Is $\left\{ u\in H^{1}\left(\Omega\right)\left|\,a<u\left(x\right) <b\:\text{a.e.}\right.\right\}$ open in $H^1(\Omega)$

Question

Let $\Omega$ bounded, open, connected and Lip. domain.Let positive constants $a,b$ such that $a<b$. Is $A = \left\{ u\in H^{1}\left(\Omega\right)\left|\,a<u\left(x\right) <b\:\text{a.e.}\right.\right\}$ open in $H^{1}\left(\Omega\right)$?

Note that:

1. $\left\{ u\in H^{1}\left(\Omega\right)\left|\,b\leq u\left(x\right)\:\text{a.e.}\right.\right\}$ and $\left\{ u\in H^{1}\left(\Omega\right)\left|\,u\left(x\right)\leq a\:\text{a.e.}\right.\right\}$ are closed sets in $H^{1}\left(\Omega\right)$: Convergence in $H^1(\Omega)$ implies convergence in $L^2(\Omega)$....Then, the complement,$A$, is open.

2. If $A$ is open, for all $u\in A$ there is $r$ such $B_{r}\left(u\right)\subset A$. Let $v\in H^{1}\left(\Omega\right)\backslash L^{\infty}\left(\Omega\right)$. Then, $u+tv\in B_{r}\left(u\right)$ for small $t$. As $A\subset L^{\infty}\left(\Omega\right)$, $u$ and $u+tv\in L^{\infty}\left(\Omega\right)$, then $v\in L^{\infty}\left(\Omega\right)$. $(\rightarrow\leftarrow)$

I don't know which, 1 or 2, is wrong.

Answer 1

The complement of $$ \{ u\in H^1: \ b\le u(x) \ a.e.\} $$ is not equal to $$ \{ u\in H^1: \ b> u(x) \ a.e.\}. $$ So (2) is right, (1) is wrong.