# A Cauchy sequence $\{x_n\}$ with infinitely many $n$ such that $x_n = c$.

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.

$$\blacksquare$$

• 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 $c$. – Robert Z Nov 5 '18 at 7:59
• @RobertZ Thanks for you help – Atif Farooq Nov 5 '18 at 8:01
• "... implying that $|x_j-c|<\epsilon, \forall j\ge M$". Since $\epsilon$ was arbitrary, this completes the argument. – Jimmy R. Nov 5 '18 at 8:02

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 $$k > N$$.
• 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. – Andreas Blass Nov 6 '18 at 0:27
• @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. – RGS Nov 6 '18 at 7:00