Is an NVS complete iff it is non meagre? Let $V$ be an NVS. We know by Baires theorem that complete $\implies$ nonmeagre. What about the converse? Can we have a non complete space that is nonmeagre or not?
For metric spaces the answer is yes we can, just disjoint union a complete space and a meagre space e.g. $\mathbb{Q}\underline{\cup} \mathbb{R}$, then the $\mathbb{Q}$ part makes it not complete and the $\mathbb{R}$ part makes it nonmeagre. But no similar tactic works for an NVS so I think the answer will be no.
 A: Each infinite dimensional Banach space $X$ contains a dense hyperplane which is of second category in itself:
Let $\{b_j: j\in J\}$ be a Hamel base of $X$, thus each $x \in X$ has a
representation
$$
x= \sum_{j \in J} \varphi_j(x)b_j
$$
with $\varphi_j(x) \not= 0$ for at most finitely many $j \in J$. For each $j \in J$ let $Y_j$ be the kernel of the linear functional $\varphi_j$. Assume by contradiction that $Y_j$ is of first category in $X$ for infinitely many $j \in J$. Let $I \subseteq J$ be a countable infinite subset such that $Y_j$ is of first category $(j \in I)$. Then for each $x \in X$ there is some $j \in I$ with $\varphi_j(x)=0$. So
$$
X=\bigcup_{j \in I} Y_j
$$
is of first category, a contradiction. Thus there are (many) $Y_j$ such that $Y_j$ is of second category in $X$. Such an $Y_j$ is then also of second category in itself. Moreover it can't be closed (otherwise it would be a closed strict subspace of $X$, hence of first category). By the same reasoning it is dense in $X$ and therefore it is not complete.
