Relating induction proof to Cantor's Theorem












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How would a proof by induction to show that $n < 2^n$ for all $n in ℤ$ be related to showing Cantor's Theorem for finite sets?



I have the proof completed however I don't quite see the relation between the two. I understand that Cantor's Theorem states that the power set of any given set will have a greater cardinality than that of the given set.



All explanations are much appreciated thank you.










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  • 1




    $begingroup$
    It's a combinatorial fact the for a set of $n$ elements, its power set has $2^n$ elements.
    $endgroup$
    – freakish
    Dec 5 '18 at 13:27












  • $begingroup$
    Appreciate it! I really was not using my head for this haha.
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37
















-1












$begingroup$


How would a proof by induction to show that $n < 2^n$ for all $n in ℤ$ be related to showing Cantor's Theorem for finite sets?



I have the proof completed however I don't quite see the relation between the two. I understand that Cantor's Theorem states that the power set of any given set will have a greater cardinality than that of the given set.



All explanations are much appreciated thank you.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's a combinatorial fact the for a set of $n$ elements, its power set has $2^n$ elements.
    $endgroup$
    – freakish
    Dec 5 '18 at 13:27












  • $begingroup$
    Appreciate it! I really was not using my head for this haha.
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37














-1












-1








-1





$begingroup$


How would a proof by induction to show that $n < 2^n$ for all $n in ℤ$ be related to showing Cantor's Theorem for finite sets?



I have the proof completed however I don't quite see the relation between the two. I understand that Cantor's Theorem states that the power set of any given set will have a greater cardinality than that of the given set.



All explanations are much appreciated thank you.










share|cite|improve this question











$endgroup$




How would a proof by induction to show that $n < 2^n$ for all $n in ℤ$ be related to showing Cantor's Theorem for finite sets?



I have the proof completed however I don't quite see the relation between the two. I understand that Cantor's Theorem states that the power set of any given set will have a greater cardinality than that of the given set.



All explanations are much appreciated thank you.







discrete-mathematics






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edited Dec 5 '18 at 13:24









Math Girl

633318




633318










asked Dec 5 '18 at 13:18









Zdravstvuyte94Zdravstvuyte94

395




395








  • 1




    $begingroup$
    It's a combinatorial fact the for a set of $n$ elements, its power set has $2^n$ elements.
    $endgroup$
    – freakish
    Dec 5 '18 at 13:27












  • $begingroup$
    Appreciate it! I really was not using my head for this haha.
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37














  • 1




    $begingroup$
    It's a combinatorial fact the for a set of $n$ elements, its power set has $2^n$ elements.
    $endgroup$
    – freakish
    Dec 5 '18 at 13:27












  • $begingroup$
    Appreciate it! I really was not using my head for this haha.
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37








1




1




$begingroup$
It's a combinatorial fact the for a set of $n$ elements, its power set has $2^n$ elements.
$endgroup$
– freakish
Dec 5 '18 at 13:27






$begingroup$
It's a combinatorial fact the for a set of $n$ elements, its power set has $2^n$ elements.
$endgroup$
– freakish
Dec 5 '18 at 13:27














$begingroup$
Appreciate it! I really was not using my head for this haha.
$endgroup$
– Zdravstvuyte94
Dec 5 '18 at 13:37




$begingroup$
Appreciate it! I really was not using my head for this haha.
$endgroup$
– Zdravstvuyte94
Dec 5 '18 at 13:37










1 Answer
1






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1












$begingroup$

If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, shoot I can't believe I didn't realize this, I must have been thinking of something else. Thank you so much!
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37











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1 Answer
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active

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active

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active

oldest

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1












$begingroup$

If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, shoot I can't believe I didn't realize this, I must have been thinking of something else. Thank you so much!
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37
















1












$begingroup$

If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, shoot I can't believe I didn't realize this, I must have been thinking of something else. Thank you so much!
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37














1












1








1





$begingroup$

If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?






share|cite|improve this answer









$endgroup$



If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 13:25









David KDavid K

53.6k342116




53.6k342116












  • $begingroup$
    Oh, shoot I can't believe I didn't realize this, I must have been thinking of something else. Thank you so much!
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37


















  • $begingroup$
    Oh, shoot I can't believe I didn't realize this, I must have been thinking of something else. Thank you so much!
    $endgroup$
    – Zdravstvuyte94
    Dec 5 '18 at 13:37
















$begingroup$
Oh, shoot I can't believe I didn't realize this, I must have been thinking of something else. Thank you so much!
$endgroup$
– Zdravstvuyte94
Dec 5 '18 at 13:37




$begingroup$
Oh, shoot I can't believe I didn't realize this, I must have been thinking of something else. Thank you so much!
$endgroup$
– Zdravstvuyte94
Dec 5 '18 at 13:37


















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