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If a bounded sequence of real numbers $(x_n)$ diverges and if a subsequence of $(x_n)$, $(x_{n_k})$ converges to $a$, show there exists another subseq.of $(x_n)$ that converges to a $b\neq a$. $\hspace{2mm}b, a\in\mathbb{R}$

My attempt: Let $(x_n)$ diverge and let $(x_{n_{k}})$ be a subsequence of $(x_n)$ such that $\lim_{k\to\infty} (x_{n_{k}})=a$, now since $(x_n)$ diverges, $\exists \varepsilon_0>0$ such that $\left |x_n-a\right |\geq \varepsilon_0$.

I'm stuck at this point, how do I go about creating another subsequence that converges to a $b\neq a$ using Bolzano Weierstrass?

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The quantifiers are missing. What you have is that "there exists some $\varepsilon_0 > 0$ such that, for all $N$, there exists $n \geq N$ for which $\lvert x_n - a \rvert \geq \varepsilon_0$."

In particular, that allows you to create a subsequence $(y_n)_{n\geq 0}=(x_{n_j})_{j\geq 0}$ of $(x_n)_{n\geq 0}$ such that $\lvert y_n - a \rvert \geq \varepsilon_0$ for all $n\geq 0$. (Can you see why?)

But then, this sequence itself is bounded, since $(x_n)_{n\geq 0}$ is. Invoking Bolzano—Weierstrass, it has a converging subsequence $(y_{n_k})_{k\geq 0}$: being a subsequence of a subsequence of $(x_n)_{n\geq 0}$, this is a subsequence of $(x_n)_{n\geq 0}$ as well. And it converges to a limit, call it $b$; but $b\neq a$, as by construction one must have $\lvert b - a \rvert \geq \varepsilon_0$.

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You know by assumption that $x_n$ doesn't converge to $a$ which means that there exists an $\epsilon>0$ and subsequence $m_n$ such that $|x_{m_n}-a|>\epsilon$. Now invoke Bolzano weierstrass on this subsequence.

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The key here is that, since $(x_n)$ DOES NOT converge to $a$, there is an $\epsilon$ such that, for \emph{infinitely many} $n$, $|x_n-a|>\epsilon$. Let $(x_{n_j})$ be the subsequence of terms that are at distance $\epsilon$ or more away from $a$, and apply BW.

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Hint:
Consider $K=\overline {(x_n)}$. Because of $(x_n)$ is bounded, $K$ is compact and $a\in K$. Now, $S=\{x_n;n\ne n_j\}_{j\in \mathbb N}$ is a sequence in $K$ and so it has a convergence subsequence. We claim that $S$ has a subsequence which is not convvergence to $a$. Else, $K=\{x_n\}\cup\{a\}$ which implies $\{x_n\}$ is convergence to $a$ contradiction.

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