Is this set open, closed, or neither?












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The preimage of an open set of a continuous function is open.
    $endgroup$
    – user295959
    Jan 24 at 18:40
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The preimage of an open set of a continuous function is open.
    $endgroup$
    – user295959
    Jan 24 at 18:40














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 18:39







Fortox

















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














  • 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










1 Answer
1






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3












$begingroup$

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.






share|cite|improve this answer









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    1 Answer
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    1






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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.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 18:42









        José Carlos SantosJosé Carlos Santos

        168k22132236




        168k22132236






























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