Bolzano-Weierstrass on $\mathbb{R}^n$ with different Norms I have been dealing with Topology recently and was told that the Bolzano-Weierstrass-Theorem still applies for every bounded sequence in $\mathbb{R}^n$, no matter the metirc, if it is induced by a norm. My question is how we can just jump from $\mathbb{R}^n$ to $\mathbb{R}^m$ dimensions while the Bolzano-Weierstrass-Theorem still holds true and why it's validity doesn't change with the Norm ($\mathbb{R}^n$, $\lVert x\rVert_2$) versus ($\mathbb{R}^n$, $\lVert x\rVert_\infty$), for example.
 A: All norms are equivalent on finite dimensional vector spaces.
That is, there are positive $c,C$ such that $$c\|x\|_1\leq \|x\|_2\leq C \|x\|_1$$ for all vectors $x$. In other words, they generate the same topology.
A: The answer to your question is in two steps. I assume you know the Bolzano-Weierstrass in $(\mathbb{R},|\cdot|)$, stating that a closed interval $[a,b]$ is compact.
First step The metric space $\left(\mathbb{R}^n,\|\cdot\|_{\infty}\right)$ is the product metric space $\prod_{i=1}^n \left(\mathbb{R},|\cdot|\right)$ with the supremum distance. Moreover, a finite product of compact topological spaces is compact, so any $\left(\prod_{i=1}^n [a_i,b_i]\right)\subset \mathbb{R}^n$ is compact for the topology of $\|\cdot\|_{\infty}$. Thus, the Bolzano-Weierstrass theorem is true for that particular norm.
Second step On $\mathbb{R}^n$, all norms are equivalent, that is for $N_1$ and $N_2$ two norms, there exists two constants $C_1$ and $C_1$ such that
\begin{align}
\forall x \in \mathbb{R}^n,C_1 N_1(x) \leqslant N_2(x) \leqslant C_2N_1(x)
\end{align}
So every norm is equivalent to $(\|\cdot\|_{\infty})$. The topologies induced are the same, and the compact subsets are the same. The particular case of step 1 show the generalized Bolzano-Weierstrass theorem.
Edit Another way to think this proof is this one: let $(p_k)$ be a bounded sequence in $\left(\mathbb{R}^n,\|\cdot\|\right)$, say $p_j = (x_1(k),x_2(k),\ldots,x_n(k))$. As all norms are equivalent, $(p_k)$ is also bounded for the $\|\cdot\|_{\infty}$ norm. Consequently, each coordinate sequence $(x_i(k))_k$ is bounded in $\mathbb{R}$.
The sequence $(x_1(k))_k$ is bounded. By the one dimensional Bolzano-Weierstrass theorem, there exists $\phi_1 : \mathbb{N} \to \mathbb{N}$ increasing such that $(x_1(\phi_1(k))_k$ is convergent to a number $x_1$.
The sequence $x_2(\phi_1(k))$ is bounded as a subsequence of $(x_2(k))$. Then there exists $\phi_2$ inscreasing such that $x_2(\phi_1(\phi_2(k)))$ converges to a number $x_2$. And so on.
Define $\phi = \phi_1\circ \phi_2\circ \cdots \circ \phi_n$. Then $p_{\phi(k)}$ has a limit $(x_1,\ldots,x_n)$ in $\mathbb{R}^n$ for the $\|\cdot\|_{\infty}$ metric. Use again the equivalence of norms to show that this is also the limit of $p_{\phi(k)}$ for the previous norm.
A: The reason why Bolzano-Weierstrass holds in every $\mathbb{R^n}$ and its independent of the norm that you have on $\mathbb{R^n}$ is based on the following facts (which are consequences of the finite dimensionality of the space).

Proposition 1 A normed linear space $(X,||.||)$ (for example $\mathbb{R^n})$ is finite dimensional if and only if its closed unit ball $B_X=\{x\in X: ||x||\leq 1\}$ is compact.

Now Proposition 1 answers to your first question, i.e. to why the Bolzano Weierstrass holds in every $\mathbb{R^n}$ indepedently of $n$. (And actually holds in every finite dimensional normed space)
Too see this, assuming we know Proposition 1 and we have a bounded sequence $(x_m)$ in $\mathbb{R^n}$ then by the boundedness of $(x_m)$ there would be a ball with center $0$ and radius $M>0$ (the bound of $x_m$) such that $x_m\in B_X(M)=\{x\in \mathbb{R^n}: ||x||\leq M\}$. By proposition 1, $B_X$ is compact, and therefore $B_X(M)$ is also compact. Hence, $x_m$ must have a convergent subsequence.
An important corollary of the Bolzano-Weierstrass is the following characterization of compactness in the context of finite dimensional normed spaces:

Corollary 1 In a finite dimensional linear space $(X,||.||)$, every subset $K$ of $X$ is compact if and only if it is closed and bounded.

Now, to your second question, to why is independently of the norm that you have in $\mathbb{R^n}$ is because of the following

Proposition 2 In every finite dimensional linear space, every two norms $||.||_1,||.||_2$ are equivalent. Meaning, that there exists two constants $C_1,C_2>0$ such that
$$C_1||x||_1\leq ||x||_2\leq C_2||x||_1$$

Now its easy to check that if a sequence converges with the usual norm $||.||_2$ in $\mathbb{R^n}$ it will also converges with any other norm.
I hope that this helps!
