Unit sphere is compact in 1-norm

  • #1
psie
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TL;DR Summary
How do I go about showing the unit sphere is compact in the ##1##-norm without using the fact that norms on ##\mathbb R^n## are equivalent?
In Introduction to Topology by Gamelin and Greene, I'm working an exercise to show the equivalence of norms in ##\mathbb R^n##. This exercise succeeds another exercise where various equivalent formulations of "equivalent norms" have been given, e.g. that two norms ##\|\cdot\|_a,\|\cdot\|_b## are equivalent iff the identity map from ##(\mathbb R^n,\|\cdot\|_a)## to ##(\mathbb R^n,\|\cdot\|_b)## is bicontinuous.

Now, in showing that all norms in ##\mathbb R^n## are equivalent, the authors show a given norm ##\|\cdot\|## is equivalent to the ##1##-norm (and then by transitivity, we have equivalence for all norms, since equivalent norms is an equivalence relation). I have already managed to understand that the identity is continuous from ##(\mathbb R^n,\|\cdot\|_1)## to ##(\mathbb R^n,\|\cdot\|)##. To show that the inverse of the identity map is continuous, the authors claim that the unit sphere in the ##1##-norm is compact. I'm getting hung up on this statement, since I don't know how to go about this without using that the norms are equivalent already. How would one show the unit sphere in the ##1##-norm is compact?

I know of Heine-Borel, but I'm not sure how and if it applies here. Any help would be very appreciated.
 
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  • #2
The unit sphere in the 1-norm is the set of points ##(x_1,\ldots,x_n)\in\mathbb{R}^n## satisfying ##|x_1|+\ldots+|x_n|=1.##

This set is bounded since ##|x_i|\leq 1## for each ##i##. It is also closed, because the map ##f:\mathbb{R}^n\to\mathbb{R}, f(x_1,\ldots,x_n)=|x_1|+\ldots+|x_n|## is continuous, and your set is the preimage of the closed set ##\{1\}.##

So, by Heine-Borel, it is compact. In the above, closed and bounded are relative to the standard (2-) norm.

It's also not hard to just directly verify that the identity map from ##\mathbb{R}^n## with the 2-norm to ##\mathbb{R}^n## with the 1-norm is continuous.
 
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  • #3
Infrared said:
So, by Heine-Borel, it is compact. In the above, closed and bounded are relative to the standard (2-) norm.
Thank you. May I ask, in what sense are closed and bounded relative to the ##2##-norm? I feel like you only used the ##1##-norm in showing that the set is closed and bounded.

So you showed the set is compact in the ##2##-norm, and since the identity map is bicontinuous between the ##1##-norm and ##2##-norm, it preserves this compact set, is that right?
 
  • #4
psie said:
Thank you. May I ask, in what sense are closed and bounded relative to the -norm? I feel like you only used the -norm in showing that the set is closed and bounded.

In a metric space ##(X,d)##, a set ##E\subseteq X## is bounded if there is a constant ##C## such that ##d(x,y)\leq C## for all ##x,y\in E.## In this case ##E## is the unit ball in the ##1## norm and ##X=\mathbb{R}^n## and ##d## is the usual metric on ##\mathbb{R}^n## (induced by the 2-norm).

So you're just trying to find a constant ##C## such that if ##x=(x_1,\ldots,x_n)## and ##y=(y_1,\ldots,y_n)## satisfy ##|x_1|+\ldots+|x_n|=1## and ##|y_1|+\ldots+|y_n|=1## then the distance from ##x## to ##y## in the usual (2-norm) sense is at most ##C.## You should work this out yourself, it's quick once you get your definitions clear.

Similarly, the map ##(x_1,\ldots,x_n)\mapsto |x_1|+\ldots+ |x_n|## is continuous in the standard sense (as is ##\{1\}## being closed). Saying that this map is continuous means the same thing as when you took multivariable calculus.


psie said:
So you showed the set is compact in the ##2##-norm, and since the identity map is bicontinuous between the ##1##-norm and ##2##-norm, it preserves this compact set, is that right?
I think you got the implication backwards- I thought showing that the identity is bicontinuous was the goal. Checking that the unit ball in the 1-norm is compact is presumably a step in your book's proof (though I don't have your book on hand). Though, it's pretty quick to directly check that the identity map is continuous in both directions without this.
 
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