For a general topological vector space, is it always true that linear projections are open maps, even the discontinuous ones? If not, are there well-known counterexamples? What about in normed spaces? Thanks in advance.
Update:
It is known when the range of the projection is finite dimensional, the projection is open. See this.
Also, I found this answer in the context of Hausdorff TVS and continuous projections. But the proof there doesn't seem to rely on Hausdorffness or continuity. So is the claim true in general, without these two restrictions?
Update2:
It seems, based on the linked answer above, the following two lemmas guarantee it's true in general.
Lemma 1. Let $X$ be a TVS, then for every $x\in X$, the translation $T_x: y\mapsto x+y$ is a homeomorphism of $X$.
Lemma 2. Let $X, Y$ be TVS, then a linear map $T: X\to Y$ is an open map iff, for every open neighborhood $U\ni 0_X$ of the origin in $X$, its image $T(U)$ contains an open neighborhood $V\ni 0_Y$ of the origin in $Y$.
Proof of Lemma 2. Suppose the condition is satisfied. Let $U'\subset X$ be any open set. For every $y\in T(U')$, there exists an $x\in U'$ such that $y=T(x)$. By lemma 1, set $U'-x$ is an open neighborhood of the origin. Hence, there exists an open neighborhood $V'\ni 0_Y$ of the origin in $Y$ such that $V'\subset T(U'-x)$. Then, $V'+y$ is an open neighborhood of $y$ such that $V'+y\subset T(U')$. Set $T(U')$ is then open. The opposite direction is trivial. $\square$
Proposition. Let $X$ be a TVS. Then, for an arbitary subspace $M\subset X$, a linear projection $P: X\to M$ satisfying $P^2=P$ is an open map, where $M$ is equipped with the subspace topology.
Proof. By lemma 2, it suffices to show, for every open neighborhood $U\ni 0$ of the origin, its image $P(U)$ contains an open neighborhood of the origin. By the definition of subspace topology, $V:=U \cap \text{ran}(P)$ is an open neighborhood of the origin in $\text{ran}(P)$. Also, for every $y\in V\subset U$, $P(y)=y$. So $V\subset P(U)$. $\square$
Are there any mistakes?