An infinite set has always a countably infinite subset?












4












$begingroup$


Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:



Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:





  1. $I$ is countably infinite: That is, to say, it has the same cardinality as $mathbb{N}$.



    $$ vert I vert = vert mathbb{N} vert $$
    Then it must be that $vert I vert leq vert mathbb{N} vert$ and $vert mathbb{N} vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbb{N} mapsto I$.



    This implies $I subseteq mathbb{N}$ and $mathbb{N} subseteq I$.



    Therefore, $mathbb{N}$ is a subset of $I$.




  2. $I$ is not countably infinite:



    Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.



    As $I setminus {i_1}$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus {i_1}$.



    Similarly, continuing we end up with a countably infinite set: $$ I_c = {i_1, i_2, i_3, dots}$$ from $I$ such that $vert I_c vert = vert mathbb{N} vert$. Then it must be that $vert I_c vert leq vert mathbb{N} vert $ and $vert mathbb{N} vert leq vert I_c vert $.



    Therefore, a countably infinite set is a subset of $I_c$.




All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe, an infinite set has always a countably infinite subset, would be a better title.
    $endgroup$
    – dmtri
    Mar 27 at 8:57






  • 3




    $begingroup$
    It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
    $endgroup$
    – 5xum
    Mar 27 at 9:11






  • 1




    $begingroup$
    Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
    $endgroup$
    – Ilmari Karonen
    Mar 27 at 10:09








  • 1




    $begingroup$
    At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbb{N}$ really a subset of $mathbb{R}$? I mean, is $3={*,*,*}$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbb{N}$.
    $endgroup$
    – Chrystomath
    Mar 27 at 18:07


















4












$begingroup$


Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:



Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:





  1. $I$ is countably infinite: That is, to say, it has the same cardinality as $mathbb{N}$.



    $$ vert I vert = vert mathbb{N} vert $$
    Then it must be that $vert I vert leq vert mathbb{N} vert$ and $vert mathbb{N} vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbb{N} mapsto I$.



    This implies $I subseteq mathbb{N}$ and $mathbb{N} subseteq I$.



    Therefore, $mathbb{N}$ is a subset of $I$.




  2. $I$ is not countably infinite:



    Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.



    As $I setminus {i_1}$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus {i_1}$.



    Similarly, continuing we end up with a countably infinite set: $$ I_c = {i_1, i_2, i_3, dots}$$ from $I$ such that $vert I_c vert = vert mathbb{N} vert$. Then it must be that $vert I_c vert leq vert mathbb{N} vert $ and $vert mathbb{N} vert leq vert I_c vert $.



    Therefore, a countably infinite set is a subset of $I_c$.




All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe, an infinite set has always a countably infinite subset, would be a better title.
    $endgroup$
    – dmtri
    Mar 27 at 8:57






  • 3




    $begingroup$
    It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
    $endgroup$
    – 5xum
    Mar 27 at 9:11






  • 1




    $begingroup$
    Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
    $endgroup$
    – Ilmari Karonen
    Mar 27 at 10:09








  • 1




    $begingroup$
    At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbb{N}$ really a subset of $mathbb{R}$? I mean, is $3={*,*,*}$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbb{N}$.
    $endgroup$
    – Chrystomath
    Mar 27 at 18:07
















4












4








4


2



$begingroup$


Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:



Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:





  1. $I$ is countably infinite: That is, to say, it has the same cardinality as $mathbb{N}$.



    $$ vert I vert = vert mathbb{N} vert $$
    Then it must be that $vert I vert leq vert mathbb{N} vert$ and $vert mathbb{N} vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbb{N} mapsto I$.



    This implies $I subseteq mathbb{N}$ and $mathbb{N} subseteq I$.



    Therefore, $mathbb{N}$ is a subset of $I$.




  2. $I$ is not countably infinite:



    Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.



    As $I setminus {i_1}$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus {i_1}$.



    Similarly, continuing we end up with a countably infinite set: $$ I_c = {i_1, i_2, i_3, dots}$$ from $I$ such that $vert I_c vert = vert mathbb{N} vert$. Then it must be that $vert I_c vert leq vert mathbb{N} vert $ and $vert mathbb{N} vert leq vert I_c vert $.



    Therefore, a countably infinite set is a subset of $I_c$.




All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.










share|cite|improve this question











$endgroup$




Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:



Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:





  1. $I$ is countably infinite: That is, to say, it has the same cardinality as $mathbb{N}$.



    $$ vert I vert = vert mathbb{N} vert $$
    Then it must be that $vert I vert leq vert mathbb{N} vert$ and $vert mathbb{N} vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbb{N} mapsto I$.



    This implies $I subseteq mathbb{N}$ and $mathbb{N} subseteq I$.



    Therefore, $mathbb{N}$ is a subset of $I$.




  2. $I$ is not countably infinite:



    Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.



    As $I setminus {i_1}$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus {i_1}$.



    Similarly, continuing we end up with a countably infinite set: $$ I_c = {i_1, i_2, i_3, dots}$$ from $I$ such that $vert I_c vert = vert mathbb{N} vert$. Then it must be that $vert I_c vert leq vert mathbb{N} vert $ and $vert mathbb{N} vert leq vert I_c vert $.



    Therefore, a countably infinite set is a subset of $I_c$.




All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.







proof-verification elementary-set-theory proof-writing alternative-proof






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 27 at 9:14







HKT

















asked Mar 27 at 8:50









HKTHKT

421317




421317












  • $begingroup$
    Maybe, an infinite set has always a countably infinite subset, would be a better title.
    $endgroup$
    – dmtri
    Mar 27 at 8:57






  • 3




    $begingroup$
    It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
    $endgroup$
    – 5xum
    Mar 27 at 9:11






  • 1




    $begingroup$
    Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
    $endgroup$
    – Ilmari Karonen
    Mar 27 at 10:09








  • 1




    $begingroup$
    At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbb{N}$ really a subset of $mathbb{R}$? I mean, is $3={*,*,*}$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbb{N}$.
    $endgroup$
    – Chrystomath
    Mar 27 at 18:07




















  • $begingroup$
    Maybe, an infinite set has always a countably infinite subset, would be a better title.
    $endgroup$
    – dmtri
    Mar 27 at 8:57






  • 3




    $begingroup$
    It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
    $endgroup$
    – 5xum
    Mar 27 at 9:11






  • 1




    $begingroup$
    Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
    $endgroup$
    – Ilmari Karonen
    Mar 27 at 10:09








  • 1




    $begingroup$
    At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbb{N}$ really a subset of $mathbb{R}$? I mean, is $3={*,*,*}$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbb{N}$.
    $endgroup$
    – Chrystomath
    Mar 27 at 18:07


















$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
Mar 27 at 8:57




$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
Mar 27 at 8:57




3




3




$begingroup$
It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
$endgroup$
– 5xum
Mar 27 at 9:11




$begingroup$
It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
$endgroup$
– 5xum
Mar 27 at 9:11




1




1




$begingroup$
Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
$endgroup$
– Ilmari Karonen
Mar 27 at 10:09






$begingroup$
Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
$endgroup$
– Ilmari Karonen
Mar 27 at 10:09






1




1




$begingroup$
At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbb{N}$ really a subset of $mathbb{R}$? I mean, is $3={*,*,*}$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbb{N}$.
$endgroup$
– Chrystomath
Mar 27 at 18:07






$begingroup$
At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbb{N}$ really a subset of $mathbb{R}$? I mean, is $3={*,*,*}$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbb{N}$.
$endgroup$
– Chrystomath
Mar 27 at 18:07












3 Answers
3






active

oldest

votes


















3












$begingroup$

Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.



The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.



Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$


    Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?




    No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.



    Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
      Counterexample is set of all irrational numbers.
      The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.






      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%2f3164234%2fan-infinite-set-has-always-a-countably-infinite-subset%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.



        The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.



        Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.



          The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.



          Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.



            The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.



            Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.






            share|cite|improve this answer









            $endgroup$



            Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.



            The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.



            Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 27 at 8:56









            Dan UznanskiDan Uznanski

            7,15821528




            7,15821528























                3












                $begingroup$


                Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?




                No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.



                Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$


                  Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?




                  No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.



                  Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$


                    Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?




                    No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.



                    Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.






                    share|cite|improve this answer









                    $endgroup$




                    Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?




                    No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.



                    Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 27 at 8:58









                    Eevee TrainerEevee Trainer

                    9,23331640




                    9,23331640























                        2












                        $begingroup$

                        In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
                        Counterexample is set of all irrational numbers.
                        The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
                          Counterexample is set of all irrational numbers.
                          The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
                            Counterexample is set of all irrational numbers.
                            The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.






                            share|cite|improve this answer









                            $endgroup$



                            In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
                            Counterexample is set of all irrational numbers.
                            The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 27 at 8:59









                            Slepecky MamutSlepecky Mamut

                            695313




                            695313






























                                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%2f3164234%2fan-infinite-set-has-always-a-countably-infinite-subset%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

                                Plaza Victoria

                                Puebla de Zaragoza

                                Musa