Id $: mathbb{R} rightarrow mathbb{ R}$ is the identity mapping then choose the correct statement












0












$begingroup$


Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.



Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.



Choose the correct statement:



$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous



$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping



My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.



Is it true ?



Any hints/solution will be appreciated.



Thank you.






general-topology







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$endgroup$








  • 2




    $begingroup$
    For (b) push forward $[0,1)$.
    $endgroup$
    – Randall
    Jan 6 at 11:56
















0












$begingroup$


Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.



Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.



Choose the correct statement:



$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous



$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping



My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.



Is it true ?



Any hints/solution will be appreciated.



Thank you.






general-topology







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    For (b) push forward $[0,1)$.
    $endgroup$
    – Randall
    Jan 6 at 11:56














0












0








0





$begingroup$


Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.



Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.



Choose the correct statement:



$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous



$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping



My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.



Is it true ?



Any hints/solution will be appreciated.



Thank you.






general-topology







share|cite|improve this question











$endgroup$




Let $τ_1$ be the topology on $mathbb R$ generated by the base $mathcal B = {[a, b):a,binmathbb{R}text{ and }a<b}$.



Let $τ_0$ be the standard topology on $mathbb{R}$ and let Id $: mathbb{R} rightarrow mathbb{ R}$ be the identity mapping.



Choose the correct statement:



$(a)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is continuous



$(b)$ Id $: (mathbb{R}, τ_1) rightarrow (mathbb R, τ_0)$ is an open mapping



My attempt I thinks both option a) and b) are correct because lower limit topology is finer then usual topology.



Is it true ?



Any hints/solution will be appreciated.



Thank you.






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 12:12









drhab

103k545136




103k545136










asked Jan 6 at 11:48









jasminejasmine

1,890418




1,890418








  • 2




    $begingroup$
    For (b) push forward $[0,1)$.
    $endgroup$
    – Randall
    Jan 6 at 11:56














  • 2




    $begingroup$
    For (b) push forward $[0,1)$.
    $endgroup$
    – Randall
    Jan 6 at 11:56








2




2




$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56




$begingroup$
For (b) push forward $[0,1)$.
$endgroup$
– Randall
Jan 6 at 11:56










3 Answers
3






active

oldest

votes


















0












$begingroup$

If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    (a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
      Draw conclusions.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.






            share|cite|improve this answer









            $endgroup$



            If $U in tau_0$ then we can write $U=cup (a_n,b_n)$ so $U=cup_n cup_j [a_n+frac 1 j,b_n)$. Hence $U in tau_1$. This proves a). Since $[0,1) in tau_1$ but $[0,1) notin tau_0$ it follows that b) is false.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 11:56









            Kavi Rama MurthyKavi Rama Murthy

            69.1k53169




            69.1k53169























                2












                $begingroup$

                (a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  (a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    (a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?






                    share|cite|improve this answer









                    $endgroup$



                    (a) is correct, but if a continuous bijective map is also open, then it is a homeomorphism. Do you think this is a homeomorphism?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 6 at 11:50









                    Matt SamuelMatt Samuel

                    38.9k63769




                    38.9k63769























                        2












                        $begingroup$

                        In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
                        Draw conclusions.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
                          Draw conclusions.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
                            Draw conclusions.






                            share|cite|improve this answer









                            $endgroup$



                            In general:$$mathsf{Id}:(X,tau)to(X,rho)text{ is open }iffmathsf{Id}:(X,rho)to(X,tau)text{ is continuous}ifftausubseteqrho$$
                            Draw conclusions.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 6 at 12:05









                            drhabdrhab

                            103k545136




                            103k545136






























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