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Let $(\mathbb{X}, ρ)$ be a metric space. Let $\{x_k\}_{k=1}^{\infty}$ be a convergent sequence in $\mathbb{X}$ with a limit $\lambda \in \mathbb{X}$. Show that the limit $\lambda$ is unique.

We know convergence means,

$\forall \frac{\epsilon}{2} > 0$ $\exists n_0 (\frac{\epsilon}{2}) \in \mathbb{N}$ such that $k \ge n_0$ $\implies$ $\rho(x_k, \lambda) < \frac{\epsilon}{2}$

Hence,

$\rho(x_j, x_k) \le \rho(x_j, \lambda) + \rho(\lambda, x_k) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

So, whenever $j,k \ge n_0(\frac{\epsilon}{2})$ $\implies \{x_k\}$ is a Cauchy sequence.

Does this make $\lambda$ unique?

Jair Taylor
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  • No. You should assume that the sequence converges to distinct limits $\lambda_0$ and $\lambda_1$ and derive a contradiction. – Brian M. Scott Apr 23 '20 at 03:59
  • @BrianM.Scott Ah okay, I'm still learning how to construct proofs. Could you please post a proof of this that I could follow? –  Apr 23 '20 at 04:06

2 Answers2

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This doesn't prove that $\lambda$ is unique. Although I think your proof has a good idea with the triangle inequality.

We want to suppose that $\lambda_1$ and $\lambda_2$ are both limits of $(x_n)_{n\geq 1}$ and we want to show that $\lambda_1=\lambda_2$. To do this, we can show that $\rho(\lambda_1,\lambda_2)=0$. To do this, we can try something like your triangle inequality trick....

Dave
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  • Could you post the proof using the triangle inequality? –  Apr 23 '20 at 04:07
  • I think it's better if you try it first and I can help you where you get stuck. Although in ADA's answer there is more detail in the approach I had in mind. – Dave Apr 23 '20 at 04:08
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Suppose $\exists a,b$ such that$ x_k \to a $ and $x_k \to b$. Then, $\forall \epsilon > 0$ $N$ such that $\rho(x_k,a), \rho(x_k,b) < \epsilon$, for all k $\geq$ N.

Then, $\rho(a,b) \leq \rho(x_k,a) + \rho(x_k,b) \leq 2 \epsilon$.

ADA
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