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 November 9th, 2013, 07:43 AM #1 Member   Joined: May 2012 Posts: 78 Thanks: 0 Countable set Hello, I have two questions: 1. Determine whether the set is countable set:[*]The set of series of natural number when there are even numbers in even places and odd numbers in odd places (eg, (3,8,5,2,1,0,...)).[*]The set of series of prime numbers (eg, (2,3,5,...) or (13,17,23,31,...)) 2. Let $A \subseteq{R}$ be infinite set of positive numbers. Let $k \in \mathbb{Z}$ that for every $B \subseteq A$: $\sum_{b \in B} b \leq k$. Prove that A is a countable set. Clue: look at the sets $A_n=\{a \in A | a>\frac{1}{n} \}=$. Thanks!
November 9th, 2013, 11:50 AM   #2
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Re: Countable set

Quote:
 Originally Posted by galc127 Hello, I have two questions: 1. Determine whether the set is countable set:[*]The set of series of natural number when there are even numbers in even places and odd numbers in odd places (eg, (3,8,5,2,1,0,...)).
For each index n, we have aleph-null choices (of odd numbers, or even numbers. Makes no difference, right?). So the cardinality is N^N = 2^N . That latter assertion requires proof, but it's standard.

Quote:
 Originally Posted by galc127 [*]The set of series of prime numbers (eg, (2,3,5,...) or (13,17,23,31,...))
For each index n, we have aleph-null choices (of prime numbers that we haven't used yet). So the cardinality is N^N = 2^N .

Quote:
 Originally Posted by galc127 2. Let $A \subseteq{R}$ be infinite set of positive numbers. Let $k \in \mathbb{Z}$ that for every $B \subseteq A$: $\sum_{b \in B} b \leq k$. Prove that A is a countable set. Clue: look at the sets $A_n=\{a \in A | a>\frac{1}{n} \}=$.
If each A_n is countable then A = Union(A_n) is countable. Therefore if A is uncountable, some A_n is uncountable and its sum therefore diverges to infinity. Since this violates the existence of k, A must be countable.

 November 9th, 2013, 12:17 PM #3 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Countable set There's a solution and discussion on question 2 here: http://math.stackexchange.com/questions ... ve-numbers Including an answer to the question of how a sum over an uncountable set B would be defined!
November 9th, 2013, 12:55 PM   #4
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Re: Countable set

Quote:
 Originally Posted by Maschke For each index n, we have aleph-null choices (of odd numbers, or even numbers. Makes no difference, right?). So the cardinality is N^N = 2^N . That latter assertion requires proof, but it's standard. For each index n, we have aleph-null choices (of prime numbers that we haven't used yet). So the cardinality is N^N = 2^N .
Thank you for your answer, but it's not very clear to me.
May you help with more clues or detailed solution?

November 9th, 2013, 01:18 PM   #5
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Re: Countable set

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 Originally Posted by galc127 [ Thank you for your answer, but it's not very clear to me. May you help with more clues or detailed solution?
I was being terse. The Stackexchange link is the same idea.

Do you see that if A is uncountable that at least one A_n must be uncountable? If you get that far, then you just need to convince yourself that the sum of an uncountable collection of numbers each of which is greater than some fixed constant 1/n must be infinite. Remember n is fixed here!

 November 9th, 2013, 01:28 PM #6 Member   Joined: May 2012 Posts: 78 Thanks: 0 Re: Countable set I understood the answer for my second question. I have problems with my first question... Unfortunately I didn't understand your answer.
November 9th, 2013, 01:43 PM   #7
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Re: Countable set

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 Originally Posted by galc127 I understood the answer for my second question. I have problems with my first question... Unfortunately I didn't understand your answer.
Can you narrow down the part you don't understand? Which of the two series in the first question?

 November 9th, 2013, 06:23 PM #8 Member   Joined: May 2012 Posts: 78 Thanks: 0 Re: Countable set You used the same logic for both series, so I have to say both... :\ It's not clear to me why $\mathbb{N}X\mathbb{N}=2^N$.
November 9th, 2013, 06:27 PM   #9
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Re: Countable set

Quote:
 Originally Posted by galc127 It's not clear to me why $\mathbb{N}X\mathbb{N}=2^N$.
It's not.

 November 9th, 2013, 06:32 PM #10 Member   Joined: May 2012 Posts: 78 Thanks: 0 Re: Countable set If it's not, may you help with the question or explain Maschke's answer? Thanks.

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