Normable topology on a finite demensional locally convex space This is a question from Royden's Real Analysis.
Show that the topology on a finite dimensional locally convex topological vector space is induced by a norm. I think this should be a well known result but I can not find any reference that gives a detailed proof of this property. And I have not seen any technique to prove a topology is normable. Any help will be appreciated.
From Royden's book, A locally convex topological vector space is a Hausdorff space, with the 3 additional properties:
1.the addition map $X\times X \to X$, $(x,y)\mapsto x+y$ is continuous
2.the scalar multiplication map $\mathbb{R}\times X \to X$, $(k,x)\mapsto kx$ is continuous
3.for any open set $O$ that contains the origin $0$, there is a convex neighborhood around $0$, denoted by $N$, which is also an element of the topological base, such that $N$ is contained in $O$.
 A: Let $E$ be the locally convex finite dimensional topological space, let $e_1,\dots,e_n$ be a basis for $E$, and let $\mathbb R^n$ be given the standard topology.
Since multiplication by a scalar and addition of vectors is continuous on $E$, it follows that the map $I:\mathbb R^n \to E$, $(x_i)_{i=1}^n \mapsto \sum_{i=1}^n x_i e_i$ is continous.
Now consider the closed unit ball $B \subset \mathbb R^n$.  Then the restriction of $I$ to $B$ is a continuous bijection that maps a compact topological space to a Hausdorff topological space.  Hence it is a homeomorphism.
So we need to show that the map $I$ is open on all of $E$.  Since $I$ is a homeomorphism from $B$ to $I(B)$, it follows that $\epsilon I(B^\circ)$ is a base of neighbourhoods of $0$ in $I(B)$.  Hence by the definition of the subspace topology, and since $E$ is locally convex, for any $\epsilon \in (0,1)$, there exists an open, convex neighbourhood $N_\epsilon$ of $0$ in $E$ such that $N_\epsilon \cap I(B) \subset \epsilon I(B^\circ)$.
Now suppose that $N_\epsilon \setminus I(B) \ne \emptyset$.  Then pick $x \in N_\epsilon \setminus I(B)$.  Then $\|I^{-1}(x)\| \ge 1$.  Since $0 \in N_\epsilon$, the line segment $[0,x] \subset N_\epsilon$.  But then $N_\epsilon$ contains elements whose preimages under $I$ have norm strictly between $1$ and $\epsilon$, contradicting that $N_\epsilon \cap I(B) \subset \epsilon I(B^\circ)$.
Hence $N_\epsilon \subset \epsilon I(B^\circ)$.
Hence any neighbourhood of $0$ in $\mathbb R^n$ maps to an open neighbourhood of $0$ in $E$.  Hence $I$ is an open map on $\mathbb R^n$.
A: I can slightly modify the proof given by@Stephen Montgomery-Smith. Then the locally convex is not necessary in the original post.
Let's cite the process in Stephen's answer. Now we acknowledge that the mapping $I$  from the closed unit ball $B \subset \mathbb R^n$ to $I(B)$ is a homeomorphism. Now we need to show that the map $I$ is an open mapping.
Since $B^\circ$ is an open neighbourhood in $B$, there is an open, balanced  neighbourhood $N$ of $0$ in $E$ such that $N \cap I(B) \subset I(B^\circ)$.
In a same way with Stephen, suppose that $N \setminus I(B) \ne \emptyset$.  Then picking $x \in N \setminus I(B)$, we have $\|I^{-1}(x)\| > 1$.  Since $N$ is balanced, $\frac{x}{\|I^{-1}~(x)\|}$ is still in $N$. Notice that $\|I^{-1}\left(\frac{x}{\|I^{-1}~(x)\|}\right)\| =1$. Then $\frac{x}{\|I^{-1}~(x)\|}$ is also in $I(B)$. It follows that $\frac{x}{\|I^{-1}~(x)\|}$ is also in $I(B^\circ)$, contradicting that $\|I^{-1}\left(\frac{x}{\|I^{-1}~(x)\|}\right)\| =1$. Hence $N \subset  I(B^\circ)$.
Moreover from this we can deduce that any neighbourhood of $0$ in $\mathbb R^n$ maps to an open neighbourhood of $0$ in $E$.
