exact definition of open sets












3












$begingroup$


What is the exact definition of open sets?
I know both definitions of open sets with respect to metric space and topological space. but what is most general definition that covers the both.










share|cite|improve this question







New contributor




mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    It would be helpful if you edited your question to include some context. For example, when you say that you know "both" definitions of open sets, what specific definitions are you referring to? Can you please reproduce those definitions in your question?
    $endgroup$
    – Xander Henderson
    41 mins ago






  • 2




    $begingroup$
    Topological spaces are more general than metric spaces in the sense that every metric space is a topological space (but not the other way around). A metric space is a topological spaces under the metric topology which just consists of open balls.
    $endgroup$
    – Dave
    36 mins ago
















3












$begingroup$


What is the exact definition of open sets?
I know both definitions of open sets with respect to metric space and topological space. but what is most general definition that covers the both.










share|cite|improve this question







New contributor




mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    It would be helpful if you edited your question to include some context. For example, when you say that you know "both" definitions of open sets, what specific definitions are you referring to? Can you please reproduce those definitions in your question?
    $endgroup$
    – Xander Henderson
    41 mins ago






  • 2




    $begingroup$
    Topological spaces are more general than metric spaces in the sense that every metric space is a topological space (but not the other way around). A metric space is a topological spaces under the metric topology which just consists of open balls.
    $endgroup$
    – Dave
    36 mins ago














3












3








3





$begingroup$


What is the exact definition of open sets?
I know both definitions of open sets with respect to metric space and topological space. but what is most general definition that covers the both.










share|cite|improve this question







New contributor




mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




What is the exact definition of open sets?
I know both definitions of open sets with respect to metric space and topological space. but what is most general definition that covers the both.







general-topology metric-spaces






share|cite|improve this question







New contributor




mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 52 mins ago









mathmania12mathmania12

163




163




New contributor




mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    $begingroup$
    It would be helpful if you edited your question to include some context. For example, when you say that you know "both" definitions of open sets, what specific definitions are you referring to? Can you please reproduce those definitions in your question?
    $endgroup$
    – Xander Henderson
    41 mins ago






  • 2




    $begingroup$
    Topological spaces are more general than metric spaces in the sense that every metric space is a topological space (but not the other way around). A metric space is a topological spaces under the metric topology which just consists of open balls.
    $endgroup$
    – Dave
    36 mins ago














  • 3




    $begingroup$
    It would be helpful if you edited your question to include some context. For example, when you say that you know "both" definitions of open sets, what specific definitions are you referring to? Can you please reproduce those definitions in your question?
    $endgroup$
    – Xander Henderson
    41 mins ago






  • 2




    $begingroup$
    Topological spaces are more general than metric spaces in the sense that every metric space is a topological space (but not the other way around). A metric space is a topological spaces under the metric topology which just consists of open balls.
    $endgroup$
    – Dave
    36 mins ago








3




3




$begingroup$
It would be helpful if you edited your question to include some context. For example, when you say that you know "both" definitions of open sets, what specific definitions are you referring to? Can you please reproduce those definitions in your question?
$endgroup$
– Xander Henderson
41 mins ago




$begingroup$
It would be helpful if you edited your question to include some context. For example, when you say that you know "both" definitions of open sets, what specific definitions are you referring to? Can you please reproduce those definitions in your question?
$endgroup$
– Xander Henderson
41 mins ago




2




2




$begingroup$
Topological spaces are more general than metric spaces in the sense that every metric space is a topological space (but not the other way around). A metric space is a topological spaces under the metric topology which just consists of open balls.
$endgroup$
– Dave
36 mins ago




$begingroup$
Topological spaces are more general than metric spaces in the sense that every metric space is a topological space (but not the other way around). A metric space is a topological spaces under the metric topology which just consists of open balls.
$endgroup$
– Dave
36 mins ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Let $X$ be a set. An open set is an element of a collection $mathcal{T}$ (whose elements are subsets of $X$), called a topology, which satisfies three conditions.





  1. $varnothing$ and $X$ are elements of $mathcal{T}$

  2. For any subcollection $mathcal{S}$ of $mathcal{T}$, the union $bigcupmathcal{S}$ is an element of $mathcal{T}$.

  3. For any finite subcollection $mathcal{S}$ of $mathcal{T}$, the intersection $bigcapmathcal{S}$ is an element of $mathcal{T}$.


Let $(M,rho)$ be a metric space. Then the metric topology $mathcal{T}_{rho}$ induced by the metric $rho$ is the collection of sets of the form $bigcupmathcal{B}$, where $mathcal{B}$ is any collection (possibly empty) of open balls. Here an open ball is a set of the form ${xin M:rho(x_0,x)<r_0}$ for fixed $x_0in M$ and for fixed positive real number $r_0$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Question: out of curiosity, can the exact same subset $S subseteq X$ be either open or not-open depending on the topology? E.g., $S in mathcal{T}_1$ but $S notin mathcal{T}_2$. Thanks
    $endgroup$
    – ted
    36 mins ago








  • 2




    $begingroup$
    @ted Yes. For example one topology $mathcal{T}_1$ could be the collection only containing $varnothing$ and $X$ and nothing else. Another topology $mathcal{T}_2$ could be the collection of all subsets. Then as long as $X$ contains at least two points, then your situation occurs.
    $endgroup$
    – Alberto Takase
    31 mins ago










  • $begingroup$
    @ted That's exactly what makes two topologies different: some sets are open in one but not the other.
    $endgroup$
    – bof
    25 mins ago



















1












$begingroup$

There is no definition of what an open set is.



There is a definition of what a Topology, and each Topology will have a class of sets that are called open but these sets can pretty much be anything we want so far as the class of sets obey certain rules of inclusion.



A metric space is a type of topological space and the definition of an open set in a metric space is such that the class of all open sets obeys the rules of inclusion required for the metric space to be considered a topology.



So the definition of a Topology is the most general definition. But every topology will have its own classification of what open sets are. These classifications of sets must obey certain rules but the actual open sets themselves need not have any consistent properties.



....



More precisely if a have a universal set $X$ then ANY $T subset P(X)$ (any class of subsets of $X$) can be called the class of all "open" sets so long as the following apply:





  1. $X$ and $emptyset$ are elements of $T$.

  2. The union of sets in $T$ will itself be in $T$

  3. Any finite intersection of sets in $T$ will itself be in $T$


As long as those rules are obeyed any set can be considered open.



If we have a metric space $X$ and we define that if a set $Asubset X$ is such: that for every point $xin A$ there will be an open ball $B_r(x)$ around $x$ so that $B_r(x) subset A$; we call such a set PEN-Oay.



Now $X$ is PEN-Oay because everything including all open balls are subsets of it. And $emptyset$ is PEN-Oay vacuously because it has no points (as every point can be said to have any property).



And we can prove but I will not that any union (even infinite unions) of PEN-Oay sets are PEN-Oay.



And we can prove any finite intersection of PEN-Oay sets are PEN-Oay.



So if being PEN-Oay satisfies all three conditions for the class of all PEN-Oay sets to be considered a class of all "open" sets.



So we can say the metric space is a Topology so that $T = {$ all "open" set$} = {$ all PEN-Oay sets$}$.






share|cite|improve this answer











$endgroup$













    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
    });


    }
    });






    mathmania12 is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3118489%2fexact-definition-of-open-sets%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









    3












    $begingroup$

    Let $X$ be a set. An open set is an element of a collection $mathcal{T}$ (whose elements are subsets of $X$), called a topology, which satisfies three conditions.





    1. $varnothing$ and $X$ are elements of $mathcal{T}$

    2. For any subcollection $mathcal{S}$ of $mathcal{T}$, the union $bigcupmathcal{S}$ is an element of $mathcal{T}$.

    3. For any finite subcollection $mathcal{S}$ of $mathcal{T}$, the intersection $bigcapmathcal{S}$ is an element of $mathcal{T}$.


    Let $(M,rho)$ be a metric space. Then the metric topology $mathcal{T}_{rho}$ induced by the metric $rho$ is the collection of sets of the form $bigcupmathcal{B}$, where $mathcal{B}$ is any collection (possibly empty) of open balls. Here an open ball is a set of the form ${xin M:rho(x_0,x)<r_0}$ for fixed $x_0in M$ and for fixed positive real number $r_0$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Question: out of curiosity, can the exact same subset $S subseteq X$ be either open or not-open depending on the topology? E.g., $S in mathcal{T}_1$ but $S notin mathcal{T}_2$. Thanks
      $endgroup$
      – ted
      36 mins ago








    • 2




      $begingroup$
      @ted Yes. For example one topology $mathcal{T}_1$ could be the collection only containing $varnothing$ and $X$ and nothing else. Another topology $mathcal{T}_2$ could be the collection of all subsets. Then as long as $X$ contains at least two points, then your situation occurs.
      $endgroup$
      – Alberto Takase
      31 mins ago










    • $begingroup$
      @ted That's exactly what makes two topologies different: some sets are open in one but not the other.
      $endgroup$
      – bof
      25 mins ago
















    3












    $begingroup$

    Let $X$ be a set. An open set is an element of a collection $mathcal{T}$ (whose elements are subsets of $X$), called a topology, which satisfies three conditions.





    1. $varnothing$ and $X$ are elements of $mathcal{T}$

    2. For any subcollection $mathcal{S}$ of $mathcal{T}$, the union $bigcupmathcal{S}$ is an element of $mathcal{T}$.

    3. For any finite subcollection $mathcal{S}$ of $mathcal{T}$, the intersection $bigcapmathcal{S}$ is an element of $mathcal{T}$.


    Let $(M,rho)$ be a metric space. Then the metric topology $mathcal{T}_{rho}$ induced by the metric $rho$ is the collection of sets of the form $bigcupmathcal{B}$, where $mathcal{B}$ is any collection (possibly empty) of open balls. Here an open ball is a set of the form ${xin M:rho(x_0,x)<r_0}$ for fixed $x_0in M$ and for fixed positive real number $r_0$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Question: out of curiosity, can the exact same subset $S subseteq X$ be either open or not-open depending on the topology? E.g., $S in mathcal{T}_1$ but $S notin mathcal{T}_2$. Thanks
      $endgroup$
      – ted
      36 mins ago








    • 2




      $begingroup$
      @ted Yes. For example one topology $mathcal{T}_1$ could be the collection only containing $varnothing$ and $X$ and nothing else. Another topology $mathcal{T}_2$ could be the collection of all subsets. Then as long as $X$ contains at least two points, then your situation occurs.
      $endgroup$
      – Alberto Takase
      31 mins ago










    • $begingroup$
      @ted That's exactly what makes two topologies different: some sets are open in one but not the other.
      $endgroup$
      – bof
      25 mins ago














    3












    3








    3





    $begingroup$

    Let $X$ be a set. An open set is an element of a collection $mathcal{T}$ (whose elements are subsets of $X$), called a topology, which satisfies three conditions.





    1. $varnothing$ and $X$ are elements of $mathcal{T}$

    2. For any subcollection $mathcal{S}$ of $mathcal{T}$, the union $bigcupmathcal{S}$ is an element of $mathcal{T}$.

    3. For any finite subcollection $mathcal{S}$ of $mathcal{T}$, the intersection $bigcapmathcal{S}$ is an element of $mathcal{T}$.


    Let $(M,rho)$ be a metric space. Then the metric topology $mathcal{T}_{rho}$ induced by the metric $rho$ is the collection of sets of the form $bigcupmathcal{B}$, where $mathcal{B}$ is any collection (possibly empty) of open balls. Here an open ball is a set of the form ${xin M:rho(x_0,x)<r_0}$ for fixed $x_0in M$ and for fixed positive real number $r_0$.






    share|cite|improve this answer











    $endgroup$



    Let $X$ be a set. An open set is an element of a collection $mathcal{T}$ (whose elements are subsets of $X$), called a topology, which satisfies three conditions.





    1. $varnothing$ and $X$ are elements of $mathcal{T}$

    2. For any subcollection $mathcal{S}$ of $mathcal{T}$, the union $bigcupmathcal{S}$ is an element of $mathcal{T}$.

    3. For any finite subcollection $mathcal{S}$ of $mathcal{T}$, the intersection $bigcapmathcal{S}$ is an element of $mathcal{T}$.


    Let $(M,rho)$ be a metric space. Then the metric topology $mathcal{T}_{rho}$ induced by the metric $rho$ is the collection of sets of the form $bigcupmathcal{B}$, where $mathcal{B}$ is any collection (possibly empty) of open balls. Here an open ball is a set of the form ${xin M:rho(x_0,x)<r_0}$ for fixed $x_0in M$ and for fixed positive real number $r_0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 35 mins ago

























    answered 49 mins ago









    Alberto TakaseAlberto Takase

    2,220619




    2,220619








    • 1




      $begingroup$
      Question: out of curiosity, can the exact same subset $S subseteq X$ be either open or not-open depending on the topology? E.g., $S in mathcal{T}_1$ but $S notin mathcal{T}_2$. Thanks
      $endgroup$
      – ted
      36 mins ago








    • 2




      $begingroup$
      @ted Yes. For example one topology $mathcal{T}_1$ could be the collection only containing $varnothing$ and $X$ and nothing else. Another topology $mathcal{T}_2$ could be the collection of all subsets. Then as long as $X$ contains at least two points, then your situation occurs.
      $endgroup$
      – Alberto Takase
      31 mins ago










    • $begingroup$
      @ted That's exactly what makes two topologies different: some sets are open in one but not the other.
      $endgroup$
      – bof
      25 mins ago














    • 1




      $begingroup$
      Question: out of curiosity, can the exact same subset $S subseteq X$ be either open or not-open depending on the topology? E.g., $S in mathcal{T}_1$ but $S notin mathcal{T}_2$. Thanks
      $endgroup$
      – ted
      36 mins ago








    • 2




      $begingroup$
      @ted Yes. For example one topology $mathcal{T}_1$ could be the collection only containing $varnothing$ and $X$ and nothing else. Another topology $mathcal{T}_2$ could be the collection of all subsets. Then as long as $X$ contains at least two points, then your situation occurs.
      $endgroup$
      – Alberto Takase
      31 mins ago










    • $begingroup$
      @ted That's exactly what makes two topologies different: some sets are open in one but not the other.
      $endgroup$
      – bof
      25 mins ago








    1




    1




    $begingroup$
    Question: out of curiosity, can the exact same subset $S subseteq X$ be either open or not-open depending on the topology? E.g., $S in mathcal{T}_1$ but $S notin mathcal{T}_2$. Thanks
    $endgroup$
    – ted
    36 mins ago






    $begingroup$
    Question: out of curiosity, can the exact same subset $S subseteq X$ be either open or not-open depending on the topology? E.g., $S in mathcal{T}_1$ but $S notin mathcal{T}_2$. Thanks
    $endgroup$
    – ted
    36 mins ago






    2




    2




    $begingroup$
    @ted Yes. For example one topology $mathcal{T}_1$ could be the collection only containing $varnothing$ and $X$ and nothing else. Another topology $mathcal{T}_2$ could be the collection of all subsets. Then as long as $X$ contains at least two points, then your situation occurs.
    $endgroup$
    – Alberto Takase
    31 mins ago




    $begingroup$
    @ted Yes. For example one topology $mathcal{T}_1$ could be the collection only containing $varnothing$ and $X$ and nothing else. Another topology $mathcal{T}_2$ could be the collection of all subsets. Then as long as $X$ contains at least two points, then your situation occurs.
    $endgroup$
    – Alberto Takase
    31 mins ago












    $begingroup$
    @ted That's exactly what makes two topologies different: some sets are open in one but not the other.
    $endgroup$
    – bof
    25 mins ago




    $begingroup$
    @ted That's exactly what makes two topologies different: some sets are open in one but not the other.
    $endgroup$
    – bof
    25 mins ago











    1












    $begingroup$

    There is no definition of what an open set is.



    There is a definition of what a Topology, and each Topology will have a class of sets that are called open but these sets can pretty much be anything we want so far as the class of sets obey certain rules of inclusion.



    A metric space is a type of topological space and the definition of an open set in a metric space is such that the class of all open sets obeys the rules of inclusion required for the metric space to be considered a topology.



    So the definition of a Topology is the most general definition. But every topology will have its own classification of what open sets are. These classifications of sets must obey certain rules but the actual open sets themselves need not have any consistent properties.



    ....



    More precisely if a have a universal set $X$ then ANY $T subset P(X)$ (any class of subsets of $X$) can be called the class of all "open" sets so long as the following apply:





    1. $X$ and $emptyset$ are elements of $T$.

    2. The union of sets in $T$ will itself be in $T$

    3. Any finite intersection of sets in $T$ will itself be in $T$


    As long as those rules are obeyed any set can be considered open.



    If we have a metric space $X$ and we define that if a set $Asubset X$ is such: that for every point $xin A$ there will be an open ball $B_r(x)$ around $x$ so that $B_r(x) subset A$; we call such a set PEN-Oay.



    Now $X$ is PEN-Oay because everything including all open balls are subsets of it. And $emptyset$ is PEN-Oay vacuously because it has no points (as every point can be said to have any property).



    And we can prove but I will not that any union (even infinite unions) of PEN-Oay sets are PEN-Oay.



    And we can prove any finite intersection of PEN-Oay sets are PEN-Oay.



    So if being PEN-Oay satisfies all three conditions for the class of all PEN-Oay sets to be considered a class of all "open" sets.



    So we can say the metric space is a Topology so that $T = {$ all "open" set$} = {$ all PEN-Oay sets$}$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      There is no definition of what an open set is.



      There is a definition of what a Topology, and each Topology will have a class of sets that are called open but these sets can pretty much be anything we want so far as the class of sets obey certain rules of inclusion.



      A metric space is a type of topological space and the definition of an open set in a metric space is such that the class of all open sets obeys the rules of inclusion required for the metric space to be considered a topology.



      So the definition of a Topology is the most general definition. But every topology will have its own classification of what open sets are. These classifications of sets must obey certain rules but the actual open sets themselves need not have any consistent properties.



      ....



      More precisely if a have a universal set $X$ then ANY $T subset P(X)$ (any class of subsets of $X$) can be called the class of all "open" sets so long as the following apply:





      1. $X$ and $emptyset$ are elements of $T$.

      2. The union of sets in $T$ will itself be in $T$

      3. Any finite intersection of sets in $T$ will itself be in $T$


      As long as those rules are obeyed any set can be considered open.



      If we have a metric space $X$ and we define that if a set $Asubset X$ is such: that for every point $xin A$ there will be an open ball $B_r(x)$ around $x$ so that $B_r(x) subset A$; we call such a set PEN-Oay.



      Now $X$ is PEN-Oay because everything including all open balls are subsets of it. And $emptyset$ is PEN-Oay vacuously because it has no points (as every point can be said to have any property).



      And we can prove but I will not that any union (even infinite unions) of PEN-Oay sets are PEN-Oay.



      And we can prove any finite intersection of PEN-Oay sets are PEN-Oay.



      So if being PEN-Oay satisfies all three conditions for the class of all PEN-Oay sets to be considered a class of all "open" sets.



      So we can say the metric space is a Topology so that $T = {$ all "open" set$} = {$ all PEN-Oay sets$}$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        There is no definition of what an open set is.



        There is a definition of what a Topology, and each Topology will have a class of sets that are called open but these sets can pretty much be anything we want so far as the class of sets obey certain rules of inclusion.



        A metric space is a type of topological space and the definition of an open set in a metric space is such that the class of all open sets obeys the rules of inclusion required for the metric space to be considered a topology.



        So the definition of a Topology is the most general definition. But every topology will have its own classification of what open sets are. These classifications of sets must obey certain rules but the actual open sets themselves need not have any consistent properties.



        ....



        More precisely if a have a universal set $X$ then ANY $T subset P(X)$ (any class of subsets of $X$) can be called the class of all "open" sets so long as the following apply:





        1. $X$ and $emptyset$ are elements of $T$.

        2. The union of sets in $T$ will itself be in $T$

        3. Any finite intersection of sets in $T$ will itself be in $T$


        As long as those rules are obeyed any set can be considered open.



        If we have a metric space $X$ and we define that if a set $Asubset X$ is such: that for every point $xin A$ there will be an open ball $B_r(x)$ around $x$ so that $B_r(x) subset A$; we call such a set PEN-Oay.



        Now $X$ is PEN-Oay because everything including all open balls are subsets of it. And $emptyset$ is PEN-Oay vacuously because it has no points (as every point can be said to have any property).



        And we can prove but I will not that any union (even infinite unions) of PEN-Oay sets are PEN-Oay.



        And we can prove any finite intersection of PEN-Oay sets are PEN-Oay.



        So if being PEN-Oay satisfies all three conditions for the class of all PEN-Oay sets to be considered a class of all "open" sets.



        So we can say the metric space is a Topology so that $T = {$ all "open" set$} = {$ all PEN-Oay sets$}$.






        share|cite|improve this answer











        $endgroup$



        There is no definition of what an open set is.



        There is a definition of what a Topology, and each Topology will have a class of sets that are called open but these sets can pretty much be anything we want so far as the class of sets obey certain rules of inclusion.



        A metric space is a type of topological space and the definition of an open set in a metric space is such that the class of all open sets obeys the rules of inclusion required for the metric space to be considered a topology.



        So the definition of a Topology is the most general definition. But every topology will have its own classification of what open sets are. These classifications of sets must obey certain rules but the actual open sets themselves need not have any consistent properties.



        ....



        More precisely if a have a universal set $X$ then ANY $T subset P(X)$ (any class of subsets of $X$) can be called the class of all "open" sets so long as the following apply:





        1. $X$ and $emptyset$ are elements of $T$.

        2. The union of sets in $T$ will itself be in $T$

        3. Any finite intersection of sets in $T$ will itself be in $T$


        As long as those rules are obeyed any set can be considered open.



        If we have a metric space $X$ and we define that if a set $Asubset X$ is such: that for every point $xin A$ there will be an open ball $B_r(x)$ around $x$ so that $B_r(x) subset A$; we call such a set PEN-Oay.



        Now $X$ is PEN-Oay because everything including all open balls are subsets of it. And $emptyset$ is PEN-Oay vacuously because it has no points (as every point can be said to have any property).



        And we can prove but I will not that any union (even infinite unions) of PEN-Oay sets are PEN-Oay.



        And we can prove any finite intersection of PEN-Oay sets are PEN-Oay.



        So if being PEN-Oay satisfies all three conditions for the class of all PEN-Oay sets to be considered a class of all "open" sets.



        So we can say the metric space is a Topology so that $T = {$ all "open" set$} = {$ all PEN-Oay sets$}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 mins ago

























        answered 24 mins ago









        fleabloodfleablood

        71k22686




        71k22686






















            mathmania12 is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            mathmania12 is a new contributor. Be nice, and check out our Code of Conduct.













            mathmania12 is a new contributor. Be nice, and check out our Code of Conduct.












            mathmania12 is a new contributor. Be nice, and check out our Code of Conduct.
















            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%2f3118489%2fexact-definition-of-open-sets%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

            GameSpot

            日野市

            Tu-95轟炸機