Example of a connected topological space with an open singleton set












0














I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:



$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.



I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.










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  • 2




    How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
    – Daniel Schepler
    Nov 21 '18 at 20:20








  • 1




    @DanielSchepler You should turn that into an answer.
    – freakish
    Nov 21 '18 at 20:36










  • Bravo! That does it. Thank you for your help.
    – Heinrich Wagner
    Nov 21 '18 at 21:52










  • As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
    – Henno Brandsma
    Nov 21 '18 at 22:24
















0














I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:



$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.



I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.










share|cite|improve this question


















  • 2




    How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
    – Daniel Schepler
    Nov 21 '18 at 20:20








  • 1




    @DanielSchepler You should turn that into an answer.
    – freakish
    Nov 21 '18 at 20:36










  • Bravo! That does it. Thank you for your help.
    – Heinrich Wagner
    Nov 21 '18 at 21:52










  • As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
    – Henno Brandsma
    Nov 21 '18 at 22:24














0












0








0







I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:



$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.



I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.










share|cite|improve this question













I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:



$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.



I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.







general-topology connectedness






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 21 '18 at 20:10









Heinrich WagnerHeinrich Wagner

24919




24919








  • 2




    How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
    – Daniel Schepler
    Nov 21 '18 at 20:20








  • 1




    @DanielSchepler You should turn that into an answer.
    – freakish
    Nov 21 '18 at 20:36










  • Bravo! That does it. Thank you for your help.
    – Heinrich Wagner
    Nov 21 '18 at 21:52










  • As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
    – Henno Brandsma
    Nov 21 '18 at 22:24














  • 2




    How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
    – Daniel Schepler
    Nov 21 '18 at 20:20








  • 1




    @DanielSchepler You should turn that into an answer.
    – freakish
    Nov 21 '18 at 20:36










  • Bravo! That does it. Thank you for your help.
    – Heinrich Wagner
    Nov 21 '18 at 21:52










  • As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
    – Henno Brandsma
    Nov 21 '18 at 22:24








2




2




How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20






How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20






1




1




@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36




@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36












Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52




Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52












As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24




As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24










1 Answer
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The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$



One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.



$X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.






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    0














    The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$



    One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
    is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.



    $X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.






    share|cite|improve this answer




























      0














      The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$



      One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
      is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.



      $X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.






      share|cite|improve this answer


























        0












        0








        0






        The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$



        One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
        is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.



        $X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.






        share|cite|improve this answer














        The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$



        One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
        is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.



        $X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 '18 at 4:43

























        answered Nov 21 '18 at 22:22









        Henno BrandsmaHenno Brandsma

        105k347114




        105k347114






























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