A Cauchy sequence is convergent over the complex plane.

I'm trying to prove that a Cauchy sequence is convergent over the complex plane. I think I have proved it, yet I would like to be sure. Any aid is greatly appreciated. Here's what I've done:

Let $\{z_n\}$ be a Cauchy sequence in $\mathbb{C} \Rightarrow \forall \varepsilon > 0 \quad \exists N \in \mathbb{N}$ such that $\forall n,m \geq N \Rightarrow ||z_n - z_m|| < \varepsilon$. Then $z_m \in B_{\varepsilon} (z_n)$. We have that $B_{\varepsilon} (z_n) \subset \overline{B_{\varepsilon} (z_n)}$. Since $\overline{B_{\varepsilon} (z_n)}$ is compact and $B_{\varepsilon} (z_n)$ has infinitely many points, then it must have a limit point, making $\{z_n\}$ convergent.

Thank you for your help.

Answer

The correct way to prove it is:

You can prove that every Cauchy sequence is bounded so $z_n$ is bounded.

Thus exists $M>0$ such that $$||z_n-0||=||z_n|| \leq M,\forall n \in \mathbb{N} \Rightarrow z_n \in cl(B(0,M))$$

Then you use your following arguments with the limit point compactness.

Here it is another proof if you want to take a look at:

Let $z_n \in \mathbb{C}$ a Cauchy sequence.

$z_n=x_n+iy_n$

We have that $$||z_m-z_n|| \rightarrow 0 \Rightarrow \sqrt{(x_n-x_m)^2+(y_n-y_m)^2} \rightarrow 0$$ as $m,n \rightarrow + \infty$

We have that $$|x_n-x_m|=\sqrt{(x_n-x_m)^2} \leq \sqrt{(x_n-x_m)^2+(y_n-y_m)^2} \rightarrow 0$$ $$|y_n-y_m|=\sqrt{(y_n-y_m)^2} \leq \sqrt{(x_n-x_m)^2+(y_n-y_m)^2} \rightarrow 0$$

Thus $x_n,y_n$ are Cauchy sequences in $\mathbb{R}$ which is complete(i.e every Cauchy sequence in $\mathbb{R}$ converges).

So exists $x_0,y_0 \in \mathbb{R}$ such that $$x_n \rightarrow x_0$$ $$y_n \rightarrow y_0$$

Thus $z_n \rightarrow x_0+i y_0$ proving that $z_n$ converges.