I'm given a metric on $(\mathbb{R}^3, d_{\mathbb{R^3}})$, where
$$
d_{\mathbb{R}^3}((x_1, x_2, x_3),(y_1, y_2, y_3))
= |x_1 – y_1| + |x_2 – y_2| + |x_3 – y_3|
$$
And then I'm asked to state and prove the Bolzano-Weierstrass Theorem for ℝ3 with this metric. So, I'm confused about this task. Is it right that I have to prove the following:
Every bounded sequence in R3 has a convergent subsequence?
I have found information for this theorem in Rn, but now I can not understand how to apply given metric.
Thank you in advance!
Best Answer
I shall call your metric simply $d$ in the sequel. As you say, we have to prove that any $d$-bounded sequence $\bigl({\bf r}_n\bigr)_{n\geq1}$ in ${\mathbb R}^3=:X$ has a convergent subsequence. Now "$d$-bounded" means that there is a point ${\bf p}\in X$ and an $M>0$ such that $d({\bf p},{\bf r}_n)\leq M$ for all $n\geq1$. We may assume WLG that ${\bf p}={\bf 0}$, i.e., that $d({\bf 0},{\bf r}_n)=|x_n|+|y_n|+|z_n|\leq M$ for all $n\geq1$.
The key idea in the proof is sieving. As $|x_n|\leq M$ for all $n\geq1$, by Bolzano's theorem for ${\mathbb R}$ the first coordinates $x_n$ of the points ${\bf r}_n$ have a convergent subsequence $j\mapsto x_{n_j}=:x'_j$ with $x'_j\to\xi$ $(j\to\infty)$. Put $y_{n_j}=:y'_j$, $z_{n_j}=:z'_j$. As $|y'_j|\leq M$ for all $j\geq1$, by Bolzano's theorem for ${\mathbb R}$ the second coordinates $y'_j$ of the points ${\bf r'}_j$ have a convergent subsequence $k\mapsto y'_{j_k}=:y''_k$ with $y''_k\to\eta$ $(k\to\infty)$. Put $x'_{j_k}=:x''_k$ and $z'_{j_k}=:z''_k$. And then the same thing a third time. In the end we have a sub-sub-sub sequence ${\bf r}'''_l=(x'''_l,y'''_l,z'''_l)$ $(l\geq 1)$ with $x'''_l\to\xi$, $\>y_l'''\to\eta$, $\>z'''_l\to\zeta$ when $l\to\infty$.
It is then easy to see that $$\lim_{l\to\infty}d\bigl({\bf r}'''_l, (\xi,\eta,\zeta)\bigr)=0\ .$$