The product topological space of two finite closed topologies is not finite closed.












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Let $X_1, X_2$ be infite sets and $tau_1$and $tau_2$ the finite closed topologies on $X_1$ and $X_2$ respectively.
Show that the product topology $tau$ on $X_1times X_2$ is not the finite closed topology.




As $X_1$ is infite then if ${x}in X$ then $X_1setminus{x}intau_1$.
Then ${x}times X_2$ is by definition an open set in the product topology space however its complement is ${x}times X_2$ is not finite as $X_2$ is infinite contradicting the fact ${x}times X_2$ is closed. Therefore ${x}times X_2$ is not open. Then the product topology cannot be the finite closed topology.



Questions:



Is this proof right? If not why not? Which are the alternatives?



Thanks in advance!






general-topology proof-verification proof-writing







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    2












    $begingroup$



    Let $X_1, X_2$ be infite sets and $tau_1$and $tau_2$ the finite closed topologies on $X_1$ and $X_2$ respectively.
    Show that the product topology $tau$ on $X_1times X_2$ is not the finite closed topology.




    As $X_1$ is infite then if ${x}in X$ then $X_1setminus{x}intau_1$.
    Then ${x}times X_2$ is by definition an open set in the product topology space however its complement is ${x}times X_2$ is not finite as $X_2$ is infinite contradicting the fact ${x}times X_2$ is closed. Therefore ${x}times X_2$ is not open. Then the product topology cannot be the finite closed topology.



    Questions:



    Is this proof right? If not why not? Which are the alternatives?



    Thanks in advance!






    general-topology proof-verification proof-writing







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$



      Let $X_1, X_2$ be infite sets and $tau_1$and $tau_2$ the finite closed topologies on $X_1$ and $X_2$ respectively.
      Show that the product topology $tau$ on $X_1times X_2$ is not the finite closed topology.




      As $X_1$ is infite then if ${x}in X$ then $X_1setminus{x}intau_1$.
      Then ${x}times X_2$ is by definition an open set in the product topology space however its complement is ${x}times X_2$ is not finite as $X_2$ is infinite contradicting the fact ${x}times X_2$ is closed. Therefore ${x}times X_2$ is not open. Then the product topology cannot be the finite closed topology.



      Questions:



      Is this proof right? If not why not? Which are the alternatives?



      Thanks in advance!






      general-topology proof-verification proof-writing







      share|cite|improve this question









      $endgroup$





      Let $X_1, X_2$ be infite sets and $tau_1$and $tau_2$ the finite closed topologies on $X_1$ and $X_2$ respectively.
      Show that the product topology $tau$ on $X_1times X_2$ is not the finite closed topology.




      As $X_1$ is infite then if ${x}in X$ then $X_1setminus{x}intau_1$.
      Then ${x}times X_2$ is by definition an open set in the product topology space however its complement is ${x}times X_2$ is not finite as $X_2$ is infinite contradicting the fact ${x}times X_2$ is closed. Therefore ${x}times X_2$ is not open. Then the product topology cannot be the finite closed topology.



      Questions:



      Is this proof right? If not why not? Which are the alternatives?



      Thanks in advance!






      general-topology proof-verification proof-writing




      general-topology proof-verification proof-writing






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 6 at 20:38









      Pedro GomesPedro Gomes

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          $begingroup$

          ${x} times X_2$ is not open, but it is closed. Since it is infinite, we are done.






          share|cite|improve this answer









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            $begingroup$

            ${x} times X_2$ is not open, but it is closed. Since it is infinite, we are done.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              ${x} times X_2$ is not open, but it is closed. Since it is infinite, we are done.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                ${x} times X_2$ is not open, but it is closed. Since it is infinite, we are done.






                share|cite|improve this answer









                $endgroup$



                ${x} times X_2$ is not open, but it is closed. Since it is infinite, we are done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 20:50









                Matt SamuelMatt Samuel

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