Countable Infinite and Uncountable Infinite sets












0












$begingroup$


Mark each statement as TRUE, FALSE, or UNKNOWN



(a) $|Bbb{R}| < aleph_1$



(b) $|Bbb{R}| = aleph_1$



(c) $|P(Bbb{R})| > aleph_1$



Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets










share|cite|improve this question









$endgroup$












  • $begingroup$
    Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:37










  • $begingroup$
    @fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
    $endgroup$
    – Viserom
    Dec 19 '18 at 6:41










  • $begingroup$
    Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:45
















0












$begingroup$


Mark each statement as TRUE, FALSE, or UNKNOWN



(a) $|Bbb{R}| < aleph_1$



(b) $|Bbb{R}| = aleph_1$



(c) $|P(Bbb{R})| > aleph_1$



Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets










share|cite|improve this question









$endgroup$












  • $begingroup$
    Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:37










  • $begingroup$
    @fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
    $endgroup$
    – Viserom
    Dec 19 '18 at 6:41










  • $begingroup$
    Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:45














0












0








0





$begingroup$


Mark each statement as TRUE, FALSE, or UNKNOWN



(a) $|Bbb{R}| < aleph_1$



(b) $|Bbb{R}| = aleph_1$



(c) $|P(Bbb{R})| > aleph_1$



Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets










share|cite|improve this question









$endgroup$




Mark each statement as TRUE, FALSE, or UNKNOWN



(a) $|Bbb{R}| < aleph_1$



(b) $|Bbb{R}| = aleph_1$



(c) $|P(Bbb{R})| > aleph_1$



Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets







infinite-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 6:31









ViseromViserom

123




123












  • $begingroup$
    Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:37










  • $begingroup$
    @fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
    $endgroup$
    – Viserom
    Dec 19 '18 at 6:41










  • $begingroup$
    Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:45


















  • $begingroup$
    Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:37










  • $begingroup$
    @fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
    $endgroup$
    – Viserom
    Dec 19 '18 at 6:41










  • $begingroup$
    Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
    $endgroup$
    – fleablood
    Dec 19 '18 at 6:45
















$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37




$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37












$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41




$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41












$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45




$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45










1 Answer
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$begingroup$

$aleph_1$ by definition is the first uncountable cardinal number.
So $|A| < aleph_1$ by definition means that $A$ is a countable set.
The reals are not countable.



We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.



By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.






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    0












    $begingroup$

    $aleph_1$ by definition is the first uncountable cardinal number.
    So $|A| < aleph_1$ by definition means that $A$ is a countable set.
    The reals are not countable.



    We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
    The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.



    By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $aleph_1$ by definition is the first uncountable cardinal number.
      So $|A| < aleph_1$ by definition means that $A$ is a countable set.
      The reals are not countable.



      We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
      The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.



      By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $aleph_1$ by definition is the first uncountable cardinal number.
        So $|A| < aleph_1$ by definition means that $A$ is a countable set.
        The reals are not countable.



        We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
        The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.



        By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.






        share|cite|improve this answer









        $endgroup$



        $aleph_1$ by definition is the first uncountable cardinal number.
        So $|A| < aleph_1$ by definition means that $A$ is a countable set.
        The reals are not countable.



        We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
        The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.



        By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 7:28









        Henno BrandsmaHenno Brandsma

        114k348124




        114k348124






























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