TITLE: Topological vector spaces (reference request)
QUESTION [13 upvotes]: In his book Topological Function Spaces Arhangel'skii says that "it is well known that every nontrivial locally convex linear topological space $X$ is homeomorphic to a space of the form $Y \times \mathbb{R}$, for some space $Y$".
I've been looking for a proof of this result but haven't found anything, which leads me to believe that this is a standard fact that everyone knows. Could any of you please help me with a reference in which this result appears?

REPLY [17 votes]: I guess that non-trivial means that the locally convex space $X$ is not endowed with trivial topology $\{\emptyset,X\}$.
This implies that $X\neq \overline{\{0\}}$ (since this closure does have the trivial topology). For $y\notin \overline{\{0\}}$,
the Hahn-Banach theorem (applied to $L=\{ty: t\in\mathbb K\}$ and the linear functional $ty\mapsto t$ which is continuous because there is a continuous seminorm $p$ on $X$ with $p(y)>0$) implies that there is a continuous linear functional $\varphi\in X'$ with $\varphi(y)=1$. Then $X$ is isomorphic (in the category of topological vector spaces which is much more than homeomorphic) to $Y\times \mathbb K$ for the kernel $Y$ of $\varphi$, an isomorphism is given by $x\mapsto (x-\varphi(x)y,\varphi(x))$.
If, for complex locally convex spaces, you insist on a homeomorphy to $Y\times \mathbb R$ identify $\mathbb C=\mathbb R\times \mathbb R$.
Since this is such a simple application of Hahn-Banach, I doubt that it is worthwhile to search for an explicit reference.

REPLY [9 votes]: This fact is a consequence of Hahn-Banach separation theorem (HBST). Such a space admits a continuous linear functional.
Let us be more specific: let $V$ be a Hausdorff HLCTVS (Hausdorff locally convex topological vector space) over $\mathbb{R}$. Let $v\in V\setminus \{0\}$ and $U$ be a convex neighbourhood of zero which does not contain $v$. From HBST, it exists a linear fonctional $\varphi$ such that $\varphi(U)=I\subset \mathbb{R}$ is an open interval and $\varphi(v)\notin I$. Then $\varphi$ is continuous and $H=\ker(\varphi)$ is closed. Then the map $p:\ V\to V$ such that $p(x):=x-\varphi(x)/\varphi(v).v$ is a continuous projection $V\to H$. The map $x\mapsto (p(x),\varphi(x))$ is an isomorphism
$V\to H\times \mathbb{R}$.
Late edit: Jochen is right, Hausdorff is too strong, non-trivial is sufficient. One has just to modify the proof above considering that, if $V$ is non-trivial $\overline{\{0\}}\neq V$, then choose your $v\notin \overline{\{0\}}$, the remainder of the proof is unchanged.
Remark 1 The converse is true in the following manner.
A LCTVS (locally convex topological vector space) $V$ can be decomposed as $V\simeq H\times \mathbb{R}$ iff its topology is not the coarsest one.
Remark 2 As regards the "reference-request" aspect of the question, you can read 
N. Bourbaki, Topological Vector Spaces, Chapters 1-5 (13 november 2002, Springer).
Where, as the treatment is systematic, you get: Non-Hausdorff LCTVS, connection with coarse topology (treated in General Topology of the same treatise) and two forms of Hahn-Banach theorem as well.