Understanding why $\mathbb{R}^2$ is complete I'm trying to understand the proof for why $\mathbb{R}^2$ is complete with respect to the standard Euclidean metric. 
If we let $(X_n)=(x_n,y_n)\in \mathbb{R}^2$ be Cauchy, then for all $\epsilon > 0$ there exists an $N\in\mathbb{N}$ such what whenever $m,n>N$ we have that $d(x_m,x_n)=\sqrt{|x_m-x_n|^2+|y_m-y_n|^2}<\epsilon$. This implies that $|x_m-x_n|<\epsilon$ and $|y_m-y_n|<\epsilon$. Since we know that $\mathbb{R}$ is complete, there exist $x$ and $y$ such that $x_n\rightarrow x$ and $y_n\rightarrow y$. 
This all makes sense to me. The next line is where I get confused. I have that "by continuity" $\lim_{n\rightarrow \infty} d(X_n,X)=\lim_{n\rightarrow \infty}\sqrt{|x_n-x|^2+|y_n-y|^2}=0$.
Why is this the case, and why does this prove the result? Are we using the continuity of the metric, and if so how does this help us? 
 A: You just need to show that $(x_n,y_n) \to (x,y)$, that is, $d((x_n,y_n),(x,y)) \to 0$.
Note that $d(a,b) \le 2 \max(|a|,|b|)$, so
$d((x_n,y_n),(x,y)) \le 2 \max (d(x_n,x), d(y_n,y))$ from which
we conclude the desired result.
A: I guess you are saying something like $(u_{n})\in{\bf{R}}^{2}$ and $u_{n}=(x_{n},y_{n})$.
So $d(u_{n},(x,y))=\sqrt{|x_{n}-x|^{2}+|y_{n}-y|^{2}}\leq\epsilon$ follows by $\sqrt{|x_{n}-x_{m}|^{2}+|y_{n}-y_{m}|^{2}}<\epsilon$ for all $m,n>N$ by taking $m\rightarrow\infty$ and the condition that $x_{m}\rightarrow x$, $y_{m}\rightarrow y$.
The continuity is like this. For fixed $n$, $x_{n},y_{n}$ are constants, then $|x_{n}-x_{m}|\rightarrow|x_{n}-x|$, $|y_{n}-y_{m}|\rightarrow|y_{n}-y|$ as $m\rightarrow\infty$. Then the continuity of $|\cdot|^{2}$ implies that $|x_{n}-x_{m}|^{2}\rightarrow|x_{n}-x|^{2}$ and $|y_{n}-y_{m}|^{2}\rightarrow|y_{n}-y|^{2}$. And the addition $(u,v)\rightarrow u+v$ is continuous and hence $|x_{n}-x_{m}|^{2}+|y_{n}-y_{m}|^{2}\rightarrow|x_{n}-x|^{2}+|y_{n}-y|^{2}$. And the $\sqrt{\cdot}$ is continuous, and the result follows in the similar way.
A: I don't like this approach, but it is correct: we can use the fact that the distance $d$ is a continuous functions from $\mathbb{R}^2\times\mathbb{R}^2$ into $\mathbb R$ and then to use the fact that therefore it maps convergent sequences into convergent sequences.
But it is much more natural to proceed as follows. Let $\bigl((x_n,y_n)\bigr)_{n\to\infty}$ be a convergent sequence in $\mathbb{R}^2$. If $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}y_n=y$ , you want to prove that $\lim_{n\to\infty}(x_n,y_n)=(x,y)$. Take $\varepsilon>0$. There are $p_x,p_y\in\mathbb N$ such that$$n\geqslant p_x\implies|x-x_n|<\frac{\varepsilon}{\sqrt2}\text{ and that }n\geqslant p_y\implies|y-y_n|<\frac{\varepsilon}{\sqrt2}.$$Let $p=\max\{p_x,p_y\}$. Then, if $n\in\mathbb N$,\begin{align}n\geqslant p\implies&n\geqslant p_x\wedge n\geqslant p_y\\\implies&|x-x_n|<\frac{\varepsilon}{\sqrt2}\wedge|y-y_n|<\frac{\varepsilon}{\sqrt2}\\\implies&\sqrt{|x-x_n|^2+|y-y_n|^2}<\varepsilon.\end{align}
