Let $V$ be a finite-dimensional normed vector space and let $S=\left\{ v\in V:\left\Vert v\right\Vert =1\right\}$ , that is, the collection of all unit vectors in $V$. Prove that $S$ is compact. Is it possible to use the Heine-Borel theorem to prove this?
Prove that the set of all unit vectors in a finite-dimensional normed vector space is compact.
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linear-algebra
general-topology
normed-spaces
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0It is closed and bounded. – 2017-01-08
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0@OpenBall, are we not? Fix a basis in $V$ and you are in $\mathbb R^{\dim X}$ with some norm. – 2017-01-08
1 Answers
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Fix an inner product in $V$. With the inner-product norm it is isometric to the Euclidean space $\mathbb{R}^n$. In finite-dimensinonal vector spaces all norms are equivalent hence they generate the same topology. As the unit sphere is closed, it is also closed as a subset of the Euclidean space. You may then use the Heine-Borel theorem.