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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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    @Analysis Fact that set of a sequence is finite does not imply neither that sequence is constant nor that this sequence converges2012-10-04

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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 $E$.