Non-trivial topology where only open sets are closedExhaustion of open sets by closed setsNatural non-trivial topology on $mathbb R$ such that there are more than $2^mathbb N$ open setswhy use open interval rather than closed interval as open sets for real line topologyClosed sets in Noetherian topologyIn the finite complement topology on $mathbbR$, is the subset $ x $ closed?Closed sets in the lower limit topology.Are closed sets in topology?For which topologies are the concepts of open and closed “interchangeable”?How to determine which sets are open in a topology?Inclusion of open sets in closed sets of Zariski topology

Is it normal that my co-workers at a fitness company criticize my food choices?

Print a physical multiplication table

"of which" is correct here?

Is there a place to find the pricing for things not mentioned in the PHB? (non-magical)

What is the Japanese sound word for the clinking of money?

What is the significance behind "40 days" that often appears in the Bible?

How to explain that I do not want to visit a country due to personal safety concern?

How to deal with taxi scam when on vacation?

Most cost effective thermostat setting: consistent temperature vs. lowest temperature possible

Have the tides ever turned twice on any open problem?

Why does overlay work only on the first tcolorbox?

Math equation in non italic font

Knife as defense against stray dogs

Is a party consisting of only a bard, a cleric, and a warlock functional long-term?

How could an airship be repaired midflight?

How could a scammer know the apps on my phone / iTunes account?

I got the following comment from a reputed math journal. What does it mean?

Recruiter wants very extensive technical details about all of my previous work

What's the meaning of a knight fighting a snail in medieval book illustrations?

Fastest way to pop N items from a large dict

What is a ^ b and (a & b) << 1?

How should I state my peer review experience in the CV?

How do I change two letters closest to a string and one letter immediately after a string using Notepad++?

Official degrees of earth’s rotation per day



Non-trivial topology where only open sets are closed


Exhaustion of open sets by closed setsNatural non-trivial topology on $mathbb R$ such that there are more than $2^mathbb N$ open setswhy use open interval rather than closed interval as open sets for real line topologyClosed sets in Noetherian topologyIn the finite complement topology on $mathbbR$, is the subset $ x $ closed?Closed sets in the lower limit topology.Are closed sets in topology?For which topologies are the concepts of open and closed “interchangeable”?How to determine which sets are open in a topology?Inclusion of open sets in closed sets of Zariski topology













1












$begingroup$


For example, on $mathbbR$ there exists trivial topology which contains only $mathbbR$ 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 $mathbbR$ 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 Iof $X$, 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 $mathbbR$ there exists trivial topology which contains only $mathbbR$ 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 $mathbbR$ 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 Iof $X$, 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 $mathbbR$ there exists trivial topology which contains only $mathbbR$ 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 $mathbbR$ 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 $mathbbR$ there exists trivial topology which contains only $mathbbR$ 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 $mathbbR$ 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 Iof $X$, 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 Iof $X$, 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 Iof $X$, 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 Iof $X$, 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 $mathcalP=X_i_iin I$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_iin I_0X_i$ 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 $mathbbR$. Put a topology on $mathbbR$ with the following open sets:
    $$
    varnothing, A, mathbbR-A, mathbbR.
    $$

    You can easily check that this always gives a topology, and a subset of $mathbbR$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbbR$ 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 $A,mathbbR-A$.
      $endgroup$
      – Arturo Magidin
      1 hour ago










    • $begingroup$
      @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbbR$. Can every topology in $X$ be made with some partitions?
      $endgroup$
      – Thom
      57 mins ago










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










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










    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%2f3151030%2fnon-trivial-topology-where-only-open-sets-are-closed%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









    2












    $begingroup$

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



    Specifically, let $mathcalP=X_i_iin I$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_iin I_0X_i$ 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 $mathcalP=X_i_iin I$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_iin I_0X_i$ 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 $mathcalP=X_i_iin I$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_iin I_0X_i$ 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 $mathcalP=X_i_iin I$ be a partition of $X$, and let $tau$ be the collection of sets of the form $cup_iin I_0X_i$ 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 11 mins ago









        Arturo MagidinArturo Magidin

        265k34590918




        265k34590918





















            3












            $begingroup$

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

            You can easily check that this always gives a topology, and a subset of $mathbbR$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbbR$ 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 $A,mathbbR-A$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbbR$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              57 mins ago










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










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















            3












            $begingroup$

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

            You can easily check that this always gives a topology, and a subset of $mathbbR$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbbR$ 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 $A,mathbbR-A$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbbR$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              57 mins ago










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










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













            3












            3








            3





            $begingroup$

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

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






            share|cite|improve this answer









            $endgroup$



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

            You can easily check that this always gives a topology, and a subset of $mathbbR$ is open if and only if it is closed. This construction generalizes to any set $X$: there's nothing special about $mathbbR$ 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 $A,mathbbR-A$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbbR$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              57 mins ago










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










            • $begingroup$
              @ArturoMagidin Fascinating. Thank you.
              $endgroup$
              – Thom
              11 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 $A,mathbbR-A$.
              $endgroup$
              – Arturo Magidin
              1 hour ago










            • $begingroup$
              @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbbR$. Can every topology in $X$ be made with some partitions?
              $endgroup$
              – Thom
              57 mins ago










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










            • $begingroup$
              @ArturoMagidin Fascinating. Thank you.
              $endgroup$
              – Thom
              11 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 $A,mathbbR-A$.
            $endgroup$
            – 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 $A,mathbbR-A$.
            $endgroup$
            – Arturo Magidin
            1 hour ago












            $begingroup$
            @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbbR$. Can every topology in $X$ be made with some partitions?
            $endgroup$
            – Thom
            57 mins ago




            $begingroup$
            @ArturoMagidin Let us denote with X set of all topologies which satisfy property in the question on $mathbbR$. Can every topology in $X$ be made with some partitions?
            $endgroup$
            – Thom
            57 mins ago












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




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












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




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

















            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%2f3151030%2fnon-trivial-topology-where-only-open-sets-are-closed%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

            Möglingen Índice Localización Historia Demografía Referencias Enlaces externos Menú de navegación48°53′18″N 9°07′45″E / 48.888333333333, 9.129166666666748°53′18″N 9°07′45″E / 48.888333333333, 9.1291666666667Sitio web oficial Mapa de Möglingen«Gemeinden in Deutschland nach Fläche, Bevölkerung und Postleitzahl am 30.09.2016»Möglingen

            Virtualbox - Configuration error: Querying “UUID” failed (VERR_CFGM_VALUE_NOT_FOUND)“VERR_SUPLIB_WORLD_WRITABLE” error when trying to installing OS in virtualboxVirtual Box Kernel errorFailed to open a seesion for the virtual machineFailed to open a session for the virtual machineUbuntu 14.04 LTS Virtualbox errorcan't use VM VirtualBoxusing virtualboxI can't run Linux-64 Bit on VirtualBoxUnable to insert the virtual optical disk (VBoxguestaddition) in virtual machine for ubuntu server in win 10VirtuaBox in Ubuntu 18.04 Issues with Win10.ISO Installation

            Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org