"Every linear mapping on a finite dimensional space is continuous" From Wiki

Every linear function on a finite-dimensional space is continuous.

I was wondering what the domain and codomain of such linear function are?
Are they any two topological vector spaces (not necessarily the same), as along as the domain is finite-dimensional? Can the codomain be a different normed space (and may not be finite-dimensional)?
I asked this because I saw elsewhere the same statement except the domain is a finite-dimensional normed space, and am also not sure if the codomain can be a different normed space (and may not be finite-dimensional).
Thanks and regards!

The special case of a linear transformations $A: \mathbb{R}^n \to \mathbb{R}^n$ being continuous leads nicely into the definition and existence of the operator norm of a matrix as proved in these notes.
To summarise that argument, if we identify $M_n(\mathbb{R})$ with $\mathbb{R^{n^2}}$, and suppose that $v \in \mathbb{R}^n$ has co-ordinates $v_j$, then by properties of the Euclidean and sup norm on $\mathbb{R}^n$ we have:
$\begin{align}||Av|| &\leq \sqrt{n} \,||Av||_{\sup} \\&= \sqrt{n}\max_i\bigg|\sum_{j}a_{ij}\,v_j\bigg|\\&\leq \sqrt{n}\max_i \sum_{j}|a_{ij}\,v_j|\\&\leq \sqrt{n} \max_i n\big(\max_j|a_{ij} v_j|\big)\\&=n\sqrt{n} \max_i \big(\max_j |a_{ij}| \max_j |v_j|\big)\\&= n\sqrt{n}\max_{i,j}|a_{ij}|||v||_{\sup}\\&\leq n \sqrt{n} \max_{i,j}|a_{ij}|||v||
\end{align}$
$\Rightarrow ||Av|| \leq C ||v||$ where $C = n\sqrt{n}\displaystyle\max_{i,j}|a_{ij}|$ is independent of $v$
So if $\varepsilon>0$ is given, choose $\delta = \dfrac{\varepsilon}{C}$ and for $v, w \in \mathbb{R}^n$ with $||v-w||< \delta$ consider
$||Av - Aw || = ||A(v-w) || \leq C ||v-w || < \delta C= \varepsilon$ from which we conclude that $A$ is uniformly continuous.
