Given a set $$G= \{c \in L^2[0,1]\ \ | \ \ c \geq 0 \ \ \text{a.e.}\}$$ Now i want to prove that $G$ is not open. In fact it does not have an interior point.
For this i have to prove that for any $c \in G$, any neighborhood of $c$ is not fully contained in $G$. i.e. any neighborhood of $c$ will always contains a point which is not in $G$. How to prove this point? Or there is any other way to show this?
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If I am not mistaken, you just have to prove that for a particular $c$, there is no neighbourhood of that particular $c$ contained in $G$. As $L^{2}[0,1]$ endowed with the $L^{2}$-norm has a metric space structure, it is enough to prove that no open ball around some $c\in G$ is contained in $G$.
Consider thus $c\equiv 0\in G$ and thus any open ball $\{f\in L^{2}[0,1]\vert \int_{0}^{1}\vert f\vert^{2}< r^{2}\}$ centered at $c\equiv 0$ and of radius $r$. Just take $f:[0,1]\to\mathbb{R}:x\mapsto f(x)=-1\mathbb{I}[0,\min\{r^{2}/2,1/2\}]$ it is clear it is contained in the neighbourhood but as $f<0$ on a set of measure $\min\{r^{2}/2,1/2\}>0$, it is not contained in $G$.
EDIT: actually, we don't even need the minimum above but it makes clearer the measure of the set on which $f$ is strictly inferior to zero.
We want to show that any neighbourhood of $c$ contains a point not in $G$. This is equivalent to finding a sequence $c_n$ of $L^2$ functions not in $G$ which converge to $c$ in the $L^2$ norm.
$c_n$ not in $G$ means that the set $\{x:c_n(x)=0\}$ has non-zero measure.
Let us consider the following: Let $A_n$ be a sequence of measurable subsets of $[0,1]$, such that $\mu(A_{n}) = \frac 1n$, and $A_{n+1} \subset A_n$. Define: $$ c_n(x) = \begin{cases} c(x) & x \notin A_n\\ 0 &x \in A_n \end{cases} $$
You can see that each $c_n$ is not in $G$, because it is zero on a non-measurable set. Use any theorem that you know (bounded convergence theorem, for example), and prove that $c_n \to c$ in $L^2[0,1]$. This means that $c \notin G ^\circ$. So $G^\circ$ is empty, of course $G$ is not open.