An infinite set has always a countably infinite subset?












3












$begingroup$


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.










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$endgroup$












  • $begingroup$
    Maybe, an infinite set has always a countably infinite subset, would be a better title.
    $endgroup$
    – dmtri
    5 hours ago






  • 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
    5 hours ago






  • 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
    4 hours ago


















3












$begingroup$


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
    5 hours ago






  • 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
    5 hours ago






  • 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
    4 hours ago
















3












3








3


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






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edited 5 hours ago







HKT

















asked 5 hours ago









HKTHKT

486317




486317












  • $begingroup$
    Maybe, an infinite set has always a countably infinite subset, would be a better title.
    $endgroup$
    – dmtri
    5 hours ago






  • 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
    5 hours ago






  • 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
    4 hours ago




















  • $begingroup$
    Maybe, an infinite set has always a countably infinite subset, would be a better title.
    $endgroup$
    – dmtri
    5 hours ago






  • 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
    5 hours ago






  • 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
    4 hours ago


















$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
5 hours ago




$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
5 hours ago




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
5 hours ago




$begingroup$
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
5 hours ago




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
4 hours ago






$begingroup$
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
4 hours ago












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





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        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 5 hours ago









            Dan UznanskiDan Uznanski

            7,13321528




            7,13321528























                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 5 hours ago









                    Eevee TrainerEevee Trainer

                    8,73431440




                    8,73431440























                        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 5 hours ago









                            Slepecky MamutSlepecky Mamut

                            695313




                            695313






























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Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029