |
add a comment |
$begingroup$
If we break apart the factors.
$(x^2+y^2) ge 0$
The origin in the set.
Or the other factor is less than $0$
$y^2 - x^2 + 1 le 0\
x^2 - y^2 ge 1$
That describes region defined by a hyperbola.
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
If we break apart the factors.
$(x^2+y^2) ge 0$
The origin in the set.
Or the other factor is less than $0$
$y^2 - x^2 + 1 le 0\
x^2 - y^2 ge 1$
That describes region defined by a hyperbola.
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
If we break apart the factors.
$(x^2+y^2) ge 0$
The origin in the set.
Or the other factor is less than $0$
$y^2 - x^2 + 1 le 0\
x^2 - y^2 ge 1$
That describes region defined by a hyperbola.
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
$endgroup$
If we break apart the factors.
$(x^2+y^2) ge 0$
The origin in the set.
Or the other factor is less than $0$
$y^2 - x^2 + 1 le 0\
x^2 - y^2 ge 1$
That describes region defined by a hyperbola.
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
answered Feb 1 at 21:24
Doug MDoug M
45.4k31954
45.4k31954
add a comment |
add a comment |
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