# Proof of compactness for sets of norm equal to one in finite-dimensional normed vector spaces

The proposition I have been trying to prove is that the set $A=\{x\in E:N(x)=1\}$ is a compact subset of the (real) finite-dimensional vector space $E$ for any norm $N:E\to \mathbb{R}$.

I am reading the proof of this proposition from a textbook of topology. Briefly, the main reasoning of the proof is as follows. There exists a norm $N^{'}(x)$ such that the metric space $(E,\rho^{'}),~\rho^{'}(x,y)=N^{'}(x-y)$, is isometric with the Euclidean metric space $(\mathbb{R}^n,|\cdot|)$. Furthermore, the set $B=\{x\in\mathbb{R}^n:|x|=1\}$ is a closed and bounded subset of $\mathbb{R}^n$, therefore it is compact. Given that compactness is a topological property and that $B$ is compact, the set $C=\{x\in E:N^{'}(x)=1\}$ is also compact.

It is further known that all the norms of $E$ are equivalent, given that $E$ is finite-dimensional. It is finally claimed in the proof that since i) $N$ and $N^{'}$ are equivalent norms, ii) compactness is a topological property and iii) $C$ is compact, $A$ is compact too.

This is all understood apart from one point; I have to prove that there exists a homomorphism $f:A\to C$. I don't see how this follows from the equivalence of norms $N$ and $N^{'}$.

I am aware that equivalence of norms implies equivalence of metrics, which means that the metric spaces $(E,\rho),~\rho(x,y)=N(x-y)$, and $(E,\rho^{'})$ have the same topology, but this hasn't helped me construct the homomorphism of interest.

I thought of defining the function $f:A\to C$ as $f(x)=\frac{x}{N^{'}(x)}$, which can be shown to be bijective, but I can't prove that it is continuous, so as to prove that it is homomorphic (in fact I am not even sure if it is continuous).

I am trying to prove this proposition without constraining the problem to inner product spaces (aka Hilbert spaces).

• Just define $f \colon (A, N) \to (C, N^\prime)$ as $f(x) \colon= x$ for all $x \in A$. To prove continuity, use the equivalence of the norms $N$ and $N^\prime$. – Saaqib Mahmood Aug 12 '14 at 14:15
• You mean to define $f$ to be the identity function? The $f$ you proposed is certainly continuous, as this follows from the fact that $(E,N)$ and $(E,N^{'})$ have the same topology. However, the identity function does not map a vector $\mathbb{x}\in A$ to a vector in $C$, since $N^{'}(x)\ne 1$ in general. – scidom Aug 12 '14 at 14:39

If $A$ has the norm $N$ and $N'$ is an equivalent norm, $N'$ is continuous. Since $N'(x)\neq 0$ for all $x\in A, f$ is continuous as well. A similar argument can be used to show the inverse is continuous.
There is an easier way however. Since $E$ is finite-dimensional, there is an isomorphism $L:E\rightarrow \mathbb{R}^n$ for some $n$. If $N$ is the norm on $E$, we can define a norm $N'$ on $\mathbb{R}^n$ by $N'(x):=N(L^{-1}(x))$ (you have to check this is a norm). Then $B=\{x\in \mathbb{R}^n:N'(x)=1\}$ is closed and bounded, since all norms on $\mathbb{R}^n$ are equivalent, hence B is compact. But then so is $A$, since $A=L^{-1}(B)$ and $L$ is an isomorphism.
• Thanks Vincent, your answer completed the proof. You are right that $f(x)=\frac{x}{N^{'}(x)}$ is continuous (it suffices to show that $x$ and $\frac{1}{N^{'}(x)}$ are continuous). It is not needed to show that the inverse is continuous, because it holds in general that if a function $f:E_1\to E_2$ is bijective and continuous and $E_1$ is compact, then $f^{-1}$ is also continuous. – scidom Aug 13 '14 at 13:14