Mixing
Is this set open, closed, or neither?
$begingroup$
Is the set $$S={(x,y,z)in mathbb{R}^3 : frac{exp(x+y^2-z)-1}{x^2+y^2-z^3} > 7}$$
open, closed, or neither (in $mathbb{R}^3$ with the Euclidian metric)?
I think I've managed to show that it is not closed by first setting $y=z=0$ and showing that there is a solution $x_* in (0,1)$ to $e^x-1=7x^2$ by using the intermediate value theorem. Then I said any sequence $(x_n)$ converging to $x_*$ that consists of elements of $(0,x_*)$ has the property that $(x_n, 0, 0) in S$ but $lim_{n to infty}(x_n,0,0) notin S$ therefore $S$ is not closed.
I'm not sure if there is a better way to do this or if $S$ is open or not.
general-topology metric-spaces
asked Jan 24 at 18:34
FortoxFortox
6718
$endgroup$
add a comment |
$begingroup$
Is the set $$S={(x,y,z)in mathbb{R}^3 : frac{exp(x+y^2-z)-1}{x^2+y^2-z^3} > 7}$$
open, closed, or neither (in $mathbb{R}^3$ with the Euclidian metric)?
I think I've managed to show that it is not closed by first setting $y=z=0$ and showing that there is a solution $x_* in (0,1)$ to $e^x-1=7x^2$ by using the intermediate value theorem. Then I said any sequence $(x_n)$ converging to $x_*$ that consists of elements of $(0,x_*)$ has the property that $(x_n, 0, 0) in S$ but $lim_{n to infty}(x_n,0,0) notin S$ therefore $S$ is not closed.
I'm not sure if there is a better way to do this or if $S$ is open or not.
general-topology metric-spaces
asked Jan 24 at 18:34
FortoxFortox
6718
$endgroup$
3
$begingroup$
The preimage of an open set of a continuous function is open.
$endgroup$
– user295959
Jan 24 at 18:40
add a comment |
$begingroup$
Is the set $$S={(x,y,z)in mathbb{R}^3 : frac{exp(x+y^2-z)-1}{x^2+y^2-z^3} > 7}$$
open, closed, or neither (in $mathbb{R}^3$ with the Euclidian metric)?
I think I've managed to show that it is not closed by first setting $y=z=0$ and showing that there is a solution $x_* in (0,1)$ to $e^x-1=7x^2$ by using the intermediate value theorem. Then I said any sequence $(x_n)$ converging to $x_*$ that consists of elements of $(0,x_*)$ has the property that $(x_n, 0, 0) in S$ but $lim_{n to infty}(x_n,0,0) notin S$ therefore $S$ is not closed.
I'm not sure if there is a better way to do this or if $S$ is open or not.
general-topology metric-spaces
asked Jan 24 at 18:34
FortoxFortox
6718
$endgroup$
Is the set $$S={(x,y,z)in mathbb{R}^3 : frac{exp(x+y^2-z)-1}{x^2+y^2-z^3} > 7}$$
open, closed, or neither (in $mathbb{R}^3$ with the Euclidian metric)?
I think I've managed to show that it is not closed by first setting $y=z=0$ and showing that there is a solution $x_* in (0,1)$ to $e^x-1=7x^2$ by using the intermediate value theorem. Then I said any sequence $(x_n)$ converging to $x_*$ that consists of elements of $(0,x_*)$ has the property that $(x_n, 0, 0) in S$ but $lim_{n to infty}(x_n,0,0) notin S$ therefore $S$ is not closed.
I'm not sure if there is a better way to do this or if $S$ is open or not.
general-topology metric-spaces
general-topology metric-spaces
asked Jan 24 at 18:34
FortoxFortox
6718
asked Jan 24 at 18:34
FortoxFortox
6718
edited Jan 24 at 18:39
Fortox
asked Jan 24 at 18:34
FortoxFortox
6718
asked Jan 24 at 18:34
FortoxFortox
6718
asked Jan 24 at 18:34
FortoxFortox
6718
6718
3
$begingroup$
The preimage of an open set of a continuous function is open.
$endgroup$
– user295959
Jan 24 at 18:40
add a comment |
3
$begingroup$
The preimage of an open set of a continuous function is open.
$endgroup$
– user295959
Jan 24 at 18:40
3
3
$begingroup$
The preimage of an open set of a continuous function is open.
$endgroup$
– user295959
Jan 24 at 18:40
$begingroup$
The preimage of an open set of a continuous function is open.
$endgroup$
– user295959
Jan 24 at 18:40
add a comment |
1 Answer
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$begingroup$
Let $W={(x,y,z)inmathbb{R}^3,|,z^3neq x^2+y^2}$. Then $W$ is an open set and if we define$$begin{array}{rccc}fcolon&W&longrightarrow&mathbb R\&(x,y,z)&mapsto&dfrac{exp(x+y^2-z)-1}{x^2+y^2-z^3},end{array}$$then, since $f$ is continuous, $f^{-1}bigl((7,infty)bigr)$ is an open subset of $W$ and therefore an oepn subset of $mathbb{R}^3$. But $S=f^{-1}bigl((7,infty)bigr)$.
It is easy to prove that $Sneqemptyset$ and that $Sneqmathbb{R}^3$. So, since $mathbb{R}^3$ is connected, it follows from the fact that $S$ is open that it cannot be closed.
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $W={(x,y,z)inmathbb{R}^3,|,z^3neq x^2+y^2}$. Then $W$ is an open set and if we define$$begin{array}{rccc}fcolon&W&longrightarrow&mathbb R\&(x,y,z)&mapsto&dfrac{exp(x+y^2-z)-1}{x^2+y^2-z^3},end{array}$$then, since $f$ is continuous, $f^{-1}bigl((7,infty)bigr)$ is an open subset of $W$ and therefore an oepn subset of $mathbb{R}^3$. But $S=f^{-1}bigl((7,infty)bigr)$.
It is easy to prove that $Sneqemptyset$ and that $Sneqmathbb{R}^3$. So, since $mathbb{R}^3$ is connected, it follows from the fact that $S$ is open that it cannot be closed.
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Let $W={(x,y,z)inmathbb{R}^3,|,z^3neq x^2+y^2}$. Then $W$ is an open set and if we define$$begin{array}{rccc}fcolon&W&longrightarrow&mathbb R\&(x,y,z)&mapsto&dfrac{exp(x+y^2-z)-1}{x^2+y^2-z^3},end{array}$$then, since $f$ is continuous, $f^{-1}bigl((7,infty)bigr)$ is an open subset of $W$ and therefore an oepn subset of $mathbb{R}^3$. But $S=f^{-1}bigl((7,infty)bigr)$.
It is easy to prove that $Sneqemptyset$ and that $Sneqmathbb{R}^3$. So, since $mathbb{R}^3$ is connected, it follows from the fact that $S$ is open that it cannot be closed.
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Let $W={(x,y,z)inmathbb{R}^3,|,z^3neq x^2+y^2}$. Then $W$ is an open set and if we define$$begin{array}{rccc}fcolon&W&longrightarrow&mathbb R\&(x,y,z)&mapsto&dfrac{exp(x+y^2-z)-1}{x^2+y^2-z^3},end{array}$$then, since $f$ is continuous, $f^{-1}bigl((7,infty)bigr)$ is an open subset of $W$ and therefore an oepn subset of $mathbb{R}^3$. But $S=f^{-1}bigl((7,infty)bigr)$.
It is easy to prove that $Sneqemptyset$ and that $Sneqmathbb{R}^3$. So, since $mathbb{R}^3$ is connected, it follows from the fact that $S$ is open that it cannot be closed.
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
Let $W={(x,y,z)inmathbb{R}^3,|,z^3neq x^2+y^2}$. Then $W$ is an open set and if we define$$begin{array}{rccc}fcolon&W&longrightarrow&mathbb R\&(x,y,z)&mapsto&dfrac{exp(x+y^2-z)-1}{x^2+y^2-z^3},end{array}$$then, since $f$ is continuous, $f^{-1}bigl((7,infty)bigr)$ is an open subset of $W$ and therefore an oepn subset of $mathbb{R}^3$. But $S=f^{-1}bigl((7,infty)bigr)$.
It is easy to prove that $Sneqemptyset$ and that $Sneqmathbb{R}^3$. So, since $mathbb{R}^3$ is connected, it follows from the fact that $S$ is open that it cannot be closed.
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 18:42
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
add a comment |
add a comment |
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3
$begingroup$
The preimage of an open set of a continuous function is open.
$endgroup$
– user295959
Jan 24 at 18:40