Finite-dimensional Hausdorff vector topologies are Euclidean

Remarks

The map $T(c_1, \ldots, c_n) = c_1 e_1 + \cdots + c_n e_n$ is continuous, because addition and scalar multiplication are continuous in $E$. For the continuity of $T^{-1}$, let $S$ be the unit sphere of $\mathbb{K}^n$; it is compact, so $T(S)$ is compact and does not contain $0$. Since $E$ is Hausdorff, there is a balanced neighbourhood $V$ of $0$ disjoint from $T(S)$. Then $T^{-1}(V)$ is balanced and misses $S$, so it lies in the open unit ball, which gives continuity of $T^{-1}$ at $0$. Example: all norms on $\R^n$ define the same topology, since each makes $\R^n$ a Hausdorff topological vector space. The Hausdorff hypothesis is needed: $\R$ with the indiscrete topology is a topological vector space. The identity map from $\R$ with the Euclidean topology to it is a continuous linear bijection, but not a homeomorphism. Consequences: every finite-dimensional subspace of a Hausdorff topological vector space is closed, and every linear map from a finite-dimensional Hausdorff topological vector space to any topological vector space is continuous.

Going deeper

For a student

This theorem explains why topology plays no role in finite-dimensional linear algebra and becomes important only in infinite dimension. It contains the familiar fact that all norms on $\R^n$ are equivalent, and it implies that finite-dimensional normed spaces are complete and that the Heine–Borel theorem holds in them. A companion result says that a Hausdorff topological vector space with a compact neighbourhood of $0$ must be finite-dimensional. Together they show that local compactness singles out the finite-dimensional spaces.