Why is Urysohn's metrization theorem so much more “popular” than the Nagata–Smirnov metrization...












5












$begingroup$


The Urysohn's metrization theorem is much better known than the Nagata–Smirnov metrization theorem. However, the former only gives a sufficient condition for metrizability whereas the latter gives a necessary and sufficient condition.



I would imagine that giving a full characterization of when a topological space is metrizable should be a huge deal, perhaps something deserving of a fields medal even. Yet it seems as though most people don't really think this way, which is why this result isn't really particularly well-known.



What's the reason for this?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    My understanding is that most space that we care about are relatively nice, i.e. second countable.
    $endgroup$
    – Jacky Chong
    Dec 28 '18 at 6:58










  • $begingroup$
    Can you give an example of a space that is straightforwardly metrizable that is not straightforwardly regular, Hausdorff, and possessing a coutably locally finite basis? (or (Bing metrization theorem) is not straightforwardly regular, $T_0$, and possessing a $sigma$-discrete basis?) I.e., when would the reverse arrow be productive?
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:01










  • $begingroup$
    @EricTowers: That is missing the point. The advantage of having a necessary and sufficient condition is not that you can use the necessary direction, but that you can use the sufficient condition in maximal generality (for any space you might want to prove is metrizable, you can hope to be able to do so using the sufficient condition).
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:08










  • $begingroup$
    @EricWofsey : You appear to be arguing my point: Urysohn's lemma gives a sufficient condition to get from properties of a space to its metrizability. The reverse arrow goes from a known metrizable space to some properties it satisfies, which appears to be a useless direction. So what use is a theorem running in the reverse direction? Or as I asked, when do you have a known metrizable space before already knowing it has the properties the reverse arrow gives you.
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:11








  • 2




    $begingroup$
    @EricTowers: No, you have totally misunderstood what I was saying. Imagine you have some space that you think is metrizable and want to prove it. You know various things about this space, and in particular you know that it's not separable. This immediately tells you that Urysohn's metrization theorem is useless to you, since its hypotheses cannot possibly hold for your space. But the Nagata-Smirnov metrization theorem may still be useful to you, since if your guess is correct and the space is metrizable, then Nagata-Smirnov is guaranteed to apply to it.
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:14
















5












$begingroup$


The Urysohn's metrization theorem is much better known than the Nagata–Smirnov metrization theorem. However, the former only gives a sufficient condition for metrizability whereas the latter gives a necessary and sufficient condition.



I would imagine that giving a full characterization of when a topological space is metrizable should be a huge deal, perhaps something deserving of a fields medal even. Yet it seems as though most people don't really think this way, which is why this result isn't really particularly well-known.



What's the reason for this?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    My understanding is that most space that we care about are relatively nice, i.e. second countable.
    $endgroup$
    – Jacky Chong
    Dec 28 '18 at 6:58










  • $begingroup$
    Can you give an example of a space that is straightforwardly metrizable that is not straightforwardly regular, Hausdorff, and possessing a coutably locally finite basis? (or (Bing metrization theorem) is not straightforwardly regular, $T_0$, and possessing a $sigma$-discrete basis?) I.e., when would the reverse arrow be productive?
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:01










  • $begingroup$
    @EricTowers: That is missing the point. The advantage of having a necessary and sufficient condition is not that you can use the necessary direction, but that you can use the sufficient condition in maximal generality (for any space you might want to prove is metrizable, you can hope to be able to do so using the sufficient condition).
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:08










  • $begingroup$
    @EricWofsey : You appear to be arguing my point: Urysohn's lemma gives a sufficient condition to get from properties of a space to its metrizability. The reverse arrow goes from a known metrizable space to some properties it satisfies, which appears to be a useless direction. So what use is a theorem running in the reverse direction? Or as I asked, when do you have a known metrizable space before already knowing it has the properties the reverse arrow gives you.
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:11








  • 2




    $begingroup$
    @EricTowers: No, you have totally misunderstood what I was saying. Imagine you have some space that you think is metrizable and want to prove it. You know various things about this space, and in particular you know that it's not separable. This immediately tells you that Urysohn's metrization theorem is useless to you, since its hypotheses cannot possibly hold for your space. But the Nagata-Smirnov metrization theorem may still be useful to you, since if your guess is correct and the space is metrizable, then Nagata-Smirnov is guaranteed to apply to it.
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:14














5












5








5


1



$begingroup$


The Urysohn's metrization theorem is much better known than the Nagata–Smirnov metrization theorem. However, the former only gives a sufficient condition for metrizability whereas the latter gives a necessary and sufficient condition.



I would imagine that giving a full characterization of when a topological space is metrizable should be a huge deal, perhaps something deserving of a fields medal even. Yet it seems as though most people don't really think this way, which is why this result isn't really particularly well-known.



What's the reason for this?










share|cite|improve this question











$endgroup$




The Urysohn's metrization theorem is much better known than the Nagata–Smirnov metrization theorem. However, the former only gives a sufficient condition for metrizability whereas the latter gives a necessary and sufficient condition.



I would imagine that giving a full characterization of when a topological space is metrizable should be a huge deal, perhaps something deserving of a fields medal even. Yet it seems as though most people don't really think this way, which is why this result isn't really particularly well-known.



What's the reason for this?







general-topology soft-question






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 6:53







MathematicsStudent1122

















asked Dec 28 '18 at 6:48









MathematicsStudent1122MathematicsStudent1122

8,96622668




8,96622668








  • 1




    $begingroup$
    My understanding is that most space that we care about are relatively nice, i.e. second countable.
    $endgroup$
    – Jacky Chong
    Dec 28 '18 at 6:58










  • $begingroup$
    Can you give an example of a space that is straightforwardly metrizable that is not straightforwardly regular, Hausdorff, and possessing a coutably locally finite basis? (or (Bing metrization theorem) is not straightforwardly regular, $T_0$, and possessing a $sigma$-discrete basis?) I.e., when would the reverse arrow be productive?
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:01










  • $begingroup$
    @EricTowers: That is missing the point. The advantage of having a necessary and sufficient condition is not that you can use the necessary direction, but that you can use the sufficient condition in maximal generality (for any space you might want to prove is metrizable, you can hope to be able to do so using the sufficient condition).
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:08










  • $begingroup$
    @EricWofsey : You appear to be arguing my point: Urysohn's lemma gives a sufficient condition to get from properties of a space to its metrizability. The reverse arrow goes from a known metrizable space to some properties it satisfies, which appears to be a useless direction. So what use is a theorem running in the reverse direction? Or as I asked, when do you have a known metrizable space before already knowing it has the properties the reverse arrow gives you.
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:11








  • 2




    $begingroup$
    @EricTowers: No, you have totally misunderstood what I was saying. Imagine you have some space that you think is metrizable and want to prove it. You know various things about this space, and in particular you know that it's not separable. This immediately tells you that Urysohn's metrization theorem is useless to you, since its hypotheses cannot possibly hold for your space. But the Nagata-Smirnov metrization theorem may still be useful to you, since if your guess is correct and the space is metrizable, then Nagata-Smirnov is guaranteed to apply to it.
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:14














  • 1




    $begingroup$
    My understanding is that most space that we care about are relatively nice, i.e. second countable.
    $endgroup$
    – Jacky Chong
    Dec 28 '18 at 6:58










  • $begingroup$
    Can you give an example of a space that is straightforwardly metrizable that is not straightforwardly regular, Hausdorff, and possessing a coutably locally finite basis? (or (Bing metrization theorem) is not straightforwardly regular, $T_0$, and possessing a $sigma$-discrete basis?) I.e., when would the reverse arrow be productive?
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:01










  • $begingroup$
    @EricTowers: That is missing the point. The advantage of having a necessary and sufficient condition is not that you can use the necessary direction, but that you can use the sufficient condition in maximal generality (for any space you might want to prove is metrizable, you can hope to be able to do so using the sufficient condition).
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:08










  • $begingroup$
    @EricWofsey : You appear to be arguing my point: Urysohn's lemma gives a sufficient condition to get from properties of a space to its metrizability. The reverse arrow goes from a known metrizable space to some properties it satisfies, which appears to be a useless direction. So what use is a theorem running in the reverse direction? Or as I asked, when do you have a known metrizable space before already knowing it has the properties the reverse arrow gives you.
    $endgroup$
    – Eric Towers
    Dec 28 '18 at 7:11








  • 2




    $begingroup$
    @EricTowers: No, you have totally misunderstood what I was saying. Imagine you have some space that you think is metrizable and want to prove it. You know various things about this space, and in particular you know that it's not separable. This immediately tells you that Urysohn's metrization theorem is useless to you, since its hypotheses cannot possibly hold for your space. But the Nagata-Smirnov metrization theorem may still be useful to you, since if your guess is correct and the space is metrizable, then Nagata-Smirnov is guaranteed to apply to it.
    $endgroup$
    – Eric Wofsey
    Dec 28 '18 at 7:14








1




1




$begingroup$
My understanding is that most space that we care about are relatively nice, i.e. second countable.
$endgroup$
– Jacky Chong
Dec 28 '18 at 6:58




$begingroup$
My understanding is that most space that we care about are relatively nice, i.e. second countable.
$endgroup$
– Jacky Chong
Dec 28 '18 at 6:58












$begingroup$
Can you give an example of a space that is straightforwardly metrizable that is not straightforwardly regular, Hausdorff, and possessing a coutably locally finite basis? (or (Bing metrization theorem) is not straightforwardly regular, $T_0$, and possessing a $sigma$-discrete basis?) I.e., when would the reverse arrow be productive?
$endgroup$
– Eric Towers
Dec 28 '18 at 7:01




$begingroup$
Can you give an example of a space that is straightforwardly metrizable that is not straightforwardly regular, Hausdorff, and possessing a coutably locally finite basis? (or (Bing metrization theorem) is not straightforwardly regular, $T_0$, and possessing a $sigma$-discrete basis?) I.e., when would the reverse arrow be productive?
$endgroup$
– Eric Towers
Dec 28 '18 at 7:01












$begingroup$
@EricTowers: That is missing the point. The advantage of having a necessary and sufficient condition is not that you can use the necessary direction, but that you can use the sufficient condition in maximal generality (for any space you might want to prove is metrizable, you can hope to be able to do so using the sufficient condition).
$endgroup$
– Eric Wofsey
Dec 28 '18 at 7:08




$begingroup$
@EricTowers: That is missing the point. The advantage of having a necessary and sufficient condition is not that you can use the necessary direction, but that you can use the sufficient condition in maximal generality (for any space you might want to prove is metrizable, you can hope to be able to do so using the sufficient condition).
$endgroup$
– Eric Wofsey
Dec 28 '18 at 7:08












$begingroup$
@EricWofsey : You appear to be arguing my point: Urysohn's lemma gives a sufficient condition to get from properties of a space to its metrizability. The reverse arrow goes from a known metrizable space to some properties it satisfies, which appears to be a useless direction. So what use is a theorem running in the reverse direction? Or as I asked, when do you have a known metrizable space before already knowing it has the properties the reverse arrow gives you.
$endgroup$
– Eric Towers
Dec 28 '18 at 7:11






$begingroup$
@EricWofsey : You appear to be arguing my point: Urysohn's lemma gives a sufficient condition to get from properties of a space to its metrizability. The reverse arrow goes from a known metrizable space to some properties it satisfies, which appears to be a useless direction. So what use is a theorem running in the reverse direction? Or as I asked, when do you have a known metrizable space before already knowing it has the properties the reverse arrow gives you.
$endgroup$
– Eric Towers
Dec 28 '18 at 7:11






2




2




$begingroup$
@EricTowers: No, you have totally misunderstood what I was saying. Imagine you have some space that you think is metrizable and want to prove it. You know various things about this space, and in particular you know that it's not separable. This immediately tells you that Urysohn's metrization theorem is useless to you, since its hypotheses cannot possibly hold for your space. But the Nagata-Smirnov metrization theorem may still be useful to you, since if your guess is correct and the space is metrizable, then Nagata-Smirnov is guaranteed to apply to it.
$endgroup$
– Eric Wofsey
Dec 28 '18 at 7:14




$begingroup$
@EricTowers: No, you have totally misunderstood what I was saying. Imagine you have some space that you think is metrizable and want to prove it. You know various things about this space, and in particular you know that it's not separable. This immediately tells you that Urysohn's metrization theorem is useless to you, since its hypotheses cannot possibly hold for your space. But the Nagata-Smirnov metrization theorem may still be useful to you, since if your guess is correct and the space is metrizable, then Nagata-Smirnov is guaranteed to apply to it.
$endgroup$
– Eric Wofsey
Dec 28 '18 at 7:14










2 Answers
2






active

oldest

votes


















5












$begingroup$

Historically, it was an important problem to find a purely topological characterisation of a space $X$ to be metrisable, without a reference to an external object like $mathbb{R}$. Urysohn was one of the first to work on this and he proved some important theorems on embedability using continuous functions on normal spaces (Urysohn's lemma). He could then characterise the separable metric spaces as those that could be embedded into $[0,1]^mathbb{N}$ (using Urysohn functions and a countable base to reduce the number of dimensions in the cube to countable). But the general problem was still open. But Urysohn's special case turned out to be already useful to find metrisable spaces in analysis, e.g.
Moreover, it can be proved easily once you covered normal spaces and Urysohn's lemma, so it occurs in many text books.



Later in the 40's and 50's people started studying open covers and refinements, paracompactness was defined and shown to be equivalent to being "fully normal". Metric spaces were proved to be paracompact that way, but there were also many non-metric paracompact spaces so one needed a strengthening of paracompactness, and this was found to be having a $sigma$-locally finite base (which by results known at that time immediately implied paracompactness) and which was then shown to be equivalent to metrisability by both Smirnov (in Russia) and Nagata (in Japan), independent of each other. Around the same time Bing (in the US) shows that having a $sigma$-discrete base (plus regular $T_0$, which we also need in Nagata-Smirnov) was also necessary and sufficient and used this to prove that a space $X$ is metrisable iff it is a developable regular Hausdorff space which is collectionwise normal (a collectionwise normal Moore space). Other characterisations were also found, using special bases most of the time. But most of that theory, though interesting, is quite extensive to fully cover in most courses, (Munkres does Nagata-Smirnov as an optional chapter, e.g.) and the problem was to show that metrisability is "intrinsically characterisable", not as a practical method to discover metric spaces, or prove concrete spaces to be metrisable.



It has been used to characterise metrisability in special classes of spaces, like (generalised) ordered spaces. It maybe deserves to be better known, as other metrisation theorems too, but it's too "deep" into general topology problems, and not as practical as Urysohn's theorem, which has an easier to check condition.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have seen the use of the Nagata-Smirnov theorem to prove that a non-compact metrizable space has an incomplete metric.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 16:38










  • $begingroup$
    My edit was for a typo ("dicover") in the last line of the 2nd-last paragraph.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 16:40



















3












$begingroup$

I think the reason for the focus on the Urysohn metrization theorem is really Urysohn's lemma, which is extremely useful and widely applicable, and the Urysohn metrization theorem is just one easy application that one might as well prove anyway. Essentially, the Nagata-Smirnov metrization theorem also uses the Urysohn lemma. Characterizing metric spaces, while interesting, is less widely applicable, and the new ideas that were required are less of a breakthrough.



In reality, most modern topologists care very little about weird spaces that they need metrization theorems to prove are metrizable. Certainly there are some who do still study these areas, but mostly topologists are focused on algebraic topology where in the sense of general topology the spaces are "boring," for example compact metric spaces, very often manifolds.






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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054639%2fwhy-is-urysohns-metrization-theorem-so-much-more-popular-than-the-nagata-smir%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









    5












    $begingroup$

    Historically, it was an important problem to find a purely topological characterisation of a space $X$ to be metrisable, without a reference to an external object like $mathbb{R}$. Urysohn was one of the first to work on this and he proved some important theorems on embedability using continuous functions on normal spaces (Urysohn's lemma). He could then characterise the separable metric spaces as those that could be embedded into $[0,1]^mathbb{N}$ (using Urysohn functions and a countable base to reduce the number of dimensions in the cube to countable). But the general problem was still open. But Urysohn's special case turned out to be already useful to find metrisable spaces in analysis, e.g.
    Moreover, it can be proved easily once you covered normal spaces and Urysohn's lemma, so it occurs in many text books.



    Later in the 40's and 50's people started studying open covers and refinements, paracompactness was defined and shown to be equivalent to being "fully normal". Metric spaces were proved to be paracompact that way, but there were also many non-metric paracompact spaces so one needed a strengthening of paracompactness, and this was found to be having a $sigma$-locally finite base (which by results known at that time immediately implied paracompactness) and which was then shown to be equivalent to metrisability by both Smirnov (in Russia) and Nagata (in Japan), independent of each other. Around the same time Bing (in the US) shows that having a $sigma$-discrete base (plus regular $T_0$, which we also need in Nagata-Smirnov) was also necessary and sufficient and used this to prove that a space $X$ is metrisable iff it is a developable regular Hausdorff space which is collectionwise normal (a collectionwise normal Moore space). Other characterisations were also found, using special bases most of the time. But most of that theory, though interesting, is quite extensive to fully cover in most courses, (Munkres does Nagata-Smirnov as an optional chapter, e.g.) and the problem was to show that metrisability is "intrinsically characterisable", not as a practical method to discover metric spaces, or prove concrete spaces to be metrisable.



    It has been used to characterise metrisability in special classes of spaces, like (generalised) ordered spaces. It maybe deserves to be better known, as other metrisation theorems too, but it's too "deep" into general topology problems, and not as practical as Urysohn's theorem, which has an easier to check condition.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have seen the use of the Nagata-Smirnov theorem to prove that a non-compact metrizable space has an incomplete metric.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:38










    • $begingroup$
      My edit was for a typo ("dicover") in the last line of the 2nd-last paragraph.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:40
















    5












    $begingroup$

    Historically, it was an important problem to find a purely topological characterisation of a space $X$ to be metrisable, without a reference to an external object like $mathbb{R}$. Urysohn was one of the first to work on this and he proved some important theorems on embedability using continuous functions on normal spaces (Urysohn's lemma). He could then characterise the separable metric spaces as those that could be embedded into $[0,1]^mathbb{N}$ (using Urysohn functions and a countable base to reduce the number of dimensions in the cube to countable). But the general problem was still open. But Urysohn's special case turned out to be already useful to find metrisable spaces in analysis, e.g.
    Moreover, it can be proved easily once you covered normal spaces and Urysohn's lemma, so it occurs in many text books.



    Later in the 40's and 50's people started studying open covers and refinements, paracompactness was defined and shown to be equivalent to being "fully normal". Metric spaces were proved to be paracompact that way, but there were also many non-metric paracompact spaces so one needed a strengthening of paracompactness, and this was found to be having a $sigma$-locally finite base (which by results known at that time immediately implied paracompactness) and which was then shown to be equivalent to metrisability by both Smirnov (in Russia) and Nagata (in Japan), independent of each other. Around the same time Bing (in the US) shows that having a $sigma$-discrete base (plus regular $T_0$, which we also need in Nagata-Smirnov) was also necessary and sufficient and used this to prove that a space $X$ is metrisable iff it is a developable regular Hausdorff space which is collectionwise normal (a collectionwise normal Moore space). Other characterisations were also found, using special bases most of the time. But most of that theory, though interesting, is quite extensive to fully cover in most courses, (Munkres does Nagata-Smirnov as an optional chapter, e.g.) and the problem was to show that metrisability is "intrinsically characterisable", not as a practical method to discover metric spaces, or prove concrete spaces to be metrisable.



    It has been used to characterise metrisability in special classes of spaces, like (generalised) ordered spaces. It maybe deserves to be better known, as other metrisation theorems too, but it's too "deep" into general topology problems, and not as practical as Urysohn's theorem, which has an easier to check condition.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have seen the use of the Nagata-Smirnov theorem to prove that a non-compact metrizable space has an incomplete metric.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:38










    • $begingroup$
      My edit was for a typo ("dicover") in the last line of the 2nd-last paragraph.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:40














    5












    5








    5





    $begingroup$

    Historically, it was an important problem to find a purely topological characterisation of a space $X$ to be metrisable, without a reference to an external object like $mathbb{R}$. Urysohn was one of the first to work on this and he proved some important theorems on embedability using continuous functions on normal spaces (Urysohn's lemma). He could then characterise the separable metric spaces as those that could be embedded into $[0,1]^mathbb{N}$ (using Urysohn functions and a countable base to reduce the number of dimensions in the cube to countable). But the general problem was still open. But Urysohn's special case turned out to be already useful to find metrisable spaces in analysis, e.g.
    Moreover, it can be proved easily once you covered normal spaces and Urysohn's lemma, so it occurs in many text books.



    Later in the 40's and 50's people started studying open covers and refinements, paracompactness was defined and shown to be equivalent to being "fully normal". Metric spaces were proved to be paracompact that way, but there were also many non-metric paracompact spaces so one needed a strengthening of paracompactness, and this was found to be having a $sigma$-locally finite base (which by results known at that time immediately implied paracompactness) and which was then shown to be equivalent to metrisability by both Smirnov (in Russia) and Nagata (in Japan), independent of each other. Around the same time Bing (in the US) shows that having a $sigma$-discrete base (plus regular $T_0$, which we also need in Nagata-Smirnov) was also necessary and sufficient and used this to prove that a space $X$ is metrisable iff it is a developable regular Hausdorff space which is collectionwise normal (a collectionwise normal Moore space). Other characterisations were also found, using special bases most of the time. But most of that theory, though interesting, is quite extensive to fully cover in most courses, (Munkres does Nagata-Smirnov as an optional chapter, e.g.) and the problem was to show that metrisability is "intrinsically characterisable", not as a practical method to discover metric spaces, or prove concrete spaces to be metrisable.



    It has been used to characterise metrisability in special classes of spaces, like (generalised) ordered spaces. It maybe deserves to be better known, as other metrisation theorems too, but it's too "deep" into general topology problems, and not as practical as Urysohn's theorem, which has an easier to check condition.






    share|cite|improve this answer











    $endgroup$



    Historically, it was an important problem to find a purely topological characterisation of a space $X$ to be metrisable, without a reference to an external object like $mathbb{R}$. Urysohn was one of the first to work on this and he proved some important theorems on embedability using continuous functions on normal spaces (Urysohn's lemma). He could then characterise the separable metric spaces as those that could be embedded into $[0,1]^mathbb{N}$ (using Urysohn functions and a countable base to reduce the number of dimensions in the cube to countable). But the general problem was still open. But Urysohn's special case turned out to be already useful to find metrisable spaces in analysis, e.g.
    Moreover, it can be proved easily once you covered normal spaces and Urysohn's lemma, so it occurs in many text books.



    Later in the 40's and 50's people started studying open covers and refinements, paracompactness was defined and shown to be equivalent to being "fully normal". Metric spaces were proved to be paracompact that way, but there were also many non-metric paracompact spaces so one needed a strengthening of paracompactness, and this was found to be having a $sigma$-locally finite base (which by results known at that time immediately implied paracompactness) and which was then shown to be equivalent to metrisability by both Smirnov (in Russia) and Nagata (in Japan), independent of each other. Around the same time Bing (in the US) shows that having a $sigma$-discrete base (plus regular $T_0$, which we also need in Nagata-Smirnov) was also necessary and sufficient and used this to prove that a space $X$ is metrisable iff it is a developable regular Hausdorff space which is collectionwise normal (a collectionwise normal Moore space). Other characterisations were also found, using special bases most of the time. But most of that theory, though interesting, is quite extensive to fully cover in most courses, (Munkres does Nagata-Smirnov as an optional chapter, e.g.) and the problem was to show that metrisability is "intrinsically characterisable", not as a practical method to discover metric spaces, or prove concrete spaces to be metrisable.



    It has been used to characterise metrisability in special classes of spaces, like (generalised) ordered spaces. It maybe deserves to be better known, as other metrisation theorems too, but it's too "deep" into general topology problems, and not as practical as Urysohn's theorem, which has an easier to check condition.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 29 '18 at 17:21

























    answered Dec 28 '18 at 23:24









    Henno BrandsmaHenno Brandsma

    113k348121




    113k348121












    • $begingroup$
      I have seen the use of the Nagata-Smirnov theorem to prove that a non-compact metrizable space has an incomplete metric.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:38










    • $begingroup$
      My edit was for a typo ("dicover") in the last line of the 2nd-last paragraph.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:40


















    • $begingroup$
      I have seen the use of the Nagata-Smirnov theorem to prove that a non-compact metrizable space has an incomplete metric.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:38










    • $begingroup$
      My edit was for a typo ("dicover") in the last line of the 2nd-last paragraph.
      $endgroup$
      – DanielWainfleet
      Dec 29 '18 at 16:40
















    $begingroup$
    I have seen the use of the Nagata-Smirnov theorem to prove that a non-compact metrizable space has an incomplete metric.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 16:38




    $begingroup$
    I have seen the use of the Nagata-Smirnov theorem to prove that a non-compact metrizable space has an incomplete metric.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 16:38












    $begingroup$
    My edit was for a typo ("dicover") in the last line of the 2nd-last paragraph.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 16:40




    $begingroup$
    My edit was for a typo ("dicover") in the last line of the 2nd-last paragraph.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 16:40











    3












    $begingroup$

    I think the reason for the focus on the Urysohn metrization theorem is really Urysohn's lemma, which is extremely useful and widely applicable, and the Urysohn metrization theorem is just one easy application that one might as well prove anyway. Essentially, the Nagata-Smirnov metrization theorem also uses the Urysohn lemma. Characterizing metric spaces, while interesting, is less widely applicable, and the new ideas that were required are less of a breakthrough.



    In reality, most modern topologists care very little about weird spaces that they need metrization theorems to prove are metrizable. Certainly there are some who do still study these areas, but mostly topologists are focused on algebraic topology where in the sense of general topology the spaces are "boring," for example compact metric spaces, very often manifolds.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      I think the reason for the focus on the Urysohn metrization theorem is really Urysohn's lemma, which is extremely useful and widely applicable, and the Urysohn metrization theorem is just one easy application that one might as well prove anyway. Essentially, the Nagata-Smirnov metrization theorem also uses the Urysohn lemma. Characterizing metric spaces, while interesting, is less widely applicable, and the new ideas that were required are less of a breakthrough.



      In reality, most modern topologists care very little about weird spaces that they need metrization theorems to prove are metrizable. Certainly there are some who do still study these areas, but mostly topologists are focused on algebraic topology where in the sense of general topology the spaces are "boring," for example compact metric spaces, very often manifolds.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        I think the reason for the focus on the Urysohn metrization theorem is really Urysohn's lemma, which is extremely useful and widely applicable, and the Urysohn metrization theorem is just one easy application that one might as well prove anyway. Essentially, the Nagata-Smirnov metrization theorem also uses the Urysohn lemma. Characterizing metric spaces, while interesting, is less widely applicable, and the new ideas that were required are less of a breakthrough.



        In reality, most modern topologists care very little about weird spaces that they need metrization theorems to prove are metrizable. Certainly there are some who do still study these areas, but mostly topologists are focused on algebraic topology where in the sense of general topology the spaces are "boring," for example compact metric spaces, very often manifolds.






        share|cite|improve this answer









        $endgroup$



        I think the reason for the focus on the Urysohn metrization theorem is really Urysohn's lemma, which is extremely useful and widely applicable, and the Urysohn metrization theorem is just one easy application that one might as well prove anyway. Essentially, the Nagata-Smirnov metrization theorem also uses the Urysohn lemma. Characterizing metric spaces, while interesting, is less widely applicable, and the new ideas that were required are less of a breakthrough.



        In reality, most modern topologists care very little about weird spaces that they need metrization theorems to prove are metrizable. Certainly there are some who do still study these areas, but mostly topologists are focused on algebraic topology where in the sense of general topology the spaces are "boring," for example compact metric spaces, very often manifolds.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 11:37









        Matt SamuelMatt Samuel

        38.8k63769




        38.8k63769






























            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%2f3054639%2fwhy-is-urysohns-metrization-theorem-so-much-more-popular-than-the-nagata-smir%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

            Wiesbaden

            Marschland

            Dieringhausen