Demonstrate $A subseteq P(cup A)$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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$A$ is a set



$P(A)=B$



$ cup A = x $



My try:



If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$



then $xsubseteq P(cup A)$



I fell like this last step is not quite right.










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    up vote
    -2
    down vote

    favorite












    $A$ is a set



    $P(A)=B$



    $ cup A = x $



    My try:



    If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$



    then $xsubseteq P(cup A)$



    I fell like this last step is not quite right.










    share|cite|improve this question

























      up vote
      -2
      down vote

      favorite









      up vote
      -2
      down vote

      favorite











      $A$ is a set



      $P(A)=B$



      $ cup A = x $



      My try:



      If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$



      then $xsubseteq P(cup A)$



      I fell like this last step is not quite right.










      share|cite|improve this question















      $A$ is a set



      $P(A)=B$



      $ cup A = x $



      My try:



      If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$



      then $xsubseteq P(cup A)$



      I fell like this last step is not quite right.







      elementary-set-theory






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      share|cite|improve this question













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      edited Sep 10 at 19:06

























      asked Sep 10 at 18:51









      Sebas Martinez Santos

      106




      106




















          1 Answer
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          Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$



          Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$






          share|cite|improve this answer




















          • But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
            – Sebas Martinez Santos
            Sep 10 at 19:17







          • 2




            @SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
            – JMoravitz
            Sep 10 at 19:33











          • Exactly, be careful Sebas that an element of A can be a set itself.
            – Riccardo Ceccon
            Sep 10 at 20:10










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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

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          up vote
          1
          down vote













          Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$



          Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$






          share|cite|improve this answer




















          • But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
            – Sebas Martinez Santos
            Sep 10 at 19:17







          • 2




            @SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
            – JMoravitz
            Sep 10 at 19:33











          • Exactly, be careful Sebas that an element of A can be a set itself.
            – Riccardo Ceccon
            Sep 10 at 20:10














          up vote
          1
          down vote













          Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$



          Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$






          share|cite|improve this answer




















          • But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
            – Sebas Martinez Santos
            Sep 10 at 19:17







          • 2




            @SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
            – JMoravitz
            Sep 10 at 19:33











          • Exactly, be careful Sebas that an element of A can be a set itself.
            – Riccardo Ceccon
            Sep 10 at 20:10












          up vote
          1
          down vote










          up vote
          1
          down vote









          Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$



          Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$






          share|cite|improve this answer












          Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$



          Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 10 at 18:59









          Riccardo Ceccon

          875320




          875320











          • But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
            – Sebas Martinez Santos
            Sep 10 at 19:17







          • 2




            @SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
            – JMoravitz
            Sep 10 at 19:33











          • Exactly, be careful Sebas that an element of A can be a set itself.
            – Riccardo Ceccon
            Sep 10 at 20:10
















          • But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
            – Sebas Martinez Santos
            Sep 10 at 19:17







          • 2




            @SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
            – JMoravitz
            Sep 10 at 19:33











          • Exactly, be careful Sebas that an element of A can be a set itself.
            – Riccardo Ceccon
            Sep 10 at 20:10















          But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
          – Sebas Martinez Santos
          Sep 10 at 19:17





          But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
          – Sebas Martinez Santos
          Sep 10 at 19:17





          2




          2




          @SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
          – JMoravitz
          Sep 10 at 19:33





          @SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
          – JMoravitz
          Sep 10 at 19:33













          Exactly, be careful Sebas that an element of A can be a set itself.
          – Riccardo Ceccon
          Sep 10 at 20:10




          Exactly, be careful Sebas that an element of A can be a set itself.
          – Riccardo Ceccon
          Sep 10 at 20:10

















           

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