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>



Maximal trivialising subspace for a vector bundle
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 $Y$.


