If $X$ is an infinite set, then $X$ with the discrete topology is not compact












0












$begingroup$


I am new to compact sets, and I had some hard time trying to solve this:



Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.



I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.



Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?



Any help is appreciated.






general-topology







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  • 1




    $begingroup$
    ${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:14








  • 1




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 3:14






  • 2




    $begingroup$
    Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 3:15












  • $begingroup$
    Thank you very much.I find it more clear now.
    $endgroup$
    – Fifi
    Dec 7 '18 at 3:18
















0












$begingroup$


I am new to compact sets, and I had some hard time trying to solve this:



Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.



I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.



Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?



Any help is appreciated.






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    ${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:14








  • 1




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 3:14






  • 2




    $begingroup$
    Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 3:15












  • $begingroup$
    Thank you very much.I find it more clear now.
    $endgroup$
    – Fifi
    Dec 7 '18 at 3:18














0












0








0





$begingroup$


I am new to compact sets, and I had some hard time trying to solve this:



Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.



I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.



Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?



Any help is appreciated.






general-topology







share|cite|improve this question











$endgroup$




I am new to compact sets, and I had some hard time trying to solve this:



Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.



I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.



Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?



Any help is appreciated.






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 19:19







Fifi

















asked Dec 7 '18 at 3:12









FifiFifi

256




256








  • 1




    $begingroup$
    ${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:14








  • 1




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 3:14






  • 2




    $begingroup$
    Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 3:15












  • $begingroup$
    Thank you very much.I find it more clear now.
    $endgroup$
    – Fifi
    Dec 7 '18 at 3:18














  • 1




    $begingroup$
    ${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:14








  • 1




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 3:14






  • 2




    $begingroup$
    Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 3:15












  • $begingroup$
    Thank you very much.I find it more clear now.
    $endgroup$
    – Fifi
    Dec 7 '18 at 3:18








1




1




$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14






$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14






1




1




$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14




$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14




2




2




$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15






$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15














$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18




$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18










1 Answer
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$begingroup$

Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.



It's an open cover because the topology is discrete and so all subsets of $X$ are open.



A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.



Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.



To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.



To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.






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    $begingroup$

    Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.



    It's an open cover because the topology is discrete and so all subsets of $X$ are open.



    A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.



    Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.



    To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.



    To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.



      It's an open cover because the topology is discrete and so all subsets of $X$ are open.



      A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.



      Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.



      To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.



      To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.



        It's an open cover because the topology is discrete and so all subsets of $X$ are open.



        A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.



        Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.



        To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.



        To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.






        share|cite|improve this answer









        $endgroup$



        Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.



        It's an open cover because the topology is discrete and so all subsets of $X$ are open.



        A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.



        Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.



        To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.



        To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Dec 7 '18 at 7:15









        Henno BrandsmaHenno Brandsma

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