Proving The Limit Is Unique Using Open Balls

Question

Prove: The Limit Is Unique

Attempt, let assume that the sequence $\{x_m\}$ has two limit namely: $p ,q$

Let $\epsilon=\frac{d(p,q)}{2}$

$p,q$ are limits of $\{x_m\}\Rightarrow$ There are $m_1\text{ and }m_2$ such that for all $n_1>m_1$ and $n_2>m_2$ let $N=\max\{n_1,n_2\}$ we get $x_{N}\in B(p,\epsilon)$ and $x_{N}\in B(q,\epsilon)$

Looking at $d(p,q)$ we get: $$d(p,q)\leq_{(1)} d(x_N,p)+d(x_N,q)<\epsilon+\epsilon=d(p,q)$$

Contradiction.

$(1)$ triangle inequality and definition of open ball

Is the proof valid?

Answer 1

Yes, it's fine. However, it's not well written:

1. You wrote “There are $m_1\text{ and }m_2$ such that for all $n_1>m_1$ and $n_2>m_2$ let $N=\max\{n_1,n_2\}$ we get $x_{N}\in B(p,\epsilon)$ and $x_{N}\in B(p,\epsilon)$ ”. There should be two sentences here: “There are $m_1$ and $m_2$ such that, for all $n_1>m_1$ and $n_2>m_2$, $d(x_{n_1},p),d(x_{n_2},q)<\epsilon$. Let $N=\max\{n_1,n_2\}$; we get $x_{N}\in B(p,\epsilon)$ and $x_{N}\in B(p,\epsilon)$.”
2. You assert that $d(p,q)\leqslant d(x_N,p)+d(x_N,q)$ by the triangle inequality and the definition of open ball. Acutally, all you need is the triangle inequality. The definition of open ball is only used later.

Answer 2

No, because you've made a typo in "we get $x_N \in B(p,\epsilon)$ and $x_N \in B(\color{red}{p},\epsilon)$", so in your proof, there's no relation between $N$ and $q$. To fix this, change $\color{red}{p}$ to $q$ so that the strict inequality following $(1)$ is valid.