Unit sphere is compact in 1-norm

[psie]

TL;DR

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.

 

[Infrared]

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.

 

[psie]

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?

 

[Infrared]

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.

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.