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Jan 22, 2022 at 14:36 history edited Calvin Khor CC BY-SA 4.0
fixing broken link
Jan 7, 2012 at 7:13 comment added Rudy the Reindeer @Fabian Nice : ) Thank you!
Jan 7, 2012 at 1:42 history edited J. M. ain't a mathematician CC BY-SA 3.0
added 15 characters in body
Jan 6, 2012 at 22:53 comment added Fabian @Matt: thank you for the remarks. I changed the answers accordingly.
Jan 6, 2012 at 22:52 history edited Fabian CC BY-SA 3.0
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Jan 5, 2012 at 14:43 comment added Rudy the Reindeer I personally find it slightly confusing to use $n$ as an index variable to sum over if $n$ also denotes the dimension of the space.
Jan 5, 2012 at 14:42 comment added Rudy the Reindeer @Fabian: Assuming that $\| x \|_\infty := \max_{i \in \{1, \dots , n\} } x_i$ I think you are missing an $n$ in the following line: $$ \| x - y \| = \dots \leq n \| x - y\|_\infty$$
May 7, 2011 at 18:47 history edited Fabian CC BY-SA 3.0
added 26 characters in body
Mar 5, 2011 at 18:07 vote accept Huy
Mar 5, 2011 at 17:00 comment added Gunnar Þór Magnússon @Huy: If you pick a basis for your space, then you've constructed a linear isomorphism between your space and $\mathbb R^n$. These are continuous, so the Heine-Borel theorem holds on your space. Alternatively, the proof of the Heine-Borel theorem should go through mostly unchanged for a finite-dimensional normed vector space.
Mar 5, 2011 at 16:37 comment added Huy This might be somewhat stupid, but is it trivial that the unit sphere is a compact set? I know Heine-Borel's theorem, but as far as I know it is only valid for subsets of $\mathbb{R}^n$.
Mar 5, 2011 at 15:02 history answered Fabian CC BY-SA 2.5