In the metric space $\mathbb{R}^3 $ with the usual metric I have the set $S=\{ (x,y,z) \in \mathbb{R}^3 | x^2+2y^2+3z^2=6 \} $. Prove $S$ is a complete metric space.
I need to prove that every Cauchy sequence in $S$ converges to a point on $S$.
First I thought about switching to spherical coordinates, but the metric becomes quite messy and hard to work with, so I attempted the following.
First consider a Cauchy sequence $(x_n,y_n,z_n ) \in S$. So for every $\epsilon >0$ and $n,m >N$ I have $$|(x_n,y_n,z_n)-(x_m,y_m,z_m)|<\epsilon$$ such that $$x_n^2+2y_n^2+3z_n^2=6$$ and $$x_m^2+2y_m^2+3z_m^2=6$$ Then since $\mathbb{R}^3$ is complete $(x_n,y_n,z_n)$ converges to a point in $\mathbb{R}^3$ so for every $\epsilon >0$ and $n >N$ we have $$|(x_n,y_n,z_n)-(c_1,c_2,c_3)|=\sqrt{(x_n-c_1)^2+(y_n-c_2)^2+(z_n-c_3)^2 } < \epsilon $$
So now I need to prove that $(c_1,c_2,c_3) \in S$ that is $$c_1^2+2c_2^2+3c_3^2=6$$ but I am stuck at this point, can you help?
To complete your proof: You have assumed that $(x_n,y_n,z_n)$ converges to some $(c_1,c_2,c_3)\in\mathbb R^3$. Note that for every $n\in\mathbb N$, $$x_n^2 + 2y_n^2 + 3z_n^2 = 6$$ Convergence in $\mathbb R^3$ implies coordinate-wise convergence (I encourage you to prove this). Hence, $(x_n,y_n,z_n) \xrightarrow{n\to\infty} (c_1,c_2,c_3)$ implies $x_n\xrightarrow{n\to\infty} c_1$, $y_n\xrightarrow{n\to\infty} c_2$ and $z_n\xrightarrow{n\to\infty} c_3$. So we have, $x_n^2\xrightarrow{n\to\infty} c_1^2$, $y_n^2\xrightarrow{n\to\infty} c_2^2$ and $z_n^2\xrightarrow{n\to\infty} c_3^2$. Taking limits as $n\to\infty$ in $x_n^2 + 2y_n^2 + 3z_n^2 = 6$, we obtain $$\lim_{n\to\infty}(x_n^2 + 2y_n^2 + 3z_n^2) = c_1^2 + 2c_2^2 + 3c_3^2 = 6$$ so $(c_1,c_2,c_3)\in S$, which is what you wanted to show.