Let $X\subset \mathcal{B}(\mathbb{R},\mathbb{R})$ be the subset of bounded functions $f:\mathbb{R}\to\mathbb{R}$ such that are discontinuous in all points. Prove that $X$ is not open (with usual metric of supremum).
I think that I found an example of a function $f\in X$ such that for every $\epsilon>0$ there exist $g\in B(f,\epsilon)\cap X^c$.
Let $f$ defined by:
$f(x)=\begin{cases} 1\ ; \text{if } x\in[-1,1]^c\cap\mathbb{Q}\\ x\ ; \text{if } x\in[-1,1]\cap\mathbb{Q}\setminus\{0\}\\ 1/2\ ; \text{if } x=0\\ -x\ ; \text{if } x\in[-1,1]\cap\mathbb{Q}^c\\ -1\ ; \text{if } x\in[-1,1]^c\cap\mathbb{Q}^c\\ \end{cases}$
$f$ clearly be in $X$ and for each $\epsilon>0$ I can define $g$ by:
$g(x)=\begin{cases} f(x)\ ; \text{if } x\in\{0\}\cup(-\epsilon,\epsilon)^c\\ 0\ ; \text{if } x\in(-\epsilon,\epsilon)\setminus\{0\}\\ \end{cases}$
By a "draw" I think that $g\in B(f,1.1\epsilon)$ and $g$ is continuous in an interval then $g\in X^c$. Therefore $X$ is not open.
Am I right?
Do you know another way to prove this fact?