Showing the closure of a compact subset need not be compactIsn't every subset of a compact space...

How can I fix this gap between bookcases I made?

Modification to Chariots for Heavy Cavalry Analogue for 4-armed race

Can you lasso down a wizard who is using the Levitate spell?

Can Medicine checks be used, with decent rolls, to completely mitigate the risk of death from ongoing damage?

How do I create uniquely male characters?

How is it possible for user's password to be changed after storage was encrypted? (on OS X, Android)

Is it possible to make sharp wind that can cut stuff from afar?

Is it possible to do 50 km distance without any previous training?

How can the DM most effectively choose 1 out of an odd number of players to be targeted by an attack or effect?

Why are 150k or 200k jobs considered good when there are 300k+ births a month?

cryptic clue: mammal sounds like relative consumer (8)

Draw simple lines in Inkscape

Why is the design of haulage companies so “special”?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

N.B. ligature in Latex

New order #4: World

Can I interfere when another PC is about to be attacked?

Where to refill my bottle in India?

DOS, create pipe for stdin/stdout of command.com(or 4dos.com) in C or Batch?

My colleague's body is amazing

Finding files for which a command fails

Should I join an office cleaning event for free?

Why Is Death Allowed In the Matrix?

How can bays and straits be determined in a procedurally generated map?



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 $mathbb{R}$ compact under the co-countable and co-finite topologies?Show that $mathbb{Q}$ 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$Bbb{R}^omega$ 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 $(mathbb{N}, T)$ 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
    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 $mathbb{N}$ 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 $(mathbb{N}, T)$ 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
    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 $mathbb{N}$ 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 $(mathbb{N}, T)$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Austin Mohr

20.8k35299




20.8k35299










asked 4 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
    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 $mathbb{N}$ 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 $A$. $mathbb{N}$ 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
    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 $mathbb{N}$ 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 $A$. $mathbb{N}$ 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
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 $A$. $mathbb{N}$ 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
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 $mathbb{N}$ 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 $mathbb{N}$ 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 $mathscr{B} equiv {B_alpha}$ 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









$endgroup$





















    1












    $begingroup$

    $A={1}$ 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









    $endgroup$














      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179012%2fshowing-the-closure-of-a-compact-subset-need-not-be-compact%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      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 $mathscr{B} equiv {B_alpha}$ 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























              1












              $begingroup$

              $A={1}$ 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









              $endgroup$


















                1












                $begingroup$

                $A={1}$ 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









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $A={1}$ 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 1 hour ago









                  Henno BrandsmaHenno Brandsma

                  115k349125




                  115k349125






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179012%2fshowing-the-closure-of-a-compact-subset-need-not-be-compact%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

                      Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

                      What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...