7.8

7.8
Let \((a_n)\) be a sequence. A point \(a\) is an \(accumulation\) \(point\) of the sequence if for ever \(\epsilon >0\) and every \(N \in \mathbb{N}\) there exists some \(n>N\) such that \(\mid a_n -a \mid < \epsilon\). Prove that if a sequence has more that one accumulation point then the sequence diverges.
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Let \((a_n)\) be a sequence in \(\mathbb{C}\) with an accumulation point \(a\). Consider a second accumulation point \(b\) where \(a \neq b\). Then we know the distance from \(a\) to \(b\) is some positive real number, ie. \(\mid a-b \mid >0\). Let \(\epsilon =\frac{\mid a-b \mid}{2}\). Then by definition of convergence, there is some \(N\) where \(n \geq N \Rightarrow \mid a_n-L \mid < \epsilon\). Also, there exists \(\mid a_n-L \mid < \epsilon\), where \(L\) is the limit of \(a_n\). Thus when \(a_n\) converges to \(L\), \(\mid a_n-b \mid > \epsilon\), which is a contradiction of the definition of convergence. Thus either \(b\) is not an accumulation point of \((a_n)\), or \((a_n)\) does not converge. Since \(b\) was assumed to be an accumulation point, then \((a_n)\) does not converge, ie. \((a_n)\) diverges.