Why is the set whose elements are a collection of sets equal to the union of that collection?












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$begingroup$


Why is {$A_i$ : $i in I$} = $bigcup$$_{i in I} A_i$?



Suppose $I = {1,2,3}$, $A_1 = {a}$, $A_2 = {b}$, and $A_3 = {c}$



Then,




  • {$A_i : i in I$} = ${A_1, A_2, A_3}$ = ${{a},{b},{c}}$


  • $bigcup$$_{i in I} A_i$ = ${a,b,c}$



It doesn't make sense to me that ${{a},{b},{c}}$ = ${a,b,c}$. These look like two different sets.






elementary-set-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This isn't true. Who said it was?
    $endgroup$
    – Randall
    Dec 22 '18 at 3:35










  • $begingroup$
    "It doesn't make sense to me that ___ ** = ** ___" Good. It shouldn't make sense to you because they aren't the same.
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:37










  • $begingroup$
    $bigcuplimits_{iin I} A_i = {x~:~exists iin I~text{such that}~xin A_i}$
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:39






  • 1




    $begingroup$
    ${A_i:iin I}nebigcup_{iin I}A_i$ but $bigcup{A_i:iin I}=bigcup_{iin I}A_i$. Could that be what's confusing you?
    $endgroup$
    – bof
    Dec 22 '18 at 4:08












  • $begingroup$
    Yes, thank you bof and the rest of you guys. It's making more sense now. $bigcup$${A_i : i in I}$ is just the short hand for $bigcup$$_{i in I} A_i$. I see it now.
    $endgroup$
    – Andres Kiani
    Dec 22 '18 at 5:20


















0












$begingroup$


Why is {$A_i$ : $i in I$} = $bigcup$$_{i in I} A_i$?



Suppose $I = {1,2,3}$, $A_1 = {a}$, $A_2 = {b}$, and $A_3 = {c}$



Then,




  • {$A_i : i in I$} = ${A_1, A_2, A_3}$ = ${{a},{b},{c}}$


  • $bigcup$$_{i in I} A_i$ = ${a,b,c}$



It doesn't make sense to me that ${{a},{b},{c}}$ = ${a,b,c}$. These look like two different sets.






elementary-set-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This isn't true. Who said it was?
    $endgroup$
    – Randall
    Dec 22 '18 at 3:35










  • $begingroup$
    "It doesn't make sense to me that ___ ** = ** ___" Good. It shouldn't make sense to you because they aren't the same.
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:37










  • $begingroup$
    $bigcuplimits_{iin I} A_i = {x~:~exists iin I~text{such that}~xin A_i}$
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:39






  • 1




    $begingroup$
    ${A_i:iin I}nebigcup_{iin I}A_i$ but $bigcup{A_i:iin I}=bigcup_{iin I}A_i$. Could that be what's confusing you?
    $endgroup$
    – bof
    Dec 22 '18 at 4:08












  • $begingroup$
    Yes, thank you bof and the rest of you guys. It's making more sense now. $bigcup$${A_i : i in I}$ is just the short hand for $bigcup$$_{i in I} A_i$. I see it now.
    $endgroup$
    – Andres Kiani
    Dec 22 '18 at 5:20
















0












0








0





$begingroup$


Why is {$A_i$ : $i in I$} = $bigcup$$_{i in I} A_i$?



Suppose $I = {1,2,3}$, $A_1 = {a}$, $A_2 = {b}$, and $A_3 = {c}$



Then,




  • {$A_i : i in I$} = ${A_1, A_2, A_3}$ = ${{a},{b},{c}}$


  • $bigcup$$_{i in I} A_i$ = ${a,b,c}$



It doesn't make sense to me that ${{a},{b},{c}}$ = ${a,b,c}$. These look like two different sets.






elementary-set-theory







share|cite|improve this question











$endgroup$




Why is {$A_i$ : $i in I$} = $bigcup$$_{i in I} A_i$?



Suppose $I = {1,2,3}$, $A_1 = {a}$, $A_2 = {b}$, and $A_3 = {c}$



Then,




  • {$A_i : i in I$} = ${A_1, A_2, A_3}$ = ${{a},{b},{c}}$


  • $bigcup$$_{i in I} A_i$ = ${a,b,c}$



It doesn't make sense to me that ${{a},{b},{c}}$ = ${a,b,c}$. These look like two different sets.






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 Dec 22 '18 at 3:37







Andres Kiani

















asked Dec 22 '18 at 3:32









Andres KianiAndres Kiani

52




52








  • 1




    $begingroup$
    This isn't true. Who said it was?
    $endgroup$
    – Randall
    Dec 22 '18 at 3:35










  • $begingroup$
    "It doesn't make sense to me that ___ ** = ** ___" Good. It shouldn't make sense to you because they aren't the same.
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:37










  • $begingroup$
    $bigcuplimits_{iin I} A_i = {x~:~exists iin I~text{such that}~xin A_i}$
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:39






  • 1




    $begingroup$
    ${A_i:iin I}nebigcup_{iin I}A_i$ but $bigcup{A_i:iin I}=bigcup_{iin I}A_i$. Could that be what's confusing you?
    $endgroup$
    – bof
    Dec 22 '18 at 4:08












  • $begingroup$
    Yes, thank you bof and the rest of you guys. It's making more sense now. $bigcup$${A_i : i in I}$ is just the short hand for $bigcup$$_{i in I} A_i$. I see it now.
    $endgroup$
    – Andres Kiani
    Dec 22 '18 at 5:20
















  • 1




    $begingroup$
    This isn't true. Who said it was?
    $endgroup$
    – Randall
    Dec 22 '18 at 3:35










  • $begingroup$
    "It doesn't make sense to me that ___ ** = ** ___" Good. It shouldn't make sense to you because they aren't the same.
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:37










  • $begingroup$
    $bigcuplimits_{iin I} A_i = {x~:~exists iin I~text{such that}~xin A_i}$
    $endgroup$
    – JMoravitz
    Dec 22 '18 at 3:39






  • 1




    $begingroup$
    ${A_i:iin I}nebigcup_{iin I}A_i$ but $bigcup{A_i:iin I}=bigcup_{iin I}A_i$. Could that be what's confusing you?
    $endgroup$
    – bof
    Dec 22 '18 at 4:08












  • $begingroup$
    Yes, thank you bof and the rest of you guys. It's making more sense now. $bigcup$${A_i : i in I}$ is just the short hand for $bigcup$$_{i in I} A_i$. I see it now.
    $endgroup$
    – Andres Kiani
    Dec 22 '18 at 5:20










1




1




$begingroup$
This isn't true. Who said it was?
$endgroup$
– Randall
Dec 22 '18 at 3:35




$begingroup$
This isn't true. Who said it was?
$endgroup$
– Randall
Dec 22 '18 at 3:35












$begingroup$
"It doesn't make sense to me that ___ ** = ** ___" Good. It shouldn't make sense to you because they aren't the same.
$endgroup$
– JMoravitz
Dec 22 '18 at 3:37




$begingroup$
"It doesn't make sense to me that ___ ** = ** ___" Good. It shouldn't make sense to you because they aren't the same.
$endgroup$
– JMoravitz
Dec 22 '18 at 3:37












$begingroup$
$bigcuplimits_{iin I} A_i = {x~:~exists iin I~text{such that}~xin A_i}$
$endgroup$
– JMoravitz
Dec 22 '18 at 3:39




$begingroup$
$bigcuplimits_{iin I} A_i = {x~:~exists iin I~text{such that}~xin A_i}$
$endgroup$
– JMoravitz
Dec 22 '18 at 3:39




1




1




$begingroup$
${A_i:iin I}nebigcup_{iin I}A_i$ but $bigcup{A_i:iin I}=bigcup_{iin I}A_i$. Could that be what's confusing you?
$endgroup$
– bof
Dec 22 '18 at 4:08






$begingroup$
${A_i:iin I}nebigcup_{iin I}A_i$ but $bigcup{A_i:iin I}=bigcup_{iin I}A_i$. Could that be what's confusing you?
$endgroup$
– bof
Dec 22 '18 at 4:08














$begingroup$
Yes, thank you bof and the rest of you guys. It's making more sense now. $bigcup$${A_i : i in I}$ is just the short hand for $bigcup$$_{i in I} A_i$. I see it now.
$endgroup$
– Andres Kiani
Dec 22 '18 at 5:20






$begingroup$
Yes, thank you bof and the rest of you guys. It's making more sense now. $bigcup$${A_i : i in I}$ is just the short hand for $bigcup$$_{i in I} A_i$. I see it now.
$endgroup$
– Andres Kiani
Dec 22 '18 at 5:20












1 Answer
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0












$begingroup$

You are right, you have shown that they are indeed not equal.



Indeed ${{1},{2}} ne { 1,2} $ since $1 in { 1,2} $ but $1 notin {{1},{2}}$.






share|cite|improve this answer









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    1 Answer
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    0












    $begingroup$

    You are right, you have shown that they are indeed not equal.



    Indeed ${{1},{2}} ne { 1,2} $ since $1 in { 1,2} $ but $1 notin {{1},{2}}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You are right, you have shown that they are indeed not equal.



      Indeed ${{1},{2}} ne { 1,2} $ since $1 in { 1,2} $ but $1 notin {{1},{2}}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You are right, you have shown that they are indeed not equal.



        Indeed ${{1},{2}} ne { 1,2} $ since $1 in { 1,2} $ but $1 notin {{1},{2}}$.






        share|cite|improve this answer









        $endgroup$



        You are right, you have shown that they are indeed not equal.



        Indeed ${{1},{2}} ne { 1,2} $ since $1 in { 1,2} $ but $1 notin {{1},{2}}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 3:36









        Siong Thye GohSiong Thye Goh

        102k1467119




        102k1467119






























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