Let be $g_i$ a sequence of functions, where for a given $x\in [a,b]$ we have the limit $\lim\limits_{i\to\infty}g_i(x)=g(x)$ (pointwise convergence) and $g_i,g:[a,b]\subseteq\mathbb{R}\to \mathbb{R}$. All members $g_i$ and $g$ are continuous. In this context I was wondering if the follwoing set is open:
$M_j:=\{x\in [a,b]~|~g_i(x)<\epsilon~\forall i>j\}$, where $\epsilon>0$.
As $M_j$ is not defined by preimage $(g_i)^{-1}(-\infty,\epsilon)$ but the intersection $\bigcap\limits_{i=j+1}^{\infty}\{(g_i)^{-1}(-\infty,\epsilon)\}<\epsilon$, I am not sure how to argue.
Any suggestions or ideas?
No.
Example: Let $g_n\colon[0,1]\to\mathbb{R}$ be $$ g_n(x)=2n\cdot\operatorname{dist}(x,\{0\}\cup[\tfrac1n,1]) $$ i.e., it is a triangular "hat" supported on $[0,\tfrac1n]$. Then $g_n\to 0$ pointwise, $0\in M_j$ for all $j$ and all $\epsilon$, but there are points arbitrarily close to $0$ that isn't in $M_j$ when $\epsilon<1$, namely $\frac1{2n}$ for all sufficiently large $n$.