Countability, separability and Lindelöf of $[-1,1]$ with another topology.

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I am trying to solve the following exercise.




Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.




Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.



If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).



Is this right or I am leaving something?










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    up vote
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    I am trying to solve the following exercise.




    Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.




    Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.



    If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).



    Is this right or I am leaving something?










    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am trying to solve the following exercise.




      Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.




      Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.



      If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).



      Is this right or I am leaving something?










      share|cite|improve this question













      I am trying to solve the following exercise.




      Study the countability axioms, separability axioms and the Lindelöf property in $(X,tau)$, where $X=[-1,1]$ and $Uintau$ if and only if $0notin U$ or $(-1,1)subset U$.




      Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.



      If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).



      Is this right or I am leaving something?







      general-topology proof-verification






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      asked Sep 2 at 10:37









      mkspk

      546617




      546617




















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          A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.



          So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.



          If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.



            So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.



            If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).






            share|cite|improve this answer


























              up vote
              2
              down vote



              accepted










              A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.



              So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.



              If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).






              share|cite|improve this answer
























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.



                So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.



                If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).






                share|cite|improve this answer














                A subset is open when we know $0 notin U$ or $(-1,1) subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.



                So already all subsets of $[-1,1]setminus 0$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $x$ for $-1 le x < 0$ and $0 < x le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.



                If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $frac12$ simultaneously, or note that $frac12$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Sep 2 at 13:23

























                answered Sep 2 at 12:44









                Henno Brandsma

                93.4k342101




                93.4k342101



























                     

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