Maximal trivialising subspace for a vector bundle Let $X$ be a locally compact Hausdorff space. Given a vector bundle $p: E\to X$, a subspace $Y$ of $X$ is called trivialising (for this bundle), if after restricting this bundle to $Y$, it is a trivial bundle. In other words, $p: p^{-1}(Y)\to Y$ is trivial. $Y$ is called maximally trivial, if it is trivial and there is no trivial subspace of $X$ which contains $Y$ as a proper subset. 

Given a point $x$ in $X$, a maximal trivialising subspace containing $x$ may
  not be unique. Does any maximal trivialising subspace must be open? Can we say
  something others about maximal trivialising subspaces?

There is no reference about this topic in the existing literatures.
A similar question has been posted on stackexchange:
https://math.stackexchange.com/questions/1581861/maximal-trivial-subspace-in-vector-bundles
 A: I claim that any maximal trivial subspace $Y$ is dense in $X$.
Otherwise, take $x\in X\setminus\operatorname{Cl}( Y) \neq\emptyset$. Since $X$ is locally compact Hausdorff, and hence regular, there are non-intersecting open neighbourhoods of $x$ and $\operatorname{Cl}(Y)$. It follows that the subspace topology on $Y\cup\{x\}$ is the disjoint union topology. Hence $E$ is trivial when restricted to $Y\cup\{x\}$ (since its trivial when restricted to each of $Y$ and $\{x\}$). Hence  $Y\cup \{x\}$ is a trivial subspace, contradicting the maximality of $Y$. 
A: In addition to Mark's argument, a maximal trivialising $Y\subset X$ is also open, if we assume that $E$ is a $\Bbbk$-vector bundle with $\Bbbk=\mathbb R$ or $\mathbb C$. So every maximal trivialising subset is open and dense, but it is not (yet) clear that every trivialising subset is contained in a maximal one.
Assume $Y$ is trivialising, but not open. Then there exists $x\in Y$ such that no neighbourhood of $x$ is contained in $Y$. We choose a trivialising neighbourhood $U$ for $E$. We have maps $\varphi\colon E|_Y\to\Bbbk^r$ and $\psi\colon E|_U\to\Bbbk^r$ such that
$$p\times\varphi\colon E|_Y\to Y\times\Bbbk^r\quad\text{and}\quad
p\times\psi\colon E|_U\to U\times\Bbbk^r$$
are vector bundle isomorphisms. Hence there exists $g\colon Y\cap U\to GL_r(\Bbbk)$ such that $\psi(e)=g(p(e))\cdot\varphi(e)$ on $E|_{Y\cap U}$.
Because $g$ is continuous and $Y\cap U$ carries the subspace topology, there exists a compact neighbourhood $K\subset U$ of $x$ and a map $\xi\colon Y\cap K\to\mathfrak{gl}_r(\Bbbk)$ such that $g|_{K\cap Y}=\exp\circ\xi$. By Urysohn's lemma, there also exists a cutoff function $\rho\colon X\to[0,1]$ with $\mathrm{supp}(\rho)\subset K$ and such that $W=\rho^{-1}(1)$ is a neighbourhood of $x$. We replace the chosen trivialisation of $E|_Y$ by $$\varphi'(e)=\exp((\rho\cdot\xi)(p(e)))\cdot\varphi(e)\;.$$
Then $\varphi'$ agrees with $\varphi$ on $Y\setminus\mathring K$,
and with $\psi$ on $W\cap Y$.
Hence, we can extend $\varphi'$ by $\psi$ on $W$, contradicting the maximality of $Y$.
