Prove that $X=\left \{ (x,y,z) \in \mathbb{R} / x^{3}+y^{3}+z^{3} \leq 1 \right \}$ is complete

Question

Let $X=\left \{ (x,y,z) \in \mathbb{R} / x^{3}+y^{3}+z^{3} \leq 1 \right \}$. Prove that $X$ is complete

My attempt

Let $(x_{n},y_{n},z_{n})_{n \in \mathbb{N}}$ be a cauchy sequence in $X$, thus $x_{n}^{3}+y_{n}^{3}+z_{n}^{3} \leq 1$ for every $n \in \mathbb{N}$.

We know that $(x_{n},y_{n},z_{n})_{n \in \mathbb{N}}$ is a cauchy sequence, so for every $\epsilon >0$, there exists $N_{1} \in \mathbb{N}$ such that $\left \| (x_{n},y_{n},z_{n}) - (x_{n'},y_{n'},z_{n'}) \right \| \leq \epsilon$, for every $n,n' \geq N$. We have to prove that $(x_{n},y_{n},z_{n})_{n \in \mathbb{N}}$ is convergent that is, for any $\epsilon > 0$ we have to find and $N$ large enough such that $\left \| (x_{n},y_{n},z_{n}) - (x_{0},y_{0},z_{0}) \right \| \leq \epsilon$, for $n \geq N$.

But I don't know what else to do. Any suggestion?

Answer 1

Let $f(x,y,z)=x^3+y^3+z^3$.

$f$ is continuous at $\mathbb R^3$ as sum of continuous functions.

and

$(-\infty,1]$ is colsed in $\mathbb R$

$\implies$

$X=f^{-1}((-\infty,1])$ is complete as a closed in the complete space $\mathbb R^3$,

Answer 2

First Method :

So let's suppose that $(x_n,y_n,z_n)$ is a Cauchy sequence then it comes from the fact each that

$\forall \epsilon >0$, there exists N such that $p,q \geq N$ then

$ ||(x_p,y_p,y_p) - (x_q,y_q,y_q)||_2 \leq \epsilon$.

But as $|x_p-x_q| \leq ||(x_p,y_p,y_p) - (x_q,y_q,y_q)||_2$, we conclude that $x_n$ is a Cauchy sequence on $mathbb{R}$, same conclusion comes for $y_n$ and $z_n$ (symmetric roles).

But $\mathbb{R}$ is complete and then $x_n$ , $y_n$ and $z_n$ converges lets note by $x,y$ and $z$ the respective limits.

$x_n^3+y_n^3+z_n^3$ is continuous expression (polynomial) in the $x_n,y_n,z_n$.

Thus we can pass to the limit in the inequation :

$x_n^3+y_n^3+z_n^3 \leq 1$ which leads to $x^3+y^3+z^3 \leq 1$.

Which ends the proof.

Second Proof (faster).

$\mathbb{R}^3$ is complete, and $\{(x,y,z) \in \mathbb{R}^3 ,x^3+y^3+z^3 \leq 1\}$ is a closed set of a complete one, then it's complete as well.

Hope it helps,

Adam