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Is the following argument correct?

Proposition. If $\{x_n\}$ is Cauchy sequence such that $x_n = c$ for infinitely many $n$, then $\lim_{n\to\infty}x_n = c$.

Proof. Let $\epsilon>0$. Since $\{x_n\}$ is a Cauchy sequence, there exists an $M\in\mathbb{N}$ such that $\forall \, j,k\ge M$, we have $|x_j-x_k|<\epsilon$. Now, since $x_n = c$ for infinitely many $c$, then surely $x_r = c$ for some $r \ge M,$ implying that $|x_j-c|<\epsilon \,\,\forall j\ge M$, completing the argument.

$\blacksquare$

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    $\begingroup$ Yes, it is correct. In similar way you can show that if there is a subsequence which converges to $c$ then the whole sequence converges to $c$. $\endgroup$ – Robert Z Nov 5 '18 at 7:59
  • $\begingroup$ @RobertZ Thanks for you help $\endgroup$ – Atif Farooq Nov 5 '18 at 8:01
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    $\begingroup$ "... implying that $|x_j-c|<\epsilon, \forall j\ge M$". Since $\epsilon$ was arbitrary, this completes the argument. $\endgroup$ – Jimmy R. Nov 5 '18 at 8:02
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As told in the comments, your proof is fine.

Usually when I am asked to prove a statement that seems rather obvious, I tend to prove it by contradiction: this usually allows me to explore the intuitive feel I have for the result.

Assume the sequence converged to $l \neq c$. Then for $\epsilon = \frac{|c - l|}{2} $ $\exists N $ s.t. $i > N \implies |x_i - l| < \epsilon $, by definition of convergence. But this is absurd, because the sequence had infinitely many terms equal to $c $ and thus infinitely many $x_k = c $ with $k > N $.

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  • $\begingroup$ This argument presupposes that the context is $\mathbb R$ or some other complete metric space, so that the sequence $(x_n)$ is guaranteed to converge and you only need to check that it can't converge to a wrong limit. But in fact the OP's proof works in any metric space (if you write $d(x,y)$ instead of $|x-y|$), even if the space isn't complete. $\endgroup$ – Andreas Blass Nov 6 '18 at 0:27
  • $\begingroup$ @AndreasBlass Indeed the OPs proof is more general, but I didn't think it was wrong to assume I was in $\mathbb R $, as the tag real-analysis suggests. $\endgroup$ – RGS Nov 6 '18 at 7:00
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    $\begingroup$ I appreciate your counsel RGS. $\endgroup$ – Atif Farooq Nov 6 '18 at 11:59

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