Proof verification: a convergent complex sequence has a unique limit.

Question

So I'm doing the exercises in Stein-Shakarchi, Vol. I and the question is as follows:

A sequence of complex numbers $\{w_n\}_{n=1}^{\infty}$ is said to converge if there exists a $w\in\Bbb{C}$ such that $$\lim_{n\to\infty}|w_n-w|=0.$$ Show that a convergent sequence of complex numbers has a unique limit.

My proof is as follows:

Let the limit of the convergent sequence $\{w_n\}_{n=1}^{\infty}$ be non-unique and let it converge to both $w$ and $w'$. Then we need to show that $w'=w$. We have $$\lim_{n\to\infty}|w_n-w|=0$$ and $$\lim_{n\to\infty}|w_n-w'|=0.$$ Since $|z|$ is real for all $z\in\Bbb{C}$, we can use the algebraic limit theorem to write $$\lim_{n\to\infty}|w_n-w|+\lim_{n\to\infty}|w_n-w'|=\lim_{n\to\infty}|w_n-w|+|w_n-w'|=0.$$ Now use the triangle inequality to get $$0=\lim_{n\to\infty}|w_n-w|+|w_n-w'|\geq\lim_{n\to\infty}|w_n-w+w_n-w'|=\lim_{n\to\infty}|2w_n-w-w'|$$ and thus we get $\lim_{n\to\infty}|w_n-\frac{w+w'}{2}|\leq 0$. But absolute value is non-negative, so $\lim_{n\to\infty}|w_n-\frac{w+w'}{2}|=0$ and $\frac{w+w'}{2}$ is also a limit of the sequence $\{w_n\}_{n=1}^{\infty}$. We thus have $w$(or $w'$) = $\frac{w+w'}{2}$ and so $w=w'$.

Is this proof correct? Since I'm assuming the sequence converges to $w$ and $w'$, I figure the last statement in the proof is valid? Am I wrong in my reasoning?

Answer 1

I've found a way to solve this more directly, without having to resort to the uniqueness of limits for $\operatorname{Re}(w_n)$ and $\operatorname{Im}(w_n)$.

Let the sequence converge to two different limits, $w$ and $w'$. Then, we have from the given definition of convergence $$\lim_{n\to\infty}|w_n-w|=\lim_{n\to\infty}|w_n-w'|=0.$$ Using the algebraic limit theorem $\lim f \pm g = \lim f \pm \lim g$ and the inequality $|a-b|\geq |a|-|b|$ for complex $a$ and $b$, we have $$0=\lim_{n\to\infty}|w_n-w|-\lim_{n\to\infty}|w_n-w'|=\lim_{n\to\infty}|w_n-w|-|w_n-w'|\leq\lim_{n\to\infty}|w_n-w-w_n+w'|$$ from which we get $|w'-w|\geq 0$. Also we have using the triangle inequality, $$0=\lim_{n\to\infty}|w_n-w|+\lim_{n\to\infty}|w_n-w'|=\lim_{n\to\infty}|w_n-w|+\lim_{n\to\infty}|w'-w_n|\geq\lim_{n\to\infty}|w_n-w+w'-w_n|$$ from which we get $|w'-w|\leq 0.$ Therefore, we must have $|w'-w|=0$ and hence $w'=w$.

Answer 2

What you wrote proves that if $w$ and $w'$ are limits of the sequence $(w_n)_{n\in\mathbb N}$, then $\frac{w+w'}2$ is also the limit of that sequence. But you cannot deduce from this that $w=\frac{w+w'}2$ or $w'=\frac{w+w'}2$, unless you know that a convergente sequence cannot have more than one limit. And that is what you want to prove.