The problem may have a very simple answer, but it is confusing me a bit now.
Let $(\mathbf{V},\lVert\cdot\rVert)$ be a finite dimensional normed vector space. A subset $\mathbf{U}$ of $\mathbf{V}$ is said to be bounded, if there is a real $M$ such that for any member $u$ of $\mathbf{U}$, we have: $\lVert u\rVert\lt M$. . Also, convergence of a sequence in $\mathbf{V}$ is defined with respect to the metric $\lVert\cdot\rVert$. Is it true that every bounded sequence of vectors in $\mathbf{V}$ admits a convergent subsequence?
If not, please give a counterexample with $\mathbf{V}$ finite dimensional.
Best Answer
I will write a little bit more details about my comment.
Take a basis $\{v_1,\ldots,v_n\}$ for $V$ and consider the isomorphism $T:\mathbb{R}^n\rightarrow V$ such that $T(e_i)=v_i$. Define a new norm on $\mathbb{R}^n$ by
$$\|x\|_{\mathbb{R}^n}=\|T(x)\|_V$$
This implies $T$ is a homeomorphism, because it takes opens balls onto open balls of the same radius, by definition of the norm on $\mathbb{R}^n$. In particular, remember that a homeomorphism takes convergent sequences in convergent sequences.
Once $T$ is a isomorphism, this is in fact a norm and, in addition, we have
$$\|v\|_V=\|T^{-1}(v)\|_{\mathbb{R}^n}$$
Take a bounded sequence $\{v_k\}_{k=1}^\infty$ on $V$. Then $x_k=T^{-1}(v_k)$ is a bounded sequence on $(\mathbb{R}^n,\|\cdot\|_{\mathbb{R}^n})$. Once all the norms on $\mathbb{R}^n$ are equivalent, this sequence is bounded with respect to the euclidean norm. Then, this sequence have a convergent subsequence, say $x_{k_j}$. Hence, $T(x_{k_j})$ is a convergent subsequence of the original $\{v_k\}_{k=1}^\infty$.