Proof on why a certain topological space is compact












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I'm studying for a test on general topology and in some old exam the following question has been asked:



For any finite subset $Ssubseteqmathbb{Z}setminus{0}$ we define $U_S:=mathbb{Z}setminus S$. Let $mathcal{T}$ be the set defined by $mathcal{T}={emptyset}cup{U_Smid Ssubseteqmathbb{Z}setminus{0} text{ finite}}$.



(a) Show that $mathcal{T}$ defines a topology on $mathbb{Z}$

(b) Show that $(mathbb{Z},mathcal{T})$ is compact



I've managed to do (a). However (b) is not really working out. Can someone give me a good tip on how to solve this, thanks in advance.






general-topology







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    0












    $begingroup$


    I'm studying for a test on general topology and in some old exam the following question has been asked:



    For any finite subset $Ssubseteqmathbb{Z}setminus{0}$ we define $U_S:=mathbb{Z}setminus S$. Let $mathcal{T}$ be the set defined by $mathcal{T}={emptyset}cup{U_Smid Ssubseteqmathbb{Z}setminus{0} text{ finite}}$.



    (a) Show that $mathcal{T}$ defines a topology on $mathbb{Z}$

    (b) Show that $(mathbb{Z},mathcal{T})$ is compact



    I've managed to do (a). However (b) is not really working out. Can someone give me a good tip on how to solve this, thanks in advance.






    general-topology







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      0



      $begingroup$


      I'm studying for a test on general topology and in some old exam the following question has been asked:



      For any finite subset $Ssubseteqmathbb{Z}setminus{0}$ we define $U_S:=mathbb{Z}setminus S$. Let $mathcal{T}$ be the set defined by $mathcal{T}={emptyset}cup{U_Smid Ssubseteqmathbb{Z}setminus{0} text{ finite}}$.



      (a) Show that $mathcal{T}$ defines a topology on $mathbb{Z}$

      (b) Show that $(mathbb{Z},mathcal{T})$ is compact



      I've managed to do (a). However (b) is not really working out. Can someone give me a good tip on how to solve this, thanks in advance.






      general-topology







      share|cite|improve this question









      $endgroup$




      I'm studying for a test on general topology and in some old exam the following question has been asked:



      For any finite subset $Ssubseteqmathbb{Z}setminus{0}$ we define $U_S:=mathbb{Z}setminus S$. Let $mathcal{T}$ be the set defined by $mathcal{T}={emptyset}cup{U_Smid Ssubseteqmathbb{Z}setminus{0} text{ finite}}$.



      (a) Show that $mathcal{T}$ defines a topology on $mathbb{Z}$

      (b) Show that $(mathbb{Z},mathcal{T})$ is compact



      I've managed to do (a). However (b) is not really working out. Can someone give me a good tip on how to solve this, thanks in advance.






      general-topology




      general-topology






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      share|cite|improve this question











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      asked Jan 11 at 11:53









      Sander KortewegSander Korteweg

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      315






















          2 Answers
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          $begingroup$

          Let ${U_i, i in I}$ be an open cover of $X$. Then some open set $U_{i_0}$ must contain $0$, this $U_{i_0}$ must be of the form $U_S$ for some finite $S subseteq mathbb{Z}setminus{0}$ by the definition of the topology.



          For each $s in S$ there must be some $U_{i_s}$ that contains $s$ (as we have a cover), and then ${U_{i_0}, U_s: s in S}$ is a finite subcover (any $x$ in $mathbb{Z}$ is in $S$ and so covered or else it’s in $U_S=U_{i_0}$) of our cover and we are done.



          Note that this topology is a weird modification of the co-finite (or finite-closed) topology on $mathbb{Z}$, modified so that all non-empty open sets contain $0$, so ${0}$ is a dense subset of $X$. $X$ is thus only $T_0$, not $T_1$ as the co-finite topology is.






          share|cite|improve this answer









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            1












            $begingroup$

            Let ${A_lambda,|,lambdainLambda}$ be an open cover of $mathbb Z$. If $A_lambda=mathbb Z$ for every $lambdainLambda$, then ${A_{lambda_0}}$ (where $lambda_0$ is some element of $Lambda$) is an open subcover. Otherwise, take $lambda_0inLambda$ such that $A_{lambda_0}neqmathbb Z$. Then $mathbb{Z}setminus A_{lambda_0}$ is finite. So…






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              2 Answers
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              2 Answers
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              active

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              2












              $begingroup$

              Let ${U_i, i in I}$ be an open cover of $X$. Then some open set $U_{i_0}$ must contain $0$, this $U_{i_0}$ must be of the form $U_S$ for some finite $S subseteq mathbb{Z}setminus{0}$ by the definition of the topology.



              For each $s in S$ there must be some $U_{i_s}$ that contains $s$ (as we have a cover), and then ${U_{i_0}, U_s: s in S}$ is a finite subcover (any $x$ in $mathbb{Z}$ is in $S$ and so covered or else it’s in $U_S=U_{i_0}$) of our cover and we are done.



              Note that this topology is a weird modification of the co-finite (or finite-closed) topology on $mathbb{Z}$, modified so that all non-empty open sets contain $0$, so ${0}$ is a dense subset of $X$. $X$ is thus only $T_0$, not $T_1$ as the co-finite topology is.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Let ${U_i, i in I}$ be an open cover of $X$. Then some open set $U_{i_0}$ must contain $0$, this $U_{i_0}$ must be of the form $U_S$ for some finite $S subseteq mathbb{Z}setminus{0}$ by the definition of the topology.



                For each $s in S$ there must be some $U_{i_s}$ that contains $s$ (as we have a cover), and then ${U_{i_0}, U_s: s in S}$ is a finite subcover (any $x$ in $mathbb{Z}$ is in $S$ and so covered or else it’s in $U_S=U_{i_0}$) of our cover and we are done.



                Note that this topology is a weird modification of the co-finite (or finite-closed) topology on $mathbb{Z}$, modified so that all non-empty open sets contain $0$, so ${0}$ is a dense subset of $X$. $X$ is thus only $T_0$, not $T_1$ as the co-finite topology is.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let ${U_i, i in I}$ be an open cover of $X$. Then some open set $U_{i_0}$ must contain $0$, this $U_{i_0}$ must be of the form $U_S$ for some finite $S subseteq mathbb{Z}setminus{0}$ by the definition of the topology.



                  For each $s in S$ there must be some $U_{i_s}$ that contains $s$ (as we have a cover), and then ${U_{i_0}, U_s: s in S}$ is a finite subcover (any $x$ in $mathbb{Z}$ is in $S$ and so covered or else it’s in $U_S=U_{i_0}$) of our cover and we are done.



                  Note that this topology is a weird modification of the co-finite (or finite-closed) topology on $mathbb{Z}$, modified so that all non-empty open sets contain $0$, so ${0}$ is a dense subset of $X$. $X$ is thus only $T_0$, not $T_1$ as the co-finite topology is.






                  share|cite|improve this answer









                  $endgroup$



                  Let ${U_i, i in I}$ be an open cover of $X$. Then some open set $U_{i_0}$ must contain $0$, this $U_{i_0}$ must be of the form $U_S$ for some finite $S subseteq mathbb{Z}setminus{0}$ by the definition of the topology.



                  For each $s in S$ there must be some $U_{i_s}$ that contains $s$ (as we have a cover), and then ${U_{i_0}, U_s: s in S}$ is a finite subcover (any $x$ in $mathbb{Z}$ is in $S$ and so covered or else it’s in $U_S=U_{i_0}$) of our cover and we are done.



                  Note that this topology is a weird modification of the co-finite (or finite-closed) topology on $mathbb{Z}$, modified so that all non-empty open sets contain $0$, so ${0}$ is a dense subset of $X$. $X$ is thus only $T_0$, not $T_1$ as the co-finite topology is.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 16:51









                  Henno BrandsmaHenno Brandsma

                  112k348120




                  112k348120























                      1












                      $begingroup$

                      Let ${A_lambda,|,lambdainLambda}$ be an open cover of $mathbb Z$. If $A_lambda=mathbb Z$ for every $lambdainLambda$, then ${A_{lambda_0}}$ (where $lambda_0$ is some element of $Lambda$) is an open subcover. Otherwise, take $lambda_0inLambda$ such that $A_{lambda_0}neqmathbb Z$. Then $mathbb{Z}setminus A_{lambda_0}$ is finite. So…






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Let ${A_lambda,|,lambdainLambda}$ be an open cover of $mathbb Z$. If $A_lambda=mathbb Z$ for every $lambdainLambda$, then ${A_{lambda_0}}$ (where $lambda_0$ is some element of $Lambda$) is an open subcover. Otherwise, take $lambda_0inLambda$ such that $A_{lambda_0}neqmathbb Z$. Then $mathbb{Z}setminus A_{lambda_0}$ is finite. So…






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let ${A_lambda,|,lambdainLambda}$ be an open cover of $mathbb Z$. If $A_lambda=mathbb Z$ for every $lambdainLambda$, then ${A_{lambda_0}}$ (where $lambda_0$ is some element of $Lambda$) is an open subcover. Otherwise, take $lambda_0inLambda$ such that $A_{lambda_0}neqmathbb Z$. Then $mathbb{Z}setminus A_{lambda_0}$ is finite. So…






                          share|cite|improve this answer









                          $endgroup$



                          Let ${A_lambda,|,lambdainLambda}$ be an open cover of $mathbb Z$. If $A_lambda=mathbb Z$ for every $lambdainLambda$, then ${A_{lambda_0}}$ (where $lambda_0$ is some element of $Lambda$) is an open subcover. Otherwise, take $lambda_0inLambda$ such that $A_{lambda_0}neqmathbb Z$. Then $mathbb{Z}setminus A_{lambda_0}$ is finite. So…







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 11 at 11:58









                          José Carlos SantosJosé Carlos Santos

                          166k22132235




                          166k22132235






























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