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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.

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    Yes, because it holds for every $\mathbb{R}^n$ with any norm (because they are all equivalent). Your space is isomorphic (then homeomorphic) to some $\mathbb{R}^n$, hence the notions of convergence on your $V$ can be viewed as convergence in this $\mathbb{R}^n$2012-05-13

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