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Non-trivial topology where only open sets are closed


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1












$begingroup$


For example, on $mathbb{R}$ there exists trivial topology which contains only $mathbb{R}$ and $emptyset$ and in that topology all open sets are closed and all closed sets are open.



Question. Does there exist non trivial topology on $mathbb{R}$ for which all open sets are closed and all closed sets are open? Further, given some general set $X$ whose number of elements is finite, could we always construct non trival topology where all open sets are closed and all closed sets are open? What if $X$ has non finite number of elements?



I hope my question is not meaningless.



Thank you for any help.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The trivial topology on a set $X$ is ${emptyset, X}$. The discrete topology on a set $X$ is $2^X$. Both have the property that you are looking for, though I am not sure if you are using the term "trivial" here in reference to the trivial topology or just to mean "simple".
    $endgroup$
    – parsiad
    1 hour ago






  • 1




    $begingroup$
    (1) The discrete topology, where all subsets are open and closed. (2) Given any partition ${X_i}_{iin I}of $X$, take as a topology the colleciton of all sets that are unions of elements of the partition; the complements are also unions of elements of the partition. This works for any set, regardless of cardinality, and any partition, regarless of size of the partition or of its constituent sets.
    $endgroup$
    – Arturo Magidin
    1 hour ago






  • 1




    $begingroup$
    Related to your question are door spaces and extremally disconnected spaces.
    $endgroup$
    – William Elliot
    1 hour ago
















1












$begingroup$


For example, on $mathbb{R}$ there exists trivial topology which contains only $mathbb{R}$ and $emptyset$ and in that topology all open sets are closed and all closed sets are open.



Question. Does there exist non trivial topology on $mathbb{R}$ for which all open sets are closed and all closed sets are open? Further, given some general set $X$ whose number of elements is finite, could we always construct non trival topology where all open sets are closed and all closed sets are open? What if $X$ has non finite number of elements?



I hope my question is not meaningless.



Thank you for any help.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The trivial topology on a set $X$ is ${emptyset, X}$. The discrete topology on a set $X$ is $2^X$. Both have the property that you are looking for, though I am not sure if you are using the term "trivial" here in reference to the trivial topology or just to mean "simple".
    $endgroup$
    – parsiad
    1 hour ago






  • 1




    $begingroup$
    (1) The discrete topology, where all subsets are open and closed. (2) Given any partition ${X_i}_{iin I}of $X$, take as a topology the colleciton of all sets that are unions of elements of the partition; the complements are also unions of elements of the partition. This works for any set, regardless of cardinality, and any partition, regarless of size of the partition or of its constituent sets.
    $endgroup$
    – Arturo Magidin
    1 hour ago






  • 1




    $begingroup$
    Related to your question are door spaces and extremally disconnected spaces.
    $endgroup$
    – William Elliot
    1 hour ago














1












1








1





$begingroup$


For example, on $mathbb{R}$ there exists trivial topology which contains only $mathbb{R}$ and $emptyset$ and in that topology all open sets are closed and all closed sets are open.



Question. Does there exist non trivial topology on $mathbb{R}$ for which all open sets are closed and all closed sets are open? Further, given some general set $X$ whose number of elements is finite, could we always construct non trival topology where all open sets are closed and all closed sets are open? What if $X$ has non finite number of elements?



I hope my question is not meaningless.



Thank you for any help.










share|cite|improve this question











$endgroup$




For example, on $mathbb{R}$ there exists trivial topology which contains only $mathbb{R}$ and $emptyset$ and in that topology all open sets are closed and all closed sets are open.



Question. Does there exist non trivial topology on $mathbb{R}$ for which all open sets are closed and all closed sets are open? Further, given some general set $X$ whose number of elements is finite, could we always construct non trival topology where all open sets are closed and all closed sets are open? What if $X$ has non finite number of elements?



I hope my question is not meaningless.



Thank you for any help.







general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









J. W. Tanner

3,2801320




3,2801320










asked 1 hour ago









ThomThom

341111




341111








  • 2




    $begingroup$
    The trivial topology on a set $X$ is ${emptyset, X}$. The discrete topology on a set $X$ is $2^X$. Both have the property that you are looking for, though I am not sure if you are using the term "trivial" here in reference to the trivial topology or just to mean "simple".
    $endgroup$
    – parsiad
    1 hour ago






  • 1




    $begingroup$
    (1) The discrete topology, where all subsets are open and closed. (2) Given any partition ${X_i}_{iin I}of $X$, take as a topology the colleciton of all sets that are unions of elements of the partition; the complements are also unions of elements of the partition. This works for any set, regardless of cardinality, and any partition, regarless of size of the partition or of its constituent sets.
    $endgroup$
    – Arturo Magidin
    1 hour ago






  • 1




    $begingroup$
    Related to your question are door spaces and extremally disconnected spaces.
    $endgroup$
    – William Elliot
    1 hour ago














  • 2




    $begingroup$
    The trivial topology on a set $X$ is ${emptyset, X}$. The discrete topology on a set $X$ is $2^X$. Both have the property that you are looking for, though I am not sure if you are using the term "trivial" here in reference to the trivial topology or just to mean "simple".
    $endgroup$
    – parsiad
    1 hour ago






  • 1




    $begingroup$
    (1) The discrete topology, where all subsets are open and closed. (2) Given any partition ${X_i}_{iin I}of $X$, take as a topology the colleciton of all sets that are unions of elements of the partition; the complements are also unions of elements of the partition. This works for any set, regardless of cardinality, and any partition, regarless of size of the partition or of its constituent sets.
    $endgroup$
    – Arturo Magidin
    1 hour ago






  • 1




    $begingroup$
    Related to your question are door spaces and extremally disconnected spaces.
    $endgroup$
    – William Elliot
    1 hour ago








2




2




$begingroup$
The trivial topology on a set $X$ is ${emptyset, X}$. The discrete topology on a set $X$ is $2^X$. Both have the property that you are looking for, though I am not sure if you are using the term "trivial" here in reference to the trivial topology or just to mean "simple".
$endgroup$
– parsiad
1 hour ago




$begingroup$
The trivial topology on a set $X$ is ${emptyset, X}$. The discrete topology on a set $X$ is $2^X$. Both have the property that you are looking for, though I am not sure if you are using the term "trivial" here in reference to the trivial topology or just to mean "simple".
$endgroup$
– parsiad
1 hour ago




1




1




$begingroup$
(1) The discrete topology, where all subsets are open and closed. (2) Given any partition ${X_i}_{iin I}of $X$, take as a topology the colleciton of all sets that are unions of elements of the partition; the complements are also unions of elements of the partition. This works for any set, regardless of cardinality, and any partition, regarless of size of the partition or of its constituent sets.
$endgroup$
– Arturo Magidin
1 hour ago




$begingroup$
(1) The discrete topology, where all subsets are open and closed. (2) Given any partition ${X_i}_{iin I}of $X$, take as a topology the colleciton of all sets that are unions of elements of the partition; the complements are also unions of elements of the partition. This works for any set, regardless of cardinality, and any partition, regarless of size of the partition or of its constituent sets.
$endgroup$
– Arturo Magidin
1 hour ago




1




1




$begingroup$
Related to your question are door spaces and extremally disconnected spaces.
$endgroup$
– William Elliot
1 hour ago




$begingroup$
Related to your question are door spaces and extremally disconnected spaces.
$endgroup$
– William Elliot
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

The topologies with this property are precisely the ones that are derived from partitions of the underlying set $X$.



Specifically, let $mathcal{P}={X_i}_{iin I}$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_{iin I_0}X_i$ for $I_0subseteq I$. Then $tau$ is a topology: the empty set corresponds to $I_0=varnothing$, the set $X$ to $I_0=I$; the union of such sets corresponds to the family indexed by the union of indices, and the intersection to the intersection of the indices. Moreover, the complement of the set corresponding to $I_0$ is the set corresponding to $I-I_0$. Thus, $tau$ has the desired property.



Now let $tau$ be any topology on $X$ with the desired property. Define an equivalence relation on $X$ by letting $xsim y$ if and only if for every $Ain tau$, $xin A$ if and only if $yin A$. Trivially, this is an equivalence relation, and so induces a partition on $X$. I claim that $tau$ is in fact, the topology induced by this partition as above.



Indeed, if $Ain tau$, then $A=cup_{xin A}[x]$, where $[x]$ is the equivalence class of $x$. Trivially $A$ is contained in the right hand side, and if $yin[x]$, then since $xin A$ then $yin A$, so we have equality.



Now, conversely, let $xin X$ and look at $[x]$. I claim that $X-[x]$ lies in $tau$. So see this, let $zin X-[x]$. Then since $znotin [x]$, there exists an open set $A_zin tau$ such that $zin A_z$ but $xnotin A_z$ (and hence, $[x]cap A_z=varnothing$). Now, $cup_{znotin[x]}A_z$ is an open set, contains every element of $X-[x]$, and intersects $[x]$ trivially because each element in the union does. That is, this open set is $A-[x]$; but since the complement of every open set is open, and $A-[x]$ is open, then $[x]$ is open. Thus, $[x]intau$.



We have then proven that every element of the partition induced by $sim$ is open, and that every open set is a union of such elements of the partition. That is, the open sets are precisely the unions of elements of the partition $X/sim$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    You can manufacture examples like this pretty easily. Let $A$ be any subset of $mathbb{R}$. Put a topology on $mathbb{R}$ with the following open sets:
    $$
    varnothing, A, mathbb{R}-A, mathbb{R}.
    $$

    You can easily check that this always gives a topology, and a subset of $mathbb{R}$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbb{R}$ here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      More generally, take any partition and take unions of elements of the partition. This example is what you get from the partition ${A,mathbb{R}-A}$.
      $endgroup$
      – Arturo Magidin
      1 hour ago










    • $begingroup$
      @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbb{R}$. Can every topology in $X$ be made with some partitions?
      $endgroup$
      – Thom
      1 hour ago










    • $begingroup$
      @Thom: Yes. I’ll write it up.
      $endgroup$
      – Arturo Magidin
      55 mins ago










    • $begingroup$
      @ArturoMagidin Fascinating. Thank you.
      $endgroup$
      – Thom
      47 mins ago











    Your Answer





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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The topologies with this property are precisely the ones that are derived from partitions of the underlying set $X$.



    Specifically, let $mathcal{P}={X_i}_{iin I}$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_{iin I_0}X_i$ for $I_0subseteq I$. Then $tau$ is a topology: the empty set corresponds to $I_0=varnothing$, the set $X$ to $I_0=I$; the union of such sets corresponds to the family indexed by the union of indices, and the intersection to the intersection of the indices. Moreover, the complement of the set corresponding to $I_0$ is the set corresponding to $I-I_0$. Thus, $tau$ has the desired property.



    Now let $tau$ be any topology on $X$ with the desired property. Define an equivalence relation on $X$ by letting $xsim y$ if and only if for every $Ain tau$, $xin A$ if and only if $yin A$. Trivially, this is an equivalence relation, and so induces a partition on $X$. I claim that $tau$ is in fact, the topology induced by this partition as above.



    Indeed, if $Ain tau$, then $A=cup_{xin A}[x]$, where $[x]$ is the equivalence class of $x$. Trivially $A$ is contained in the right hand side, and if $yin[x]$, then since $xin A$ then $yin A$, so we have equality.



    Now, conversely, let $xin X$ and look at $[x]$. I claim that $X-[x]$ lies in $tau$. So see this, let $zin X-[x]$. Then since $znotin [x]$, there exists an open set $A_zin tau$ such that $zin A_z$ but $xnotin A_z$ (and hence, $[x]cap A_z=varnothing$). Now, $cup_{znotin[x]}A_z$ is an open set, contains every element of $X-[x]$, and intersects $[x]$ trivially because each element in the union does. That is, this open set is $A-[x]$; but since the complement of every open set is open, and $A-[x]$ is open, then $[x]$ is open. Thus, $[x]intau$.



    We have then proven that every element of the partition induced by $sim$ is open, and that every open set is a union of such elements of the partition. That is, the open sets are precisely the unions of elements of the partition $X/sim$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The topologies with this property are precisely the ones that are derived from partitions of the underlying set $X$.



      Specifically, let $mathcal{P}={X_i}_{iin I}$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_{iin I_0}X_i$ for $I_0subseteq I$. Then $tau$ is a topology: the empty set corresponds to $I_0=varnothing$, the set $X$ to $I_0=I$; the union of such sets corresponds to the family indexed by the union of indices, and the intersection to the intersection of the indices. Moreover, the complement of the set corresponding to $I_0$ is the set corresponding to $I-I_0$. Thus, $tau$ has the desired property.



      Now let $tau$ be any topology on $X$ with the desired property. Define an equivalence relation on $X$ by letting $xsim y$ if and only if for every $Ain tau$, $xin A$ if and only if $yin A$. Trivially, this is an equivalence relation, and so induces a partition on $X$. I claim that $tau$ is in fact, the topology induced by this partition as above.



      Indeed, if $Ain tau$, then $A=cup_{xin A}[x]$, where $[x]$ is the equivalence class of $x$. Trivially $A$ is contained in the right hand side, and if $yin[x]$, then since $xin A$ then $yin A$, so we have equality.



      Now, conversely, let $xin X$ and look at $[x]$. I claim that $X-[x]$ lies in $tau$. So see this, let $zin X-[x]$. Then since $znotin [x]$, there exists an open set $A_zin tau$ such that $zin A_z$ but $xnotin A_z$ (and hence, $[x]cap A_z=varnothing$). Now, $cup_{znotin[x]}A_z$ is an open set, contains every element of $X-[x]$, and intersects $[x]$ trivially because each element in the union does. That is, this open set is $A-[x]$; but since the complement of every open set is open, and $A-[x]$ is open, then $[x]$ is open. Thus, $[x]intau$.



      We have then proven that every element of the partition induced by $sim$ is open, and that every open set is a union of such elements of the partition. That is, the open sets are precisely the unions of elements of the partition $X/sim$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The topologies with this property are precisely the ones that are derived from partitions of the underlying set $X$.



        Specifically, let $mathcal{P}={X_i}_{iin I}$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_{iin I_0}X_i$ for $I_0subseteq I$. Then $tau$ is a topology: the empty set corresponds to $I_0=varnothing$, the set $X$ to $I_0=I$; the union of such sets corresponds to the family indexed by the union of indices, and the intersection to the intersection of the indices. Moreover, the complement of the set corresponding to $I_0$ is the set corresponding to $I-I_0$. Thus, $tau$ has the desired property.



        Now let $tau$ be any topology on $X$ with the desired property. Define an equivalence relation on $X$ by letting $xsim y$ if and only if for every $Ain tau$, $xin A$ if and only if $yin A$. Trivially, this is an equivalence relation, and so induces a partition on $X$. I claim that $tau$ is in fact, the topology induced by this partition as above.



        Indeed, if $Ain tau$, then $A=cup_{xin A}[x]$, where $[x]$ is the equivalence class of $x$. Trivially $A$ is contained in the right hand side, and if $yin[x]$, then since $xin A$ then $yin A$, so we have equality.



        Now, conversely, let $xin X$ and look at $[x]$. I claim that $X-[x]$ lies in $tau$. So see this, let $zin X-[x]$. Then since $znotin [x]$, there exists an open set $A_zin tau$ such that $zin A_z$ but $xnotin A_z$ (and hence, $[x]cap A_z=varnothing$). Now, $cup_{znotin[x]}A_z$ is an open set, contains every element of $X-[x]$, and intersects $[x]$ trivially because each element in the union does. That is, this open set is $A-[x]$; but since the complement of every open set is open, and $A-[x]$ is open, then $[x]$ is open. Thus, $[x]intau$.



        We have then proven that every element of the partition induced by $sim$ is open, and that every open set is a union of such elements of the partition. That is, the open sets are precisely the unions of elements of the partition $X/sim$.






        share|cite|improve this answer









        $endgroup$



        The topologies with this property are precisely the ones that are derived from partitions of the underlying set $X$.



        Specifically, let $mathcal{P}={X_i}_{iin I}$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_{iin I_0}X_i$ for $I_0subseteq I$. Then $tau$ is a topology: the empty set corresponds to $I_0=varnothing$, the set $X$ to $I_0=I$; the union of such sets corresponds to the family indexed by the union of indices, and the intersection to the intersection of the indices. Moreover, the complement of the set corresponding to $I_0$ is the set corresponding to $I-I_0$. Thus, $tau$ has the desired property.



        Now let $tau$ be any topology on $X$ with the desired property. Define an equivalence relation on $X$ by letting $xsim y$ if and only if for every $Ain tau$, $xin A$ if and only if $yin A$. Trivially, this is an equivalence relation, and so induces a partition on $X$. I claim that $tau$ is in fact, the topology induced by this partition as above.



        Indeed, if $Ain tau$, then $A=cup_{xin A}[x]$, where $[x]$ is the equivalence class of $x$. Trivially $A$ is contained in the right hand side, and if $yin[x]$, then since $xin A$ then $yin A$, so we have equality.



        Now, conversely, let $xin X$ and look at $[x]$. I claim that $X-[x]$ lies in $tau$. So see this, let $zin X-[x]$. Then since $znotin [x]$, there exists an open set $A_zin tau$ such that $zin A_z$ but $xnotin A_z$ (and hence, $[x]cap A_z=varnothing$). Now, $cup_{znotin[x]}A_z$ is an open set, contains every element of $X-[x]$, and intersects $[x]$ trivially because each element in the union does. That is, this open set is $A-[x]$; but since the complement of every open set is open, and $A-[x]$ is open, then $[x]$ is open. Thus, $[x]intau$.



        We have then proven that every element of the partition induced by $sim$ is open, and that every open set is a union of such elements of the partition. That is, the open sets are precisely the unions of elements of the partition $X/sim$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 47 mins ago









        Arturo MagidinArturo Magidin

        265k34590918




        265k34590918























            3












            $begingroup$

            You can manufacture examples like this pretty easily. Let $A$ be any subset of $mathbb{R}$. Put a topology on $mathbb{R}$ with the following open sets:
            $$
            varnothing, A, mathbb{R}-A, mathbb{R}.
            $$

            You can easily check that this always gives a topology, and a subset of $mathbb{R}$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbb{R}$ here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              More generally, take any partition and take unions of elements of the partition. This example is what you get from the partition ${A,mathbb{R}-A}$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbb{R}$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              1 hour ago










            • $begingroup$
              @Thom: Yes. I’ll write it up.
              $endgroup$
              – Arturo Magidin
              55 mins ago










            • $begingroup$
              @ArturoMagidin Fascinating. Thank you.
              $endgroup$
              – Thom
              47 mins ago
















            3












            $begingroup$

            You can manufacture examples like this pretty easily. Let $A$ be any subset of $mathbb{R}$. Put a topology on $mathbb{R}$ with the following open sets:
            $$
            varnothing, A, mathbb{R}-A, mathbb{R}.
            $$

            You can easily check that this always gives a topology, and a subset of $mathbb{R}$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbb{R}$ here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              More generally, take any partition and take unions of elements of the partition. This example is what you get from the partition ${A,mathbb{R}-A}$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbb{R}$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              1 hour ago










            • $begingroup$
              @Thom: Yes. I’ll write it up.
              $endgroup$
              – Arturo Magidin
              55 mins ago










            • $begingroup$
              @ArturoMagidin Fascinating. Thank you.
              $endgroup$
              – Thom
              47 mins ago














            3












            3








            3





            $begingroup$

            You can manufacture examples like this pretty easily. Let $A$ be any subset of $mathbb{R}$. Put a topology on $mathbb{R}$ with the following open sets:
            $$
            varnothing, A, mathbb{R}-A, mathbb{R}.
            $$

            You can easily check that this always gives a topology, and a subset of $mathbb{R}$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbb{R}$ here.






            share|cite|improve this answer









            $endgroup$



            You can manufacture examples like this pretty easily. Let $A$ be any subset of $mathbb{R}$. Put a topology on $mathbb{R}$ with the following open sets:
            $$
            varnothing, A, mathbb{R}-A, mathbb{R}.
            $$

            You can easily check that this always gives a topology, and a subset of $mathbb{R}$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbb{R}$ here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            RandallRandall

            10.6k11431




            10.6k11431












            • $begingroup$
              More generally, take any partition and take unions of elements of the partition. This example is what you get from the partition ${A,mathbb{R}-A}$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbb{R}$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              1 hour ago










            • $begingroup$
              @Thom: Yes. I’ll write it up.
              $endgroup$
              – Arturo Magidin
              55 mins ago










            • $begingroup$
              @ArturoMagidin Fascinating. Thank you.
              $endgroup$
              – Thom
              47 mins ago


















            • $begingroup$
              More generally, take any partition and take unions of elements of the partition. This example is what you get from the partition ${A,mathbb{R}-A}$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbb{R}$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              1 hour ago










            • $begingroup$
              @Thom: Yes. I’ll write it up.
              $endgroup$
              – Arturo Magidin
              55 mins ago










            • $begingroup$
              @ArturoMagidin Fascinating. Thank you.
              $endgroup$
              – Thom
              47 mins ago
















            $begingroup$
            More generally, take any partition and take unions of elements of the partition. This example is what you get from the partition ${A,mathbb{R}-A}$.
            $endgroup$
            – Arturo Magidin
            1 hour ago




            $begingroup$
            More generally, take any partition and take unions of elements of the partition. This example is what you get from the partition ${A,mathbb{R}-A}$.
            $endgroup$
            – Arturo Magidin
            1 hour ago












            $begingroup$
            @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbb{R}$. Can every topology in $X$ be made with some partitions?
            $endgroup$
            – Thom
            1 hour ago




            $begingroup$
            @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbb{R}$. Can every topology in $X$ be made with some partitions?
            $endgroup$
            – Thom
            1 hour ago












            $begingroup$
            @Thom: Yes. I’ll write it up.
            $endgroup$
            – Arturo Magidin
            55 mins ago




            $begingroup$
            @Thom: Yes. I’ll write it up.
            $endgroup$
            – Arturo Magidin
            55 mins ago












            $begingroup$
            @ArturoMagidin Fascinating. Thank you.
            $endgroup$
            – Thom
            47 mins ago




            $begingroup$
            @ArturoMagidin Fascinating. Thank you.
            $endgroup$
            – Thom
            47 mins ago


















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