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

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 to help you ask effective homework-related questions. –  Arthur Fischer Oct 4 '12 at 17:40
What you can say about sequence $\{(-1)^n \}_{n=1}^{\infty}$? –  M. Strochyk Oct 4 '12 at 17:41
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. –  user29999 Oct 4 '12 at 17:44
@ArthurFischer thank you for the info –  HarveyMudd Oct 4 '12 at 17:53
@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 ? –  HarveyMudd Oct 4 '12 at 17:54
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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 $E$.