Determine whether $S cap T$ is closed and bounded












0












$begingroup$


Let $$S={xin mathbb{R} : text{ } x^6-x^5 leq 100}$$



$$T={x^2-2x : text{ } xin (0, infty)}$$



Determine wether $ S cap T$ is closed and bounded?



My Try



$S$ is bounded hence $ S cap T$ is also bounded.



$S=[a,b]$ where $a<-1$ because -1 satisfy the equation given. and $T=(-1, infty)$ hence $ S cap T=[-1,b]$



Hence $ S cap T$ is closed and bounded



However I am not satisfied with my reasoning.
Can anyone tell me the proper solution.






real-analysis calculus analysis elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why is $T=(-2,infty)$? And if it were the case, would there not be $S cap T=(-2,b]$?
    $endgroup$
    – Mindlack
    Dec 18 '18 at 15:49










  • $begingroup$
    In the title you ask for open or closed, in the body you ask for closed and bounded. Which is it?
    $endgroup$
    – Christoph
    Dec 18 '18 at 15:49










  • $begingroup$
    Note that $T=[-1,infty)$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:10












  • $begingroup$
    @cristoph Its closed and bounded
    $endgroup$
    – Rakesh Bhatt
    Dec 18 '18 at 16:19










  • $begingroup$
    You have to justify why $S$ is a bounded interval, and your formula for $T$ is incorrect, take $x=1$ in the defining set.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:21


















0












$begingroup$


Let $$S={xin mathbb{R} : text{ } x^6-x^5 leq 100}$$



$$T={x^2-2x : text{ } xin (0, infty)}$$



Determine wether $ S cap T$ is closed and bounded?



My Try



$S$ is bounded hence $ S cap T$ is also bounded.



$S=[a,b]$ where $a<-1$ because -1 satisfy the equation given. and $T=(-1, infty)$ hence $ S cap T=[-1,b]$



Hence $ S cap T$ is closed and bounded



However I am not satisfied with my reasoning.
Can anyone tell me the proper solution.






real-analysis calculus analysis elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why is $T=(-2,infty)$? And if it were the case, would there not be $S cap T=(-2,b]$?
    $endgroup$
    – Mindlack
    Dec 18 '18 at 15:49










  • $begingroup$
    In the title you ask for open or closed, in the body you ask for closed and bounded. Which is it?
    $endgroup$
    – Christoph
    Dec 18 '18 at 15:49










  • $begingroup$
    Note that $T=[-1,infty)$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:10












  • $begingroup$
    @cristoph Its closed and bounded
    $endgroup$
    – Rakesh Bhatt
    Dec 18 '18 at 16:19










  • $begingroup$
    You have to justify why $S$ is a bounded interval, and your formula for $T$ is incorrect, take $x=1$ in the defining set.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:21
















0












0








0





$begingroup$


Let $$S={xin mathbb{R} : text{ } x^6-x^5 leq 100}$$



$$T={x^2-2x : text{ } xin (0, infty)}$$



Determine wether $ S cap T$ is closed and bounded?



My Try



$S$ is bounded hence $ S cap T$ is also bounded.



$S=[a,b]$ where $a<-1$ because -1 satisfy the equation given. and $T=(-1, infty)$ hence $ S cap T=[-1,b]$



Hence $ S cap T$ is closed and bounded



However I am not satisfied with my reasoning.
Can anyone tell me the proper solution.






real-analysis calculus analysis elementary-set-theory







share|cite|improve this question











$endgroup$




Let $$S={xin mathbb{R} : text{ } x^6-x^5 leq 100}$$



$$T={x^2-2x : text{ } xin (0, infty)}$$



Determine wether $ S cap T$ is closed and bounded?



My Try



$S$ is bounded hence $ S cap T$ is also bounded.



$S=[a,b]$ where $a<-1$ because -1 satisfy the equation given. and $T=(-1, infty)$ hence $ S cap T=[-1,b]$



Hence $ S cap T$ is closed and bounded



However I am not satisfied with my reasoning.
Can anyone tell me the proper solution.






real-analysis calculus analysis elementary-set-theory




real-analysis calculus analysis elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 16:17







Rakesh Bhatt

















asked Dec 18 '18 at 15:46









Rakesh BhattRakesh Bhatt

952214




952214












  • $begingroup$
    Why is $T=(-2,infty)$? And if it were the case, would there not be $S cap T=(-2,b]$?
    $endgroup$
    – Mindlack
    Dec 18 '18 at 15:49










  • $begingroup$
    In the title you ask for open or closed, in the body you ask for closed and bounded. Which is it?
    $endgroup$
    – Christoph
    Dec 18 '18 at 15:49










  • $begingroup$
    Note that $T=[-1,infty)$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:10












  • $begingroup$
    @cristoph Its closed and bounded
    $endgroup$
    – Rakesh Bhatt
    Dec 18 '18 at 16:19










  • $begingroup$
    You have to justify why $S$ is a bounded interval, and your formula for $T$ is incorrect, take $x=1$ in the defining set.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:21




















  • $begingroup$
    Why is $T=(-2,infty)$? And if it were the case, would there not be $S cap T=(-2,b]$?
    $endgroup$
    – Mindlack
    Dec 18 '18 at 15:49










  • $begingroup$
    In the title you ask for open or closed, in the body you ask for closed and bounded. Which is it?
    $endgroup$
    – Christoph
    Dec 18 '18 at 15:49










  • $begingroup$
    Note that $T=[-1,infty)$.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:10












  • $begingroup$
    @cristoph Its closed and bounded
    $endgroup$
    – Rakesh Bhatt
    Dec 18 '18 at 16:19










  • $begingroup$
    You have to justify why $S$ is a bounded interval, and your formula for $T$ is incorrect, take $x=1$ in the defining set.
    $endgroup$
    – copper.hat
    Dec 18 '18 at 16:21


















$begingroup$
Why is $T=(-2,infty)$? And if it were the case, would there not be $S cap T=(-2,b]$?
$endgroup$
– Mindlack
Dec 18 '18 at 15:49




$begingroup$
Why is $T=(-2,infty)$? And if it were the case, would there not be $S cap T=(-2,b]$?
$endgroup$
– Mindlack
Dec 18 '18 at 15:49












$begingroup$
In the title you ask for open or closed, in the body you ask for closed and bounded. Which is it?
$endgroup$
– Christoph
Dec 18 '18 at 15:49




$begingroup$
In the title you ask for open or closed, in the body you ask for closed and bounded. Which is it?
$endgroup$
– Christoph
Dec 18 '18 at 15:49












$begingroup$
Note that $T=[-1,infty)$.
$endgroup$
– copper.hat
Dec 18 '18 at 16:10






$begingroup$
Note that $T=[-1,infty)$.
$endgroup$
– copper.hat
Dec 18 '18 at 16:10














$begingroup$
@cristoph Its closed and bounded
$endgroup$
– Rakesh Bhatt
Dec 18 '18 at 16:19




$begingroup$
@cristoph Its closed and bounded
$endgroup$
– Rakesh Bhatt
Dec 18 '18 at 16:19












$begingroup$
You have to justify why $S$ is a bounded interval, and your formula for $T$ is incorrect, take $x=1$ in the defining set.
$endgroup$
– copper.hat
Dec 18 '18 at 16:21






$begingroup$
You have to justify why $S$ is a bounded interval, and your formula for $T$ is incorrect, take $x=1$ in the defining set.
$endgroup$
– copper.hat
Dec 18 '18 at 16:21












1 Answer
1






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1












$begingroup$

Let $s(x) = x^6-x^5$, since $s$ is continuous, $S=s^{-1} ((-infty, 100])$ is closed.



Since $s(x) to infty$ as $|x| to infty$ we see that $S$ is bounded, hence compact.



Let $t(x) = x^2-2x = x(x-2)$, differentiating shows that $t$ has a $min$ of $-1$. at $x=1$,
and we see that $t(x) to infty$ as $x to infty$. Hence $T=[-1,infty)$, which is
closed.



Hence $T cap S$ is closed.






share|cite|improve this answer









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    $begingroup$

    Let $s(x) = x^6-x^5$, since $s$ is continuous, $S=s^{-1} ((-infty, 100])$ is closed.



    Since $s(x) to infty$ as $|x| to infty$ we see that $S$ is bounded, hence compact.



    Let $t(x) = x^2-2x = x(x-2)$, differentiating shows that $t$ has a $min$ of $-1$. at $x=1$,
    and we see that $t(x) to infty$ as $x to infty$. Hence $T=[-1,infty)$, which is
    closed.



    Hence $T cap S$ is closed.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $s(x) = x^6-x^5$, since $s$ is continuous, $S=s^{-1} ((-infty, 100])$ is closed.



      Since $s(x) to infty$ as $|x| to infty$ we see that $S$ is bounded, hence compact.



      Let $t(x) = x^2-2x = x(x-2)$, differentiating shows that $t$ has a $min$ of $-1$. at $x=1$,
      and we see that $t(x) to infty$ as $x to infty$. Hence $T=[-1,infty)$, which is
      closed.



      Hence $T cap S$ is closed.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $s(x) = x^6-x^5$, since $s$ is continuous, $S=s^{-1} ((-infty, 100])$ is closed.



        Since $s(x) to infty$ as $|x| to infty$ we see that $S$ is bounded, hence compact.



        Let $t(x) = x^2-2x = x(x-2)$, differentiating shows that $t$ has a $min$ of $-1$. at $x=1$,
        and we see that $t(x) to infty$ as $x to infty$. Hence $T=[-1,infty)$, which is
        closed.



        Hence $T cap S$ is closed.






        share|cite|improve this answer









        $endgroup$



        Let $s(x) = x^6-x^5$, since $s$ is continuous, $S=s^{-1} ((-infty, 100])$ is closed.



        Since $s(x) to infty$ as $|x| to infty$ we see that $S$ is bounded, hence compact.



        Let $t(x) = x^2-2x = x(x-2)$, differentiating shows that $t$ has a $min$ of $-1$. at $x=1$,
        and we see that $t(x) to infty$ as $x to infty$. Hence $T=[-1,infty)$, which is
        closed.



        Hence $T cap S$ is closed.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 16:26









        copper.hatcopper.hat

        127k559160




        127k559160






























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