Formula for Cardinality of Finite Sets

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I recently found this formula. How do I prove it?



Let $S$ be a finite set which contains no repeats (every element is unique).



Then $$|S|=sum_xin Sdelta_xx$$
Where $delta_ij$ is the Kronecker delta (https://en.wikipedia.org/wiki/Kronecker_delta).



Edit:
The formula is now $$|S|=sum_xin S1$$
Which is pretty intuitive, so I don't need help on a proof anymore.







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  • What exactly is giving you trouble? Do you have a specific definition of $|S|$ in mind?
    – Sambo
    Aug 24 at 2:48










  • I just don't know 1. whether or not its right for finite |S| and 2. how to prove it, assuming its right for finite |S|
    – clathratus
    Aug 24 at 2:51







  • 1




    To me, the statement seems almost tautological
    – Sambo
    Aug 24 at 2:54






  • 2




    Sets have no repeated elements anyway. What is the context here? Measure theory? The real line? Anything else? Also, do you know the counting measure?
    – Asaf Karagila♦
    Aug 24 at 2:55






  • 1




    I'm voting to close this question because the OP has edited it to provide an answer.
    – Ethan Bolker
    Aug 24 at 12:08














up vote
1
down vote

favorite












I recently found this formula. How do I prove it?



Let $S$ be a finite set which contains no repeats (every element is unique).



Then $$|S|=sum_xin Sdelta_xx$$
Where $delta_ij$ is the Kronecker delta (https://en.wikipedia.org/wiki/Kronecker_delta).



Edit:
The formula is now $$|S|=sum_xin S1$$
Which is pretty intuitive, so I don't need help on a proof anymore.







share|cite|improve this question






















  • What exactly is giving you trouble? Do you have a specific definition of $|S|$ in mind?
    – Sambo
    Aug 24 at 2:48










  • I just don't know 1. whether or not its right for finite |S| and 2. how to prove it, assuming its right for finite |S|
    – clathratus
    Aug 24 at 2:51







  • 1




    To me, the statement seems almost tautological
    – Sambo
    Aug 24 at 2:54






  • 2




    Sets have no repeated elements anyway. What is the context here? Measure theory? The real line? Anything else? Also, do you know the counting measure?
    – Asaf Karagila♦
    Aug 24 at 2:55






  • 1




    I'm voting to close this question because the OP has edited it to provide an answer.
    – Ethan Bolker
    Aug 24 at 12:08












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I recently found this formula. How do I prove it?



Let $S$ be a finite set which contains no repeats (every element is unique).



Then $$|S|=sum_xin Sdelta_xx$$
Where $delta_ij$ is the Kronecker delta (https://en.wikipedia.org/wiki/Kronecker_delta).



Edit:
The formula is now $$|S|=sum_xin S1$$
Which is pretty intuitive, so I don't need help on a proof anymore.







share|cite|improve this question














I recently found this formula. How do I prove it?



Let $S$ be a finite set which contains no repeats (every element is unique).



Then $$|S|=sum_xin Sdelta_xx$$
Where $delta_ij$ is the Kronecker delta (https://en.wikipedia.org/wiki/Kronecker_delta).



Edit:
The formula is now $$|S|=sum_xin S1$$
Which is pretty intuitive, so I don't need help on a proof anymore.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 4:25









Andrés E. Caicedo

63.3k7153238




63.3k7153238










asked Aug 24 at 2:42









clathratus

356




356











  • What exactly is giving you trouble? Do you have a specific definition of $|S|$ in mind?
    – Sambo
    Aug 24 at 2:48










  • I just don't know 1. whether or not its right for finite |S| and 2. how to prove it, assuming its right for finite |S|
    – clathratus
    Aug 24 at 2:51







  • 1




    To me, the statement seems almost tautological
    – Sambo
    Aug 24 at 2:54






  • 2




    Sets have no repeated elements anyway. What is the context here? Measure theory? The real line? Anything else? Also, do you know the counting measure?
    – Asaf Karagila♦
    Aug 24 at 2:55






  • 1




    I'm voting to close this question because the OP has edited it to provide an answer.
    – Ethan Bolker
    Aug 24 at 12:08
















  • What exactly is giving you trouble? Do you have a specific definition of $|S|$ in mind?
    – Sambo
    Aug 24 at 2:48










  • I just don't know 1. whether or not its right for finite |S| and 2. how to prove it, assuming its right for finite |S|
    – clathratus
    Aug 24 at 2:51







  • 1




    To me, the statement seems almost tautological
    – Sambo
    Aug 24 at 2:54






  • 2




    Sets have no repeated elements anyway. What is the context here? Measure theory? The real line? Anything else? Also, do you know the counting measure?
    – Asaf Karagila♦
    Aug 24 at 2:55






  • 1




    I'm voting to close this question because the OP has edited it to provide an answer.
    – Ethan Bolker
    Aug 24 at 12:08















What exactly is giving you trouble? Do you have a specific definition of $|S|$ in mind?
– Sambo
Aug 24 at 2:48




What exactly is giving you trouble? Do you have a specific definition of $|S|$ in mind?
– Sambo
Aug 24 at 2:48












I just don't know 1. whether or not its right for finite |S| and 2. how to prove it, assuming its right for finite |S|
– clathratus
Aug 24 at 2:51





I just don't know 1. whether or not its right for finite |S| and 2. how to prove it, assuming its right for finite |S|
– clathratus
Aug 24 at 2:51





1




1




To me, the statement seems almost tautological
– Sambo
Aug 24 at 2:54




To me, the statement seems almost tautological
– Sambo
Aug 24 at 2:54




2




2




Sets have no repeated elements anyway. What is the context here? Measure theory? The real line? Anything else? Also, do you know the counting measure?
– Asaf Karagila♦
Aug 24 at 2:55




Sets have no repeated elements anyway. What is the context here? Measure theory? The real line? Anything else? Also, do you know the counting measure?
– Asaf Karagila♦
Aug 24 at 2:55




1




1




I'm voting to close this question because the OP has edited it to provide an answer.
– Ethan Bolker
Aug 24 at 12:08




I'm voting to close this question because the OP has edited it to provide an answer.
– Ethan Bolker
Aug 24 at 12:08















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