Showing the closure of a compact subset need not be compact












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



Say we have the topological space $(mathbb{N}, T)$ 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 $overline{A} = Bbb N$



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)










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  • 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 $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    5 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
    5 hours ago






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    $begingroup$
    I'm not sure what you mean, but you should use this definition of compactness. I've included an argument for why $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    5 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 $(mathbb{N}, T)$ 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 $overline{A} = Bbb N$



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 $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    5 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
    5 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 $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    5 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 $(mathbb{N}, T)$ 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 $overline{A} = Bbb N$



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 $(mathbb{N}, T)$ 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 $overline{A} = Bbb N$



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






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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 $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    5 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
    5 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 $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    5 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 $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
    $endgroup$
    – parsiad
    5 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
    5 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 $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
    $endgroup$
    – parsiad
    5 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










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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 $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
5 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 $A$. $mathbb{N}$ is not compact in your topology because the infinite cover ${ {1, ldots, n} }_{n=1}^infty$ of $mathbb{N}$ cannot be reduce to a finite subcover.
$endgroup$
– parsiad
5 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
5 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
5 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 $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
5 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 $mathbb{N}$ is not compact in my comment above that you can revisit after understanding the linked definition.
$endgroup$
– parsiad
5 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
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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 $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
Let $k$ be a member of $A$.
Since $mathscr{B}$ 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 $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
This shows that $mathbb{N}$ 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, $overline{A} = mathbb{N}$.



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.






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



    $overline{A} = mathbb N$ 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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    • $begingroup$
      To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
      $endgroup$
      – DanielWainfleet
      8 mins ago














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    2 Answers
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    2 Answers
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    $begingroup$


    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 $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
    Let $k$ be a member of $A$.
    Since $mathscr{B}$ 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 $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
    Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
    Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
    This shows that $mathbb{N}$ 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, $overline{A} = mathbb{N}$.



    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 $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
      Let $k$ be a member of $A$.
      Since $mathscr{B}$ 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 $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
      Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
      Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
      This shows that $mathbb{N}$ 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, $overline{A} = mathbb{N}$.



      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 $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
        Let $k$ be a member of $A$.
        Since $mathscr{B}$ 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 $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
        Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
        Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
        This shows that $mathbb{N}$ 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, $overline{A} = mathbb{N}$.



        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 $mathscr{B} equiv {B_alpha}$ be an open cover of $A$.
        Let $k$ be a member of $A$.
        Since $mathscr{B}$ 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 $mathscr{C} equiv {C_n}_{n=1}^infty$ of $mathbb{N}$.
        Let $mathscr{C}^prime$ be a finite subcollection of $mathscr{C}$.
        Note that the set $bigcup_{C in mathscr{C}^prime} C$ has a maximum element (call it $N$) and hence this subcollection does not cover $mathbb{N}$ (because none of its members contain $N+1$).
        This shows that $mathbb{N}$ 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, $overline{A} = mathbb{N}$.



        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.



            $overline{A} = mathbb N$ 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









            $endgroup$













            • $begingroup$
              To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
              $endgroup$
              – DanielWainfleet
              8 mins ago


















            2












            $begingroup$

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



            $overline{A} = mathbb N$ 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









            $endgroup$













            • $begingroup$
              To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
              $endgroup$
              – DanielWainfleet
              8 mins ago
















            2












            2








            2





            $begingroup$

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



            $overline{A} = mathbb N$ 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.



            $overline{A} = mathbb N$ 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 2 hours ago









            Henno BrandsmaHenno Brandsma

            115k349125




            115k349125












            • $begingroup$
              To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
              $endgroup$
              – DanielWainfleet
              8 mins ago




















            • $begingroup$
              To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
              $endgroup$
              – DanielWainfleet
              8 mins ago


















            $begingroup$
            To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
            $endgroup$
            – DanielWainfleet
            8 mins ago






            $begingroup$
            To the proposer: No infinite $Bsubset Bbb N$ is compact in this topology because ${{1,b}: bin B}$ is an open cover of $B$ with no finite sub-cover. And any finite subset (in any topological space) is compact. So in this topology on $Bbb N,$ a subset is compact iff it is finite. So if $1in Csubset Bbb N$ and $C$ is finite then $C$ is compact but $overline C=Bbb N$ is not compact.
            $endgroup$
            – DanielWainfleet
            8 mins ago




















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