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April 4th, 2015, 01:43 PM   #1
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Finite and Infinite Sets

Let \[A\] be an infinite set. For each \[n\in \mathbb{N}\] shows that there is a subset \[B\subset A\], such that \[card(B)=n\].
Note: \[card(B)=cardinality\]
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April 4th, 2015, 02:37 PM   #2
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Surely you use induction.
  • build a subset $S$ of cardinality 1, which we can do because $A$ has at least one element.
  • if we have a subset $S$ of cardinality $k$, we can show that the complement of $S$ in $A$ is not empty (indeed, it isn't finite)
  • so we can take $S$ and add an element from its complement to make a new subset of $A$ with cardinality $k+1$.
Thanks from Luiz
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April 14th, 2015, 03:06 PM   #3
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Quote:
Originally Posted by v8archie View Post
Surely you use induction.
  • build a subset $S$ of cardinality 1, which we can do because $A$ has at least one element.
  • if we have a subset $S$ of cardinality $k$, we can show that the complement of $S$ in $A$ is not empty (indeed, it isn't finite)
  • so we can take $S$ and add an element from its complement to make a new subset of $A$ with cardinality $k+1$.
You could make full demo?
Using all the steps because you can not yet.
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April 16th, 2015, 05:43 AM   #4
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Did you not understand what v8archie said? He told you exactly how to do this problem. What is your difficulty?
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April 22nd, 2015, 08:04 AM   #5
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Guy suck at statements by mathematical induction ... you do to me ??????????
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April 22nd, 2015, 10:15 AM   #6
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I set out the steps above. Try it, you might even succeed!
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