Proving $mathbb{N}$ is not compact












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$begingroup$


In $mathbb{N}$, we define the topology defined by $$T=emptyset cup{{0,1,2...,n} :space nin mathbb{N}}cup mathbb{N}$$



Now I want to prove that $(mathbb{N},T)$ is not compact.



Suppose $(mathbb{N},T)$ is compact, then there exists a collecion of open sets $U_i$ so that $mathbb{N}=bigcuplimits_{i=1}^{n} U_{i}$. $U_i$ are of the form ${0,1,2,3,...,i}$ then there exists $U_j>U_i$ for all $i=1,...,n$ $ineq j$. Now take $Min mathbb{N}$ so that $M>x$ for all $xin U_j$ . This $M$ exists because $U_j$ is a finite subset of $mathbb{N}$ and therefore is bounded. Then $Min mathbb{N}$ however $Mnotinbigcuplimits_{i=1}^{infty} U_{i}$. By contradiction I have proven that $(mathbb{N},T)$ is not compact.



Is this proof correct? I was doubting if I could take $mathbb{N}$ as a $U_i$



EDIT



Second try:



Take the open cover $bigcuplimits_{i=1}^{infty} U_{i}=mathbb{N}$ where $U_i={1,2,...,i}$ then by any $ninmathbb{N}$, the open subcover $bigcuplimits_{i=1}^{n} U_{i}$ does not contain $mathbb{N}$ and therefore $(mathbb{N},T)$ is not compact.






general-topology proof-verification compactness







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$endgroup$

















    0












    $begingroup$


    In $mathbb{N}$, we define the topology defined by $$T=emptyset cup{{0,1,2...,n} :space nin mathbb{N}}cup mathbb{N}$$



    Now I want to prove that $(mathbb{N},T)$ is not compact.



    Suppose $(mathbb{N},T)$ is compact, then there exists a collecion of open sets $U_i$ so that $mathbb{N}=bigcuplimits_{i=1}^{n} U_{i}$. $U_i$ are of the form ${0,1,2,3,...,i}$ then there exists $U_j>U_i$ for all $i=1,...,n$ $ineq j$. Now take $Min mathbb{N}$ so that $M>x$ for all $xin U_j$ . This $M$ exists because $U_j$ is a finite subset of $mathbb{N}$ and therefore is bounded. Then $Min mathbb{N}$ however $Mnotinbigcuplimits_{i=1}^{infty} U_{i}$. By contradiction I have proven that $(mathbb{N},T)$ is not compact.



    Is this proof correct? I was doubting if I could take $mathbb{N}$ as a $U_i$



    EDIT



    Second try:



    Take the open cover $bigcuplimits_{i=1}^{infty} U_{i}=mathbb{N}$ where $U_i={1,2,...,i}$ then by any $ninmathbb{N}$, the open subcover $bigcuplimits_{i=1}^{n} U_{i}$ does not contain $mathbb{N}$ and therefore $(mathbb{N},T)$ is not compact.






    general-topology proof-verification compactness







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In $mathbb{N}$, we define the topology defined by $$T=emptyset cup{{0,1,2...,n} :space nin mathbb{N}}cup mathbb{N}$$



      Now I want to prove that $(mathbb{N},T)$ is not compact.



      Suppose $(mathbb{N},T)$ is compact, then there exists a collecion of open sets $U_i$ so that $mathbb{N}=bigcuplimits_{i=1}^{n} U_{i}$. $U_i$ are of the form ${0,1,2,3,...,i}$ then there exists $U_j>U_i$ for all $i=1,...,n$ $ineq j$. Now take $Min mathbb{N}$ so that $M>x$ for all $xin U_j$ . This $M$ exists because $U_j$ is a finite subset of $mathbb{N}$ and therefore is bounded. Then $Min mathbb{N}$ however $Mnotinbigcuplimits_{i=1}^{infty} U_{i}$. By contradiction I have proven that $(mathbb{N},T)$ is not compact.



      Is this proof correct? I was doubting if I could take $mathbb{N}$ as a $U_i$



      EDIT



      Second try:



      Take the open cover $bigcuplimits_{i=1}^{infty} U_{i}=mathbb{N}$ where $U_i={1,2,...,i}$ then by any $ninmathbb{N}$, the open subcover $bigcuplimits_{i=1}^{n} U_{i}$ does not contain $mathbb{N}$ and therefore $(mathbb{N},T)$ is not compact.






      general-topology proof-verification compactness







      share|cite|improve this question











      $endgroup$




      In $mathbb{N}$, we define the topology defined by $$T=emptyset cup{{0,1,2...,n} :space nin mathbb{N}}cup mathbb{N}$$



      Now I want to prove that $(mathbb{N},T)$ is not compact.



      Suppose $(mathbb{N},T)$ is compact, then there exists a collecion of open sets $U_i$ so that $mathbb{N}=bigcuplimits_{i=1}^{n} U_{i}$. $U_i$ are of the form ${0,1,2,3,...,i}$ then there exists $U_j>U_i$ for all $i=1,...,n$ $ineq j$. Now take $Min mathbb{N}$ so that $M>x$ for all $xin U_j$ . This $M$ exists because $U_j$ is a finite subset of $mathbb{N}$ and therefore is bounded. Then $Min mathbb{N}$ however $Mnotinbigcuplimits_{i=1}^{infty} U_{i}$. By contradiction I have proven that $(mathbb{N},T)$ is not compact.



      Is this proof correct? I was doubting if I could take $mathbb{N}$ as a $U_i$



      EDIT



      Second try:



      Take the open cover $bigcuplimits_{i=1}^{infty} U_{i}=mathbb{N}$ where $U_i={1,2,...,i}$ then by any $ninmathbb{N}$, the open subcover $bigcuplimits_{i=1}^{n} U_{i}$ does not contain $mathbb{N}$ and therefore $(mathbb{N},T)$ is not compact.






      general-topology proof-verification compactness




      general-topology proof-verification compactness






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      share|cite|improve this question













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      edited Dec 13 '18 at 16:04







      John Keeper

















      asked Dec 13 '18 at 15:31









      John KeeperJohn Keeper

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          2 Answers
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          $begingroup$

          No, it is not correct. Asserting that $mathbb N$ is compact means that for every family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$, there is a finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. So, assertint that $mathbb N$ is not compact means that there is a family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$ for which there is no finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. This is indeed true. Take $Lambda=mathbb N$ and, for each $ninLambda$, $U_n={0,1,2,ldots,n}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I have added a much shorter proof, is it correct now?
            $endgroup$
            – John Keeper
            Dec 13 '18 at 16:04










          • $begingroup$
            Yes, that proof is correct.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 16:06










          • $begingroup$
            I have to disagree that the 2nd proof is correct. It has the right idea, but falls short of actually proving in two details: Why does $bigcup_{i=0}^n U_i$ not contain $Bbb N$? And you have only argued that certain finite subsets of ${U_i}$ (those with every $U_i$ for $i le n$) are not covers. But you need to show it for all finite subsets of ${U_i}$. Both of these are easy to answer, and maybe you (John Keeper) were just abbreviating. But you will want to fill in the missing details before the proof is correct.
            $endgroup$
            – Paul Sinclair
            Dec 14 '18 at 0:22



















          1












          $begingroup$

          Your proof isn't quite correct. If you assume that $mathbb N$ is compact in this topology, you have to start by assuming that any open cover has a finite subcover. Since not every set in the topology is of the form ${0,1,2,ldots,n}$, you can't assume every set in the cover is of this form, hence your second sentence is invalid.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            1












            $begingroup$

            No, it is not correct. Asserting that $mathbb N$ is compact means that for every family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$, there is a finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. So, assertint that $mathbb N$ is not compact means that there is a family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$ for which there is no finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. This is indeed true. Take $Lambda=mathbb N$ and, for each $ninLambda$, $U_n={0,1,2,ldots,n}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I have added a much shorter proof, is it correct now?
              $endgroup$
              – John Keeper
              Dec 13 '18 at 16:04










            • $begingroup$
              Yes, that proof is correct.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 16:06










            • $begingroup$
              I have to disagree that the 2nd proof is correct. It has the right idea, but falls short of actually proving in two details: Why does $bigcup_{i=0}^n U_i$ not contain $Bbb N$? And you have only argued that certain finite subsets of ${U_i}$ (those with every $U_i$ for $i le n$) are not covers. But you need to show it for all finite subsets of ${U_i}$. Both of these are easy to answer, and maybe you (John Keeper) were just abbreviating. But you will want to fill in the missing details before the proof is correct.
              $endgroup$
              – Paul Sinclair
              Dec 14 '18 at 0:22
















            1












            $begingroup$

            No, it is not correct. Asserting that $mathbb N$ is compact means that for every family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$, there is a finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. So, assertint that $mathbb N$ is not compact means that there is a family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$ for which there is no finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. This is indeed true. Take $Lambda=mathbb N$ and, for each $ninLambda$, $U_n={0,1,2,ldots,n}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I have added a much shorter proof, is it correct now?
              $endgroup$
              – John Keeper
              Dec 13 '18 at 16:04










            • $begingroup$
              Yes, that proof is correct.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 16:06










            • $begingroup$
              I have to disagree that the 2nd proof is correct. It has the right idea, but falls short of actually proving in two details: Why does $bigcup_{i=0}^n U_i$ not contain $Bbb N$? And you have only argued that certain finite subsets of ${U_i}$ (those with every $U_i$ for $i le n$) are not covers. But you need to show it for all finite subsets of ${U_i}$. Both of these are easy to answer, and maybe you (John Keeper) were just abbreviating. But you will want to fill in the missing details before the proof is correct.
              $endgroup$
              – Paul Sinclair
              Dec 14 '18 at 0:22














            1












            1








            1





            $begingroup$

            No, it is not correct. Asserting that $mathbb N$ is compact means that for every family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$, there is a finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. So, assertint that $mathbb N$ is not compact means that there is a family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$ for which there is no finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. This is indeed true. Take $Lambda=mathbb N$ and, for each $ninLambda$, $U_n={0,1,2,ldots,n}$.






            share|cite|improve this answer









            $endgroup$



            No, it is not correct. Asserting that $mathbb N$ is compact means that for every family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$, there is a finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. So, assertint that $mathbb N$ is not compact means that there is a family $(U_lambda)_{lambdainLambda}$ of open subsets of $mathbb N$ whose union contains $mathbb N$ for which there is no finite subset $F$ of $Lambda$ such that $mathbb{N}subsetbigcup_{lambdain F}U_lambda$. This is indeed true. Take $Lambda=mathbb N$ and, for each $ninLambda$, $U_n={0,1,2,ldots,n}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 15:37









            José Carlos SantosJosé Carlos Santos

            164k22131234




            164k22131234












            • $begingroup$
              I have added a much shorter proof, is it correct now?
              $endgroup$
              – John Keeper
              Dec 13 '18 at 16:04










            • $begingroup$
              Yes, that proof is correct.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 16:06










            • $begingroup$
              I have to disagree that the 2nd proof is correct. It has the right idea, but falls short of actually proving in two details: Why does $bigcup_{i=0}^n U_i$ not contain $Bbb N$? And you have only argued that certain finite subsets of ${U_i}$ (those with every $U_i$ for $i le n$) are not covers. But you need to show it for all finite subsets of ${U_i}$. Both of these are easy to answer, and maybe you (John Keeper) were just abbreviating. But you will want to fill in the missing details before the proof is correct.
              $endgroup$
              – Paul Sinclair
              Dec 14 '18 at 0:22


















            • $begingroup$
              I have added a much shorter proof, is it correct now?
              $endgroup$
              – John Keeper
              Dec 13 '18 at 16:04










            • $begingroup$
              Yes, that proof is correct.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 16:06










            • $begingroup$
              I have to disagree that the 2nd proof is correct. It has the right idea, but falls short of actually proving in two details: Why does $bigcup_{i=0}^n U_i$ not contain $Bbb N$? And you have only argued that certain finite subsets of ${U_i}$ (those with every $U_i$ for $i le n$) are not covers. But you need to show it for all finite subsets of ${U_i}$. Both of these are easy to answer, and maybe you (John Keeper) were just abbreviating. But you will want to fill in the missing details before the proof is correct.
              $endgroup$
              – Paul Sinclair
              Dec 14 '18 at 0:22
















            $begingroup$
            I have added a much shorter proof, is it correct now?
            $endgroup$
            – John Keeper
            Dec 13 '18 at 16:04




            $begingroup$
            I have added a much shorter proof, is it correct now?
            $endgroup$
            – John Keeper
            Dec 13 '18 at 16:04












            $begingroup$
            Yes, that proof is correct.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 16:06




            $begingroup$
            Yes, that proof is correct.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 16:06












            $begingroup$
            I have to disagree that the 2nd proof is correct. It has the right idea, but falls short of actually proving in two details: Why does $bigcup_{i=0}^n U_i$ not contain $Bbb N$? And you have only argued that certain finite subsets of ${U_i}$ (those with every $U_i$ for $i le n$) are not covers. But you need to show it for all finite subsets of ${U_i}$. Both of these are easy to answer, and maybe you (John Keeper) were just abbreviating. But you will want to fill in the missing details before the proof is correct.
            $endgroup$
            – Paul Sinclair
            Dec 14 '18 at 0:22




            $begingroup$
            I have to disagree that the 2nd proof is correct. It has the right idea, but falls short of actually proving in two details: Why does $bigcup_{i=0}^n U_i$ not contain $Bbb N$? And you have only argued that certain finite subsets of ${U_i}$ (those with every $U_i$ for $i le n$) are not covers. But you need to show it for all finite subsets of ${U_i}$. Both of these are easy to answer, and maybe you (John Keeper) were just abbreviating. But you will want to fill in the missing details before the proof is correct.
            $endgroup$
            – Paul Sinclair
            Dec 14 '18 at 0:22











            1












            $begingroup$

            Your proof isn't quite correct. If you assume that $mathbb N$ is compact in this topology, you have to start by assuming that any open cover has a finite subcover. Since not every set in the topology is of the form ${0,1,2,ldots,n}$, you can't assume every set in the cover is of this form, hence your second sentence is invalid.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Your proof isn't quite correct. If you assume that $mathbb N$ is compact in this topology, you have to start by assuming that any open cover has a finite subcover. Since not every set in the topology is of the form ${0,1,2,ldots,n}$, you can't assume every set in the cover is of this form, hence your second sentence is invalid.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Your proof isn't quite correct. If you assume that $mathbb N$ is compact in this topology, you have to start by assuming that any open cover has a finite subcover. Since not every set in the topology is of the form ${0,1,2,ldots,n}$, you can't assume every set in the cover is of this form, hence your second sentence is invalid.






                share|cite|improve this answer









                $endgroup$



                Your proof isn't quite correct. If you assume that $mathbb N$ is compact in this topology, you have to start by assuming that any open cover has a finite subcover. Since not every set in the topology is of the form ${0,1,2,ldots,n}$, you can't assume every set in the cover is of this form, hence your second sentence is invalid.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 15:37









                AweyganAweygan

                14.4k21441




                14.4k21441






























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