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 May 16th, 2017, 01:11 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2 Infinite Subset A subset $S$ of $N$ is infinite if and only if A) $S$ is not bounded below. B) $S$ is not bounded above. C) $\displaystyle \ni$ $n_0$ $\displaystyle \epsilon$ $N$ such that $\displaystyle \forall$ $n$ $\displaystyle \geq$ $n_0$ , $n$ $\displaystyle \epsilon$ $S$. D) $\displaystyle \forall$ $a$ $\displaystyle \epsilon$ $S$, $\displaystyle \ni$ $x$ $\displaystyle \epsilon$ $N$ such that $x < a$. As $N$ is an infinite set which is not bounded above, I'm thinking as it's subset should also possess the same property to be an Infinite set. Correct me If I'm wrong
May 17th, 2017, 11:21 PM   #2
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Quote:
 Originally Posted by Lalitha183 A subset $S$ of $N$ is infinite if and only if A) $S$ is not bounded below. B) $S$ is not bounded above. C) $\displaystyle \ni$ $n_0$ $\displaystyle \epsilon$ $N$ such that $\displaystyle \forall$ $n$ $\displaystyle \geq$ $n_0$ , $n$ $\displaystyle \epsilon$ $S$. D) $\displaystyle \forall$ $a$ $\displaystyle \epsilon$ $S$, $\displaystyle \ni$ $x$ $\displaystyle \epsilon$ $N$ such that $x < a$. As $N$ is an infinite set which is not bounded above, I'm thinking as it's subset should also possess the same property to be an Infinite set. Correct me If I'm wrong
Please some one check and let me know If I'm wrong
Thanks

May 18th, 2017, 05:15 AM   #3
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Quote:
 Originally Posted by Lalitha183 A subset $S$ of $N$ is infinite if and only if A) $S$ is not bounded below.
All subsets of N, whether finite of infinite, have 0 as a lower bound. This is wrong.

Quote:
 B) $S$ is not bounded above.
Yes, that is correct. If n were an upper bound on set S then S would have no more than n members so not be infinite.

Quote:
 C) $\displaystyle \ni$ $n_0$ $\displaystyle \epsilon$ $N$ such that $\displaystyle \forall$ $n$ $\displaystyle \geq$ $n_0$ , $n$ $\displaystyle \epsilon$ $S$.
Consider the set of all even positive integers. It is an infinite set but does not have this property. This is wrong.

Quote:
 D) $\displaystyle \forall$ $a$ $\displaystyle \epsilon$ $S$, $\displaystyle \ni$ $x$ $\displaystyle \epsilon$ $N$ such that $x < a$.
Now, consider the odd positive integers. 1 is in that set but there is no positive integer less than 1. This is wrong.

Quote:
 As $N$ is an infinite set which is not bounded above, I'm thinking as it's subset should also possess the same property to be an Infinite set. Correct me If I'm wrong
Not all properties!

Last edited by Country Boy; May 18th, 2017 at 05:18 AM.

 May 18th, 2017, 05:29 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,151 Thanks: 2390 Math Focus: Mainly analysis and algebra I think you are correct.

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