Is $B=\{f \in C_b(\Bbb R, \Bbb R)| f(x) > 0 \rbrace $ open and what is its interior? Problem: Let $B=\{f \in C_b(\Bbb R, \Bbb R)| f(x) > 0$ for all $ x \in \Bbb R \}$ where $C_b$ are functions that are both continuous and bounded. Is $B$ open and what is the interior of $B$?
My attempt:
$B$ is not open. Take $f(x)=1/n$, $g(x)=-1/n$ constant functions where $n \in \Bbb N$ then considering $\Bbb R$ as a metric space with the standard metric and $d(f,g)= sup \{d(f(x),g(x))| x \in A \}$ we have for any $\epsilon >0$ that 
$|d(f,g)| = |2/n|$ which can be made arbitrarily small so that any epsilon ball around $f$ will contain a function that is not always positive, hence $B$ cannot be open.
Now as for the interior I have no idea where to start on this and am in need of a hint. Thanks.
 A: Here's one way to look at this: let $f\in B$ be such that $\inf(f)=0$. That means, that for any $\epsilon>0$, there is an $x_\epsilon\in\mathbb{R}$ with $0<f(x_\epsilon)<\epsilon$. If we let $f_\epsilon(x)=f(x)-\epsilon$, then $f_\epsilon(x_\epsilon)<0$, and so $f_\epsilon$ is not in $B$. It is clearly in $C_b(\mathbb{R}, \mathbb{R})$. Since $d(f,f_\epsilon)=\epsilon$, we see there are points arbitrarily close to $f$ that are not in $B$.  Thus no neighborhood of $f$ is contained in $B$, and so $f\notin\operatorname{int}(B)$.
On the other hand, if $\inf(f)=\delta>0$, then if $d(f,g)<\delta/2$, we have 
\begin{align}
g(x) &= f(x) - \left(f(x) - g(x)\right) \\
&\ge f(x) - |f(x) - g(x)|\\
&\ge \delta - \delta/2\\
&> 0
\end{align}
so $B_{\delta/2}(f)\subset B$.
A: Let $f(x) = 1/(1+x^2).$ Then $f\in B.$ For $n=1,2,\dots,$ define functions $g_n$ in $C_b\setminus B$ as follows: $g_n(x) = f(x), x\le n,$ $g_n(x)$ is the linear function on $[n,n+1]$ whose graph connects $(n,f(n))$ and $(n+1,0),$ and $g_n(x) = 0, x\ge n+1.$ Note that
$$\|g_n-f\|_{C_b} \le f(n)\,\, \text {for all } n.$$
Thus $(g_n)$ is a sequence in $C_b\setminus B$ that converges to $f$ in $C_b.$ This proves that no $B(f,r ),r>0,$ is contained in $B.$ Thus $B$ is not open in $C_b.$
As for the interior of $B:$ Consider $\{f\in B: \inf_{\mathbb R} f>0\}.$ 
