Showing the closure of a compact subset need not be compactIsn't every subset of a compact space compact?Example on closure of a subset of a subspace of a topological space in Munkres's TopologyCompact subset of a non compact topological spaceWhy subspace of a compact space not compactIs $mathbbR$ compact under the co-countable and co-finite topologies?Show that $mathbbQ$ is not locally compact with a characterization of local compactnessIs the closure of a compact set compact?A topological space is locally compact then here is an open base at each point has all of its set with compact closure$BbbR^omega$ is not Locally CompactShowing a subset of $Bbb R^2$ is compact with the relative topology

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Showing the closure of a compact subset need not be compact


Isn't every subset of a compact space compact?Example on closure of a subset of a subspace of a topological space in Munkres's TopologyCompact subset of a non compact topological spaceWhy subspace of a compact space not compactIs $mathbbR$ compact under the co-countable and co-finite topologies?Show that $mathbbQ$ is not locally compact with a characterization of local compactnessIs the closure of a compact set compact?A topological space is locally compact then here is an open base at each point has all of its set with compact closure$BbbR^omega$ is not Locally CompactShowing a subset of $Bbb R^2$ is compact with the relative topology













1












$begingroup$


Could someone tell me if my line of reasoning is correct here:



Say we have the topological space $(mathbbN, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.



I want to show that the closure of a compact set in this topology need not be compact.



Let $A equiv 1, ...., n$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overlineA = Bbb N$



If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)










share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    Your reasoning for $A$ being compact seems incorrect. $A = 1, ldots, n$ is compact because given an open cover $ B_alpha $ of $A$, we can find $B_alpha_k ni k$ for each $k in A$ so that $ B_alpha_k _k=1^n$ covers $A$. $mathbbN$ is not compact in your topology because the infinite cover $ 1, ldots, n _n=1^infty$ of $mathbbN$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover $B_alpha$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
    $endgroup$
    – can'tcauchy
    4 hours ago






  • 1




    $begingroup$
    I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbbN$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A=1,...n as being able to be covered by the set 1,....n in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
    $endgroup$
    – can'tcauchy
    4 hours ago











  • $begingroup$
    Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
    $endgroup$
    – parsiad
    4 hours ago
















1












$begingroup$


Could someone tell me if my line of reasoning is correct here:



Say we have the topological space $(mathbbN, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.



I want to show that the closure of a compact set in this topology need not be compact.



Let $A equiv 1, ...., n$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overlineA = Bbb N$



If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)










share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    Your reasoning for $A$ being compact seems incorrect. $A = 1, ldots, n$ is compact because given an open cover $ B_alpha $ of $A$, we can find $B_alpha_k ni k$ for each $k in A$ so that $ B_alpha_k _k=1^n$ covers $A$. $mathbbN$ is not compact in your topology because the infinite cover $ 1, ldots, n _n=1^infty$ of $mathbbN$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover $B_alpha$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
    $endgroup$
    – can'tcauchy
    4 hours ago






  • 1




    $begingroup$
    I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbbN$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A=1,...n as being able to be covered by the set 1,....n in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
    $endgroup$
    – can'tcauchy
    4 hours ago











  • $begingroup$
    Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
    $endgroup$
    – parsiad
    4 hours ago














1












1








1





$begingroup$


Could someone tell me if my line of reasoning is correct here:



Say we have the topological space $(mathbbN, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.



I want to show that the closure of a compact set in this topology need not be compact.



Let $A equiv 1, ...., n$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overlineA = Bbb N$



If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)










share|cite|improve this question











$endgroup$




Could someone tell me if my line of reasoning is correct here:



Say we have the topological space $(mathbbN, T)$ comprising of the empty set together with all subsets of $Bbb N$ that contain the element $1$.



I want to show that the closure of a compact set in this topology need not be compact.



Let $A equiv 1, ...., n$.
Then, this set is compact as it can be covered by a single open set in $T$.
Its limit points are every number in $Bbb N$ which is not 1.
Therefore, its closure is $overlineA = Bbb N$



If the above is true then I get confused because it seems that $Bbb N$ is an open set in $T$ so therefore can it not also be covered with a single open set in T and so would be compact as well? (I know intuitively that compact, being in some sense a measure of 'smallness', would indicate that $Bbb N$ shouldn't be compact but I don't see how to get that line of reasoning using the properties of this topology)







general-topology proof-writing compactness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Austin Mohr

20.8k35299




20.8k35299










asked 5 hours ago









can'tcauchycan'tcauchy

1,023417




1,023417







  • 3




    $begingroup$
    Your reasoning for $A$ being compact seems incorrect. $A = 1, ldots, n$ is compact because given an open cover $ B_alpha $ of $A$, we can find $B_alpha_k ni k$ for each $k in A$ so that $ B_alpha_k _k=1^n$ covers $A$. $mathbbN$ is not compact in your topology because the infinite cover $ 1, ldots, n _n=1^infty$ of $mathbbN$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover $B_alpha$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
    $endgroup$
    – can'tcauchy
    4 hours ago






  • 1




    $begingroup$
    I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbbN$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A=1,...n as being able to be covered by the set 1,....n in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
    $endgroup$
    – can'tcauchy
    4 hours ago











  • $begingroup$
    Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
    $endgroup$
    – parsiad
    4 hours ago













  • 3




    $begingroup$
    Your reasoning for $A$ being compact seems incorrect. $A = 1, ldots, n$ is compact because given an open cover $ B_alpha $ of $A$, we can find $B_alpha_k ni k$ for each $k in A$ so that $ B_alpha_k _k=1^n$ covers $A$. $mathbbN$ is not compact in your topology because the infinite cover $ 1, ldots, n _n=1^infty$ of $mathbbN$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover $B_alpha$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
    $endgroup$
    – can'tcauchy
    4 hours ago






  • 1




    $begingroup$
    I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbbN$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    4 hours ago











  • $begingroup$
    @parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A=1,...n as being able to be covered by the set 1,....n in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
    $endgroup$
    – can'tcauchy
    4 hours ago











  • $begingroup$
    Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
    $endgroup$
    – parsiad
    4 hours ago








3




3




$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = 1, ldots, n$ is compact because given an open cover $ B_alpha $ of $A$, we can find $B_alpha_k ni k$ for each $k in A$ so that $ B_alpha_k _k=1^n$ covers $A$. $mathbbN$ is not compact in your topology because the infinite cover $ 1, ldots, n _n=1^infty$ of $mathbbN$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
4 hours ago





$begingroup$
Your reasoning for $A$ being compact seems incorrect. $A = 1, ldots, n$ is compact because given an open cover $ B_alpha $ of $A$, we can find $B_alpha_k ni k$ for each $k in A$ so that $ B_alpha_k _k=1^n$ covers $A$. $mathbbN$ is not compact in your topology because the infinite cover $ 1, ldots, n _n=1^infty$ of $mathbbN$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
4 hours ago













$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover $B_alpha$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
$endgroup$
– can'tcauchy
4 hours ago




$begingroup$
@parsiad Ah okay , so I should think of each one of the points in A needing to be covered by an open set in T rather than the set itself being able to be fully covered by a single set. So then using that fact I could then use the rest of my argument to say that $Bbb N$ is not compact as given an open cover $B_alpha$ of $Bbb N$ then for each $nin Bbb N$ we can cover it with a set from T but we need infinitely many to do it for all points so it's not compact ?
$endgroup$
– can'tcauchy
4 hours ago




1




1




$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbbN$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
4 hours ago





$begingroup$
I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbbN$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
4 hours ago













$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A=1,...n as being able to be covered by the set 1,....n in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago





$begingroup$
@parsiad what you said in your comment is what I meant. Thanks ! But maybe the part I was unclear about was that I think I had been considering the set A=1,...n as being able to be covered by the set 1,....n in T, but as it says in the definition we have to think about A as being covered by points in a collection of open sets within A, and this can be reduced to a finite number of open sets which do this . Are the limit point and closure part of my argument is fine ,correct ?
$endgroup$
– can'tcauchy
4 hours ago













$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago





$begingroup$
Yes, I think that part of your argument is correct. I posted a detailed answer anyways.
$endgroup$
– parsiad
4 hours ago











2 Answers
2






active

oldest

votes


















4












$begingroup$


Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
We call $A$ compact if for each collection $ B_alpha subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection $B_alpha_1, ldots, B_alpha_n$ such that $B_alpha_1 cup cdots cup B_alpha_n supset A$.



Remark. We call the original collection an open cover and the subcollection a finite subcover.




Consider the topology in your original question.



Let $A equiv 1, ldots, n$ and $mathscrB equiv B_alpha$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscrB$ is a cover of $A$, we can find $alpha_k$ such that $k in B_alpha_k$.
Therefore, $B_alpha_k_k=1^n$ covers $A$, and hence $A$ is compact.



Next, let $C_n equiv 1,ldots,n$ and consider the cover $mathscrC equiv C_n_n=1^infty$ of $mathbbN$.
Let $mathscrC^prime$ be a finite subcollection of $mathscrC$.
Note that the set $bigcup_C in mathscrC^prime C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbbN$ (because none of its members contain $N+1$).
This shows that $mathbbN$ is not compact.



Next, let $n > 1$.
Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
Therefore, $overlineA = mathbbN$.



In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
This is only possible for non-Hausdorff spaces.
Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    $A=1$ is compact as any cover of it has a one-element subcover.



    $overlineA = mathbb N$ which is not compact, as witnessed by the open cover $$1,2,1,3,1,4,ldots, 1,n, ldots$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.






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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$


      Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
      We call $A$ compact if for each collection $ B_alpha subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection $B_alpha_1, ldots, B_alpha_n$ such that $B_alpha_1 cup cdots cup B_alpha_n supset A$.



      Remark. We call the original collection an open cover and the subcollection a finite subcover.




      Consider the topology in your original question.



      Let $A equiv 1, ldots, n$ and $mathscrB equiv B_alpha$ be an open cover of $A$.
      Let $k$ be a member of $A$.
      Since $mathscrB$ is a cover of $A$, we can find $alpha_k$ such that $k in B_alpha_k$.
      Therefore, $B_alpha_k_k=1^n$ covers $A$, and hence $A$ is compact.



      Next, let $C_n equiv 1,ldots,n$ and consider the cover $mathscrC equiv C_n_n=1^infty$ of $mathbbN$.
      Let $mathscrC^prime$ be a finite subcollection of $mathscrC$.
      Note that the set $bigcup_C in mathscrC^prime C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbbN$ (because none of its members contain $N+1$).
      This shows that $mathbbN$ is not compact.



      Next, let $n > 1$.
      Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
      Therefore, $overlineA = mathbbN$.



      In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
      This is only possible for non-Hausdorff spaces.
      Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.






      share|cite|improve this answer









      $endgroup$

















        4












        $begingroup$


        Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
        We call $A$ compact if for each collection $ B_alpha subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection $B_alpha_1, ldots, B_alpha_n$ such that $B_alpha_1 cup cdots cup B_alpha_n supset A$.



        Remark. We call the original collection an open cover and the subcollection a finite subcover.




        Consider the topology in your original question.



        Let $A equiv 1, ldots, n$ and $mathscrB equiv B_alpha$ be an open cover of $A$.
        Let $k$ be a member of $A$.
        Since $mathscrB$ is a cover of $A$, we can find $alpha_k$ such that $k in B_alpha_k$.
        Therefore, $B_alpha_k_k=1^n$ covers $A$, and hence $A$ is compact.



        Next, let $C_n equiv 1,ldots,n$ and consider the cover $mathscrC equiv C_n_n=1^infty$ of $mathbbN$.
        Let $mathscrC^prime$ be a finite subcollection of $mathscrC$.
        Note that the set $bigcup_C in mathscrC^prime C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbbN$ (because none of its members contain $N+1$).
        This shows that $mathbbN$ is not compact.



        Next, let $n > 1$.
        Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
        Therefore, $overlineA = mathbbN$.



        In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
        This is only possible for non-Hausdorff spaces.
        Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.






        share|cite|improve this answer









        $endgroup$















          4












          4








          4





          $begingroup$


          Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
          We call $A$ compact if for each collection $ B_alpha subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection $B_alpha_1, ldots, B_alpha_n$ such that $B_alpha_1 cup cdots cup B_alpha_n supset A$.



          Remark. We call the original collection an open cover and the subcollection a finite subcover.




          Consider the topology in your original question.



          Let $A equiv 1, ldots, n$ and $mathscrB equiv B_alpha$ be an open cover of $A$.
          Let $k$ be a member of $A$.
          Since $mathscrB$ is a cover of $A$, we can find $alpha_k$ such that $k in B_alpha_k$.
          Therefore, $B_alpha_k_k=1^n$ covers $A$, and hence $A$ is compact.



          Next, let $C_n equiv 1,ldots,n$ and consider the cover $mathscrC equiv C_n_n=1^infty$ of $mathbbN$.
          Let $mathscrC^prime$ be a finite subcollection of $mathscrC$.
          Note that the set $bigcup_C in mathscrC^prime C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbbN$ (because none of its members contain $N+1$).
          This shows that $mathbbN$ is not compact.



          Next, let $n > 1$.
          Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
          Therefore, $overlineA = mathbbN$.



          In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
          This is only possible for non-Hausdorff spaces.
          Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.






          share|cite|improve this answer









          $endgroup$




          Definition. Let $A subset X$ where $(X, tau)$ is a topological space.
          We call $A$ compact if for each collection $ B_alpha subset tau$ such that $cup_alpha B_alpha supset A$, we can find a finite subcollection $B_alpha_1, ldots, B_alpha_n$ such that $B_alpha_1 cup cdots cup B_alpha_n supset A$.



          Remark. We call the original collection an open cover and the subcollection a finite subcover.




          Consider the topology in your original question.



          Let $A equiv 1, ldots, n$ and $mathscrB equiv B_alpha$ be an open cover of $A$.
          Let $k$ be a member of $A$.
          Since $mathscrB$ is a cover of $A$, we can find $alpha_k$ such that $k in B_alpha_k$.
          Therefore, $B_alpha_k_k=1^n$ covers $A$, and hence $A$ is compact.



          Next, let $C_n equiv 1,ldots,n$ and consider the cover $mathscrC equiv C_n_n=1^infty$ of $mathbbN$.
          Let $mathscrC^prime$ be a finite subcollection of $mathscrC$.
          Note that the set $bigcup_C in mathscrC^prime C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbbN$ (because none of its members contain $N+1$).
          This shows that $mathbbN$ is not compact.



          Next, let $n > 1$.
          Since any open set in your topology must have 1 as a member, it follows that $n$ is a limit point of $A$.
          Therefore, $overlineA = mathbbN$.



          In summary, you have just found an example of a topology for which the closure of a compact set is not necessarily compact.
          This is only possible for non-Hausdorff spaces.
          Indeed, your topology is non-Hausdorff since any two non-empty neighbourhoods are not disjoint because they both contain the point 1.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          parsiadparsiad

          18.7k32453




          18.7k32453





















              2












              $begingroup$

              $A=1$ is compact as any cover of it has a one-element subcover.



              $overlineA = mathbb N$ which is not compact, as witnessed by the open cover $$1,2,1,3,1,4,ldots, 1,n, ldots$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                $A=1$ is compact as any cover of it has a one-element subcover.



                $overlineA = mathbb N$ which is not compact, as witnessed by the open cover $$1,2,1,3,1,4,ldots, 1,n, ldots$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  $A=1$ is compact as any cover of it has a one-element subcover.



                  $overlineA = mathbb N$ which is not compact, as witnessed by the open cover $$1,2,1,3,1,4,ldots, 1,n, ldots$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.






                  share|cite|improve this answer









                  $endgroup$



                  $A=1$ is compact as any cover of it has a one-element subcover.



                  $overlineA = mathbb N$ which is not compact, as witnessed by the open cover $$1,2,1,3,1,4,ldots, 1,n, ldots$$ of $mathbb N$ from which we cannot omit a member (or it wouldn't cover), so has no finite subcover.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Henno BrandsmaHenno Brandsma

                  115k349125




                  115k349125



























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