Every perfect set has cardinality $2^{aleph_0}$?












0












$begingroup$


It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:




Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.




This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?



I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...






general-topology descriptive-set-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The proof is by showing there is a "Cantor set" in any such perfect sets
    $endgroup$
    – Paul Plummer
    Dec 31 '18 at 1:07










  • $begingroup$
    In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
    $endgroup$
    – spaceisdarkgreen
    Dec 31 '18 at 1:19


















0












$begingroup$


It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:




Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.




This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?



I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...






general-topology descriptive-set-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The proof is by showing there is a "Cantor set" in any such perfect sets
    $endgroup$
    – Paul Plummer
    Dec 31 '18 at 1:07










  • $begingroup$
    In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
    $endgroup$
    – spaceisdarkgreen
    Dec 31 '18 at 1:19
















0












0








0





$begingroup$


It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:




Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.




This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?



I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...






general-topology descriptive-set-theory







share|cite|improve this question











$endgroup$




It is well known that perfect sets in $mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:




Every perfect set in $mathbb{R}^n$ has cardinality $2^{aleph_0}$.




This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?



I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...






general-topology descriptive-set-theory




general-topology descriptive-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 18:57









Andrés E. Caicedo

65.8k8160251




65.8k8160251










asked Dec 31 '18 at 0:58









ColescuColescu

3,26711136




3,26711136








  • 2




    $begingroup$
    The proof is by showing there is a "Cantor set" in any such perfect sets
    $endgroup$
    – Paul Plummer
    Dec 31 '18 at 1:07










  • $begingroup$
    In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
    $endgroup$
    – spaceisdarkgreen
    Dec 31 '18 at 1:19
















  • 2




    $begingroup$
    The proof is by showing there is a "Cantor set" in any such perfect sets
    $endgroup$
    – Paul Plummer
    Dec 31 '18 at 1:07










  • $begingroup$
    In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
    $endgroup$
    – spaceisdarkgreen
    Dec 31 '18 at 1:19










2




2




$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07




$begingroup$
The proof is by showing there is a "Cantor set" in any such perfect sets
$endgroup$
– Paul Plummer
Dec 31 '18 at 1:07












$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19






$begingroup$
In answer to the question about whether it generalizes, the "natural setting" is Polish spaces en.wikipedia.org/wiki/Polish_space en.wikipedia.org/wiki/Perfect_set_property
$endgroup$
– spaceisdarkgreen
Dec 31 '18 at 1:19












1 Answer
1






active

oldest

votes


















4












$begingroup$

Yes, it does not require the continuum hypothesis to prove.



Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.



Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.



Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).




It might seem like we used the axiom of choice here, in two places:




  • Choosing $l(X)$ and $r(X)$, for a perfect set $X$.


  • Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.



However, we don't actually need AC here:




  • If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...


  • There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).







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%2f3057339%2fevery-perfect-set-has-cardinality-2-aleph-0%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Yes, it does not require the continuum hypothesis to prove.



    Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.



    Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.



    Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).




    It might seem like we used the axiom of choice here, in two places:




    • Choosing $l(X)$ and $r(X)$, for a perfect set $X$.


    • Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.



    However, we don't actually need AC here:




    • If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...


    • There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).







    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Yes, it does not require the continuum hypothesis to prove.



      Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.



      Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.



      Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).




      It might seem like we used the axiom of choice here, in two places:




      • Choosing $l(X)$ and $r(X)$, for a perfect set $X$.


      • Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.



      However, we don't actually need AC here:




      • If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...


      • There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).







      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Yes, it does not require the continuum hypothesis to prove.



        Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.



        Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.



        Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).




        It might seem like we used the axiom of choice here, in two places:




        • Choosing $l(X)$ and $r(X)$, for a perfect set $X$.


        • Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.



        However, we don't actually need AC here:




        • If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...


        • There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).







        share|cite|improve this answer









        $endgroup$



        Yes, it does not require the continuum hypothesis to prove.



        Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.



        Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)supseteq r(l(P))supseteq l(r(l(P)))supseteq r(l(r(l(P))))supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.



        Now just check that if $fnot=g$ we have $p_fnot=p_g$ (HINT: $r(X)cap l(X)=emptyset$ ...).




        It might seem like we used the axiom of choice here, in two places:




        • Choosing $l(X)$ and $r(X)$, for a perfect set $X$.


        • Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.



        However, we don't actually need AC here:




        • If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $Xcap (-infty, p]$ and $Xcap [q,infty)$ are each perfect. But $mathbb{Q}^2$ is well-orderable ...


        • There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 1:08









        Noah SchweberNoah Schweber

        127k10151293




        127k10151293






























            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%2f3057339%2fevery-perfect-set-has-cardinality-2-aleph-0%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more