State and prove the Bolzano-Weierstrass Theorem for $\mathbb{R}^3$ with metric?

Question

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!

Answer 1

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 $$\lim_{l\to\infty}d\bigl({\bf r}'''_l, (\xi,\eta,\zeta)\bigr)=0\ .$$

Answer 2

Answering this is easier if one has some basic understanding of functional analysis. First note that $\mathbb{R} ^3$ is a linear space. The norm $||x||_1:=|x_1|+|x_2|+|x_3|$ induces the given metric. Now in any finite dimensional normed linear space any two norms are equivalent i. e there exists constants $M_1,M_2>0$ s.t $||x||_1 \leq M_1||x||_2$ and $||x||_2 \leq M_2||x||_1$, where $||\cdot||_2$ is our well known euclidean norm. So if $y_n$ is a sequence bounded w.r.t $||\cdot||_1$ then its also bounded w.r.t $||\cdot||_2$ and so it has a convergent subseqence w.r.t $||\cdot||_2$. This subseqence is now convergent w.r.t $||\cdot||_1$ by equivalence of the norms.