We define $ (\mathbb{R}^m, \|.\|)$ to be a finite dimensional normed vector space with $ \|.\|$ is defined to be any norm in $ \mathbb{R^m}$. Let $S = \lbrace x \in \mathbb{R}^m: \| x\| = 1 \rbrace.$ Prove that $S$ is compact in $ (\mathbb{R}^m, \|.\|).$
The compactness of the unit sphere in finite dimensional normed vector space
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3have you heard of Heine-Borel? – 2012-03-07
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0@Sulaiman: Since you are happy with the answers for this questions, I suggest you accept one of them by ticking it. It makes this site much more organised. While I'm at it, I also suggest to you to accept the (good) answers to your other questions. – 2012-03-07
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0@Michalis Thank you but I already did. In fact I ticked up both answers if this is not a problem. – 2012-03-07
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0@Sulaiman: What you did is upvote. For the OP there is also another option, that marks a question as "answered". It is the tick right below the up/downvote buttons. It is important that you mark answered questions, as the users of this site will be able to concentrate on the unanswered ones. – 2012-03-07
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0Thank you @Michalis! sure! done!! – 2012-03-07
2 Answers
You can use induction on $m$ and properties of $\mathbb R$ to show compacity using sequential compactness, which means the same thing for metric spaces. Now consider the norm induced on the space $\mathbb R \cong \mathbb R \times \{ 0 \} \times \dots \times \{ 0 \}$ viewed as a sub-metric-space of $\mathbb R^m$, and also consider the sequence $^1 x_n$ induced by putting all the other components but the first equal to $0$. Therefore the first component is a sequence of real numbers. Since in the reals, every metric is equivalent to the absolute value metric in the following sense $$ \forall (\mathbb R,d), \quad \exists c_1, c_2 > 0 \quad s.t. \quad \forall x,y \in \mathbb R, \quad c_1 d(x,y) \le |x-y| \le c_2 d(x,y). $$ One can deduce that the Bolzano-Weierstrass theorem also holds if we replace $| \cdot |$ by the induced metric from the norm in $\mathbb R^m$. Since the sequence $x_n$ is bounded, the sequence $^1x_n$ is also bounded in $\mathbb R$. Therefore there exists a subsequence of the sequence $x_n$ such that the first component converges. Repeat this procedure with the other components $^kx_n$ with $1 \le k \le n$, and you will get a subsequence that converges component by component, hence converges. This gives you for every sequence an element $x$ and a subsequence for which $x_n \to x$. Since $\| x_n \| = 1$ for every $n$, clearly $\|x \| = 1$, so that your subsequence converges in $S$ and we are done.
That is one way to do it ; if you have seen theorems in class that might help, perhaps they might make this less complicated.
Hope that helps,
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1Thanks, this answered my question. Nice work. – 2012-03-07
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0Is there any reference for this proof? – 2012-03-12
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0I just wrote it ; so I guess you could call me the reference. – 2012-03-12
Use the Bolzano-Weierstrass theorem:
Since all the norms on $\mathbb{R}^m$ are equivalent, your subset will be closed and bounded in the euclidian norm $||\cdot||_2$, and hence compact.
Here is an exercise I found, that shows that all norms on $\mathbb{R}^m$ are equivalent: http://math.bu.edu/people/paul/771/equivalent_norms.pdf
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0The compacity depends on the metric space, thus on the *metric* ; your way is one way to go, but it's not complete yet ; you need to show that compacity is a property invariant by *equivalent metrics induced by norms*. – 2012-03-07
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0The equivalence essentially relies on the fact that equivalent metrics induce open balls that can be included in one another. – 2012-03-07
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0@Patrick Da Silva: I don't understand what you mean. I want to use the Bolzano-Weierstrass theorem, thus it is enough to show that $S$ is closed and bounded in the euclidian norm. Showing that the norms are equivalent implies this, as equivalent norms induce the same topology. – 2012-03-07
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0@Patrick Da Silva , elaborating on what Michalis has said, we don't have to go into open balls being included in each other - since equivalent norms induce the same topology, compactness is invariant since in the "open coverings" definition of compactness, the only property of the sets involved that are used are openness. – 2012-03-07
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0@Ragib Zaman : I don't know why but I heard somewhere that equivalent norms don't induce the same topology... I was kind of tired at the moment so I thought it required some thinking and just thought I'd notice there's still work to do. (I think the fact that equivalent norms induce the same topology goes with the fact that balls are included in each other.) – 2012-03-08
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0@PatrickDaSilva It just depends on how you define "equivalent norm". Most references seem to use the definition [here](http://planetmath.org/encyclopedia/EquivalentNorms.html) and properties 2 & 3 are the relevant results. I like to use the definition that it induces the same topology because I get to the use same definition for equivalent metrics and equivalent topology and because topological results come out immediately (such as in this question.) – 2012-03-09
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0@Michalis It is in fact true that for *any* finite dimensional vector space, all norms (if any exist) are equivalent. This is one reason Functional Analysis rarely studies finite dimensional spaces. – 2012-03-09
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1@RagibZaman: This is not generally true, it depends on your base field. It is true if you have a complete valued field like $\mathbb{R},\mathbb{C}$ or $\mathbb{Q}_p$, but fails for example if you look at vector spaces over $\mathbb{Q}$ (e.g. number fields). – 2012-03-09
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1@Patrick Da Silva: You are right, proving that the balls are included in each other you can show that two equivalent norms (with the "inequality"-definition) induce the same topology, now I understand your remark. – 2012-03-09
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1@Michalis Sorry, I should have remembered that! Every normed vector space I've been studying lately has been over $\mathbb{R}$ or $\mathbb{C}$. – 2012-03-09