Cardinality of ${emptyset, mathcal{P}(emptyset), {emptyset} } $












8












$begingroup$


This is an old exam question from "Diskrete Mathematik" at ETH Zurich



$mathcal{P}(A)$ denotes the Powerset, which is the set of all subsets of A:



$mathcal{P}(A) := {S|Ssubseteq A }$.



I believe that ${emptyset, mathcal{P}(emptyset), {emptyset} } $ since
$mathcal{P}(emptyset) = {emptyset} $ can be reduced to ${emptyset, {emptyset} } $ and thus the cardinality of said set would be 2. However, the inofficial solution is 3.






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Who wrote this unofficial solution? Unofficial solutions have the problem that they might contain typos. I mean, also official solutions can contain typos, but that's another story.
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:08






  • 2




    $begingroup$
    Can you be absolutely, totally, 100% sure that $mathcal{P}(X)$ denotes the power set of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jan 16 '18 at 16:09










  • $begingroup$
    @Daniel: Good point, in some old set theory texts, it sometimes appeared as $S(X)$ actually. (I guess "subsets"...)
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:10












  • $begingroup$
    Your answer, $2$, is correct.
    $endgroup$
    – Andreas Blass
    Jan 16 '18 at 18:31










  • $begingroup$
    @DanielFischer Yes
    $endgroup$
    – Waterhouse
    Jan 26 '18 at 12:30
















8












$begingroup$


This is an old exam question from "Diskrete Mathematik" at ETH Zurich



$mathcal{P}(A)$ denotes the Powerset, which is the set of all subsets of A:



$mathcal{P}(A) := {S|Ssubseteq A }$.



I believe that ${emptyset, mathcal{P}(emptyset), {emptyset} } $ since
$mathcal{P}(emptyset) = {emptyset} $ can be reduced to ${emptyset, {emptyset} } $ and thus the cardinality of said set would be 2. However, the inofficial solution is 3.






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Who wrote this unofficial solution? Unofficial solutions have the problem that they might contain typos. I mean, also official solutions can contain typos, but that's another story.
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:08






  • 2




    $begingroup$
    Can you be absolutely, totally, 100% sure that $mathcal{P}(X)$ denotes the power set of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jan 16 '18 at 16:09










  • $begingroup$
    @Daniel: Good point, in some old set theory texts, it sometimes appeared as $S(X)$ actually. (I guess "subsets"...)
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:10












  • $begingroup$
    Your answer, $2$, is correct.
    $endgroup$
    – Andreas Blass
    Jan 16 '18 at 18:31










  • $begingroup$
    @DanielFischer Yes
    $endgroup$
    – Waterhouse
    Jan 26 '18 at 12:30














8












8








8


1



$begingroup$


This is an old exam question from "Diskrete Mathematik" at ETH Zurich



$mathcal{P}(A)$ denotes the Powerset, which is the set of all subsets of A:



$mathcal{P}(A) := {S|Ssubseteq A }$.



I believe that ${emptyset, mathcal{P}(emptyset), {emptyset} } $ since
$mathcal{P}(emptyset) = {emptyset} $ can be reduced to ${emptyset, {emptyset} } $ and thus the cardinality of said set would be 2. However, the inofficial solution is 3.






elementary-set-theory







share|cite|improve this question











$endgroup$




This is an old exam question from "Diskrete Mathematik" at ETH Zurich



$mathcal{P}(A)$ denotes the Powerset, which is the set of all subsets of A:



$mathcal{P}(A) := {S|Ssubseteq A }$.



I believe that ${emptyset, mathcal{P}(emptyset), {emptyset} } $ since
$mathcal{P}(emptyset) = {emptyset} $ can be reduced to ${emptyset, {emptyset} } $ and thus the cardinality of said set would be 2. However, the inofficial solution is 3.






elementary-set-theory




elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 '18 at 16:20







Waterhouse

















asked Jan 16 '18 at 16:04









WaterhouseWaterhouse

1218




1218












  • $begingroup$
    Who wrote this unofficial solution? Unofficial solutions have the problem that they might contain typos. I mean, also official solutions can contain typos, but that's another story.
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:08






  • 2




    $begingroup$
    Can you be absolutely, totally, 100% sure that $mathcal{P}(X)$ denotes the power set of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jan 16 '18 at 16:09










  • $begingroup$
    @Daniel: Good point, in some old set theory texts, it sometimes appeared as $S(X)$ actually. (I guess "subsets"...)
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:10












  • $begingroup$
    Your answer, $2$, is correct.
    $endgroup$
    – Andreas Blass
    Jan 16 '18 at 18:31










  • $begingroup$
    @DanielFischer Yes
    $endgroup$
    – Waterhouse
    Jan 26 '18 at 12:30


















  • $begingroup$
    Who wrote this unofficial solution? Unofficial solutions have the problem that they might contain typos. I mean, also official solutions can contain typos, but that's another story.
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:08






  • 2




    $begingroup$
    Can you be absolutely, totally, 100% sure that $mathcal{P}(X)$ denotes the power set of $X$?
    $endgroup$
    – Daniel Fischer♦
    Jan 16 '18 at 16:09










  • $begingroup$
    @Daniel: Good point, in some old set theory texts, it sometimes appeared as $S(X)$ actually. (I guess "subsets"...)
    $endgroup$
    – Asaf Karagila♦
    Jan 16 '18 at 16:10












  • $begingroup$
    Your answer, $2$, is correct.
    $endgroup$
    – Andreas Blass
    Jan 16 '18 at 18:31










  • $begingroup$
    @DanielFischer Yes
    $endgroup$
    – Waterhouse
    Jan 26 '18 at 12:30
















$begingroup$
Who wrote this unofficial solution? Unofficial solutions have the problem that they might contain typos. I mean, also official solutions can contain typos, but that's another story.
$endgroup$
– Asaf Karagila♦
Jan 16 '18 at 16:08




$begingroup$
Who wrote this unofficial solution? Unofficial solutions have the problem that they might contain typos. I mean, also official solutions can contain typos, but that's another story.
$endgroup$
– Asaf Karagila♦
Jan 16 '18 at 16:08




2




2




$begingroup$
Can you be absolutely, totally, 100% sure that $mathcal{P}(X)$ denotes the power set of $X$?
$endgroup$
– Daniel Fischer♦
Jan 16 '18 at 16:09




$begingroup$
Can you be absolutely, totally, 100% sure that $mathcal{P}(X)$ denotes the power set of $X$?
$endgroup$
– Daniel Fischer♦
Jan 16 '18 at 16:09












$begingroup$
@Daniel: Good point, in some old set theory texts, it sometimes appeared as $S(X)$ actually. (I guess "subsets"...)
$endgroup$
– Asaf Karagila♦
Jan 16 '18 at 16:10






$begingroup$
@Daniel: Good point, in some old set theory texts, it sometimes appeared as $S(X)$ actually. (I guess "subsets"...)
$endgroup$
– Asaf Karagila♦
Jan 16 '18 at 16:10














$begingroup$
Your answer, $2$, is correct.
$endgroup$
– Andreas Blass
Jan 16 '18 at 18:31




$begingroup$
Your answer, $2$, is correct.
$endgroup$
– Andreas Blass
Jan 16 '18 at 18:31












$begingroup$
@DanielFischer Yes
$endgroup$
– Waterhouse
Jan 26 '18 at 12:30




$begingroup$
@DanielFischer Yes
$endgroup$
– Waterhouse
Jan 26 '18 at 12:30










1 Answer
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$begingroup$

The power set of a finite set with cardinality $x$ has cardinality $2^x$. As you pointed out, ${emptyset, mathcal{P}(emptyset), {emptyset} }$ can be simplified down to ${emptyset, {emptyset} }$, which is equivalent to $mathcal{P}({emptyset})$. Using the property I described initially, we end up with,



$$
2^{|{emptyset}|} = 2^1 = 2.
$$



Therefore, the correct solution is $2$ and not $3$ as the unofficial solution suggested.






share|cite|improve this answer









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    $begingroup$

    The power set of a finite set with cardinality $x$ has cardinality $2^x$. As you pointed out, ${emptyset, mathcal{P}(emptyset), {emptyset} }$ can be simplified down to ${emptyset, {emptyset} }$, which is equivalent to $mathcal{P}({emptyset})$. Using the property I described initially, we end up with,



    $$
    2^{|{emptyset}|} = 2^1 = 2.
    $$



    Therefore, the correct solution is $2$ and not $3$ as the unofficial solution suggested.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The power set of a finite set with cardinality $x$ has cardinality $2^x$. As you pointed out, ${emptyset, mathcal{P}(emptyset), {emptyset} }$ can be simplified down to ${emptyset, {emptyset} }$, which is equivalent to $mathcal{P}({emptyset})$. Using the property I described initially, we end up with,



      $$
      2^{|{emptyset}|} = 2^1 = 2.
      $$



      Therefore, the correct solution is $2$ and not $3$ as the unofficial solution suggested.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The power set of a finite set with cardinality $x$ has cardinality $2^x$. As you pointed out, ${emptyset, mathcal{P}(emptyset), {emptyset} }$ can be simplified down to ${emptyset, {emptyset} }$, which is equivalent to $mathcal{P}({emptyset})$. Using the property I described initially, we end up with,



        $$
        2^{|{emptyset}|} = 2^1 = 2.
        $$



        Therefore, the correct solution is $2$ and not $3$ as the unofficial solution suggested.






        share|cite|improve this answer









        $endgroup$



        The power set of a finite set with cardinality $x$ has cardinality $2^x$. As you pointed out, ${emptyset, mathcal{P}(emptyset), {emptyset} }$ can be simplified down to ${emptyset, {emptyset} }$, which is equivalent to $mathcal{P}({emptyset})$. Using the property I described initially, we end up with,



        $$
        2^{|{emptyset}|} = 2^1 = 2.
        $$



        Therefore, the correct solution is $2$ and not $3$ as the unofficial solution suggested.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 16:39









        EdOverflowEdOverflow

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