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.
discrete-mathematics
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add a comment |
$begingroup$
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.
discrete-mathematics
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1
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It's a combinatorial fact the for a set of $n$ elements, its power set has $2^n$ elements.
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– freakish
Dec 5 '18 at 13:27
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Appreciate it! I really was not using my head for this haha.
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– Zdravstvuyte94
Dec 5 '18 at 13:37
add a comment |
$begingroup$
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.
discrete-mathematics
$endgroup$
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
discrete-mathematics
edited Dec 5 '18 at 13:24
Math Girl
633318
633318
asked Dec 5 '18 at 13:18
Zdravstvuyte94Zdravstvuyte94
395
395
1
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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
add a comment |
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
add a comment |
1 Answer
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If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?
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$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?
$endgroup$
If a finite set $S$ has cardinality $n,$ what is the cardinality of the power set of $S$?
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
add a comment |
$begingroup$
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
add a comment |
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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