# Under a finite dimensional norm-induced metric space, is every bounded sequence has a converging subsequence?

What I know is that every bounded sequence in $\Bbb R^n$ has a converging sub sequence regardless of the chosen norm, since this is true under the Euclidean norm, and all norms on $\Bbb R^n$ are equivalent. My question is

Is every bounded sequence has a converging sub sequence for any norm-induced metric space?

If not, a counter-example is appreciated. Thanks!

• Sequentially compact=compact in a metric space, so your first claim is wrong. Do you mean to say complete? – Hayden Jan 22 '17 at 15:37
• @Hayden I am new to this stuff and I am a bit confused right now. So any bounded sequence has a converging sub sequence in $\Bbb R^n$ with respect to the euclidean norm? Since all norms on $\Bbb R^n$ are equivalent, then the sequence is still bounded w.r.t. other norms, and the sub sequence is still convergent w.r.t. those other norms? Then we say $\Bbb R^n$ is sequentially compact regardless of the chosen norm? Where am I going wrong...? – River Jan 22 '17 at 15:50
• For $\mathbb R^n$ to be sequentially compact, every sequence, not only the bounded ones would need to have a convergent subsequence. That is the definition of sequential compactness. However $\mathbb R^n$ is locally compact, i.e. each point has a compact neighborhood. From this it follows that each bounded sequence has a convergent subsequence. Is that probably what you mean? – Tim B. Jan 22 '17 at 15:52
• @LeBtz Thanks a lot. I misunderstood the concept. Sorry. I edited the question. – River Jan 22 '17 at 15:56
• As a real vector space $\mathbb{R}^n$ is complete for any norm, so yes any bounded sequence has a convergent sub-sequence. Now with $d(a,b) = |\arctan(b)-\arctan(a)|$, $\mathbb{R}$ is a metric space, but it is not complete because the sequence $u_n = n$ is Cauchy but its limit is $+\infty \not \in \mathbb{R}$ – reuns Jan 22 '17 at 16:10

I assume that by finite dimensional norm-induced metric space you mean a vector space in the first place because otherwise the concept of a norm or dimension doesn't make any sense.

If $(V, \|\cdot\|)$ is any $n$-dimensional normed $\mathbb F$-Vector space, where $\mathbb F\in\{\mathbb R,\mathbb C\}$ and $\{b_1,...,b_n\}$ is a basis of $V$, then the unique linear map $\varphi$ that maps $b_i$ to $e_i$ where $\{e_1,...,e_n\}$ is the standard base of $\mathbb F^n$ is an isomorphism. It is easily checked that $|\cdot|$ defined by $|x| := \|\varphi^{-1}(x)\|$ defines a norm on $\mathbb F^n$. For this norm it is true that each bounded sequence has a convergent subsequence. If $(x_n)_{n\in\mathbb N}$ is a bounded sequence in $V$, then $(\varphi(x_n))_{n\in\mathbb N}$ is bounded in $(\mathbb F^n, |\cdot|)$, so it has a convergent subsequence $(\varphi(x_{n_k}))_k$, but then $(x_{n_k})_k$ must converge in $V$ because $\varphi^{-1}$ is continuous.