My Math Forum Finite and Infinite Sets

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 April 4th, 2015, 01:43 PM #1 Member   Joined: Mar 2015 From: Brasil Posts: 90 Thanks: 4 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: $card(B)=cardinality$
 April 4th, 2015, 02:37 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra 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
April 14th, 2015, 03:06 PM   #3
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Joined: Mar 2015
From: Brasil

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Quote:
 Originally Posted by v8archie 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.

 April 16th, 2015, 05:43 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Did you not understand what v8archie said? He told you exactly how to do this problem. What is your difficulty?
 April 22nd, 2015, 08:04 AM #5 Member   Joined: Mar 2015 From: Brasil Posts: 90 Thanks: 4 Guy suck at statements by mathematical induction ... you do to me ??????????
 April 22nd, 2015, 10:15 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra I set out the steps above. Try it, you might even succeed!

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