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if the sequence is finite, then if it converges to some real number that will be in the set..

For this we have to add if limit exists!

We can take 1,1,1,1,.... sequence which has a finite set of 1. Thus its limit is 1 or say seq. converges to 1. This satisfies the a) condition but for the formal proof should I use the definition of countability of sets ? Can you help me out with this ?

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    Welcome to math.SE! Please consider taking the time to familiarise yourself with the [faq] to learn some of the common practices here. In addition, as this question appears to be homework, please consider reading [this page](http://meta.math.stackexchange.com/q/1803/8348) to help you ask effective homework-related questions.2012-10-04
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    What you can say about sequence $\{(-1)^n \}_{n=1}^{\infty}$?2012-10-04
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    You need explain better what do you want. For example, if the sequence can assume the values in finite set $\{-1,1\}$ the sequence $ -1,1-1,1 \cdots$ don't have limit.2012-10-04
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    @ArthurFischer thank you for the info2012-10-04
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    @M.Strochyk yeah for the -1, 1, -1, 1,... case it doesn't hold. But for the 1,1,1,1,... case which method should I use ?2012-10-04
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    @Analysis Fact that set of a sequence is finite does not imply neither that sequence is constant nor that this sequence converges2012-10-04

1 Answers 1

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a) Let $F$ be a non-empty finite subset of $\mathbb{R}$. Let $\{a_n\}$ be a sequence in $F$. Suppose lim $a_n = a$. Suppose $a$ does not belong to $F$. Let $\epsilon =$ min $\{|a - b|\colon b \in F\}$. Since $a$ does not belong to $F$, $\epsilon > 0$. Since lim $a_n = a$, there exists $n$ such that $|a - a_n| < \epsilon$. This is a contradiction.

b) Let $E = \{1/n\colon n = 1, 2, \dots\}$. Then lim $1/n = 0$ does not belong to $E$.