So in the text Functional Analysis by Walter Rudin the proposition 1.21 states
If $n$ is a positive integer and $Y$ is an $n$-dimensional subspace of a complex topological vector space $X$, then every isomorphism of $C^n$ onto $Y$ is a homeomorphism and $Y$ is a closed set
So if $f:\Bbb R^k\rightarrow\Bbb R^n$ for $k\le n$ is a monomorphism (i.e. a linear injective map) then $f\big[\Bbb R^k\big]$ is a $k$-linear subspace of $\Bbb R^n$ and in particular the above theorem implies that $f$ is a homemorphism between $\Bbb R^k$ and $f\big[\Bbb R^k\big]$ because any monomorphism is an isomorphism onto its image: so finally we conclude that $f$ is a closed map because $f\big[\Bbb R^k\big]$ is closed in $\Bbb R^n$ and so any closed set in $f\big[\Bbb R^k\big]$ is even a closed set in $\Bbb R^n$. Obviously the theorem is enunciated for complex topological vector spaces but when I studied it sometimes ago it seemed to me that it was even valid for real topological vector spaces but I can not be sure now about. So are my argumentations correct? Otherwise is $f$ a closed map? So could someone help me, please?
Best Answer
The result Rudin states for complex vector spaces also holds for real ones (IIRC, the local compactness of the field is the important thing, not the fact that $\Bbb C$ is algebraically closed or something along those lines, see Bourbaki's text I think).
So indeed if $f: \Bbb R^k \to \Bbb R^n$ is a linear monomorphism, it is a closed map and $f[\Bbb R^k]$ is some $k$-dimensional linear subspace of $\Bbb R^n$ (which are always closed anyway, we don't need heavy theorems for that, as it's the intersection of zero-sets of a functional (which are all continuous on $\Bbb R^n$)).