Bolzano-Weierstrass on $\mathbb{R}^n$ with different Norms

Question

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.

Answer 1

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!