Closed set in topological space generated by sets of the form [a, b).












2












$begingroup$


Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.



My effort:



We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?






real-analysis general-topology







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.



    My effort:



    We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?






    real-analysis general-topology







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.



      My effort:



      We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?






      real-analysis general-topology







      share|cite|improve this question









      $endgroup$




      Let $tau$ be the topology on $mathbb{R}$ generated by $B = {[a, b)subset mathbb{R}: -infty<a<b<infty}$. Then the set ${xin mathbb{R}: 4sin^2xleq 1}cup big{frac{pi}{2}big}$ is closed in $(mathbb{R}, tau)$.



      My effort:



      We know that $4sin^2xleq 1$ whenever $-frac{1}{2}leq sin x leq frac{1}{2}$, i.e. $x in big[-frac{pi}{6}, frac{pi}{6}big]cup big[-frac{11pi}{6}, frac{13pi}{6}big]cupcdots$. How to proceed further?






      real-analysis general-topology




      real-analysis general-topology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      Mittal GMittal G

      1,250516




      1,250516






















          2 Answers
          2






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          6












          $begingroup$

          Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know it! And…?
            $endgroup$
            – José Carlos Santos
            4 hours ago










          • $begingroup$
            Sorry! I got it! $(a,b)$ is open in the above topology also.
            $endgroup$
            – Ajay Kumar Nair
            4 hours ago










          • $begingroup$
            Indeed. That's important.
            $endgroup$
            – José Carlos Santos
            4 hours ago



















          2












          $begingroup$

          One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:



          $$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$



          And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I know it! And…?
              $endgroup$
              – José Carlos Santos
              4 hours ago










            • $begingroup$
              Sorry! I got it! $(a,b)$ is open in the above topology also.
              $endgroup$
              – Ajay Kumar Nair
              4 hours ago










            • $begingroup$
              Indeed. That's important.
              $endgroup$
              – José Carlos Santos
              4 hours ago
















            6












            $begingroup$

            Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I know it! And…?
              $endgroup$
              – José Carlos Santos
              4 hours ago










            • $begingroup$
              Sorry! I got it! $(a,b)$ is open in the above topology also.
              $endgroup$
              – Ajay Kumar Nair
              4 hours ago










            • $begingroup$
              Indeed. That's important.
              $endgroup$
              – José Carlos Santos
              4 hours ago














            6












            6








            6





            $begingroup$

            Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.






            share|cite|improve this answer









            $endgroup$



            Since $left{fracpi2right}$ is closed in $(mathbb{R},tau)$, all you need to do is to prove that$$cdotscupleft[-fracpi6,fracpi6right]cupleft[-frac{11pi}{6}, frac{13pi}{6}right]cupcdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(mathbb{R},tau)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            José Carlos SantosJosé Carlos Santos

            164k22131234




            164k22131234












            • $begingroup$
              I know it! And…?
              $endgroup$
              – José Carlos Santos
              4 hours ago










            • $begingroup$
              Sorry! I got it! $(a,b)$ is open in the above topology also.
              $endgroup$
              – Ajay Kumar Nair
              4 hours ago










            • $begingroup$
              Indeed. That's important.
              $endgroup$
              – José Carlos Santos
              4 hours ago


















            • $begingroup$
              I know it! And…?
              $endgroup$
              – José Carlos Santos
              4 hours ago










            • $begingroup$
              Sorry! I got it! $(a,b)$ is open in the above topology also.
              $endgroup$
              – Ajay Kumar Nair
              4 hours ago










            • $begingroup$
              Indeed. That's important.
              $endgroup$
              – José Carlos Santos
              4 hours ago
















            $begingroup$
            I know it! And…?
            $endgroup$
            – José Carlos Santos
            4 hours ago




            $begingroup$
            I know it! And…?
            $endgroup$
            – José Carlos Santos
            4 hours ago












            $begingroup$
            Sorry! I got it! $(a,b)$ is open in the above topology also.
            $endgroup$
            – Ajay Kumar Nair
            4 hours ago




            $begingroup$
            Sorry! I got it! $(a,b)$ is open in the above topology also.
            $endgroup$
            – Ajay Kumar Nair
            4 hours ago












            $begingroup$
            Indeed. That's important.
            $endgroup$
            – José Carlos Santos
            4 hours ago




            $begingroup$
            Indeed. That's important.
            $endgroup$
            – José Carlos Santos
            4 hours ago











            2












            $begingroup$

            One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:



            $$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$



            And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:



              $$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$



              And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:



                $$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$



                And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.






                share|cite|improve this answer









                $endgroup$



                One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:



                $$(a, b) = cup_{n in Bbb N} [a-frac{1}{n}, b).$$



                And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $tau$, so we're done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Robert ShoreRobert Shore

                1,49915




                1,49915






























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