Are these three sets open? $M=$ space metric
$R=$ real set with usual metric.
$\mathcal{B}(X;R)$ the set of bounded functions with $\sup$ metric, this is, $d(f,g)=\sup.\{d(f(x),g(x));x\in X\}$
$\mathcal{C}_0([a,b];R)$ the set of continuos funtions with $\sup$ metric also.
The sets $A,B,C$ are open?
a) $M=\mathcal{B}(X;R)$, $A=A_a=\{f:X\to R : f(a)>0$ for $a\in X$ fixed$\}$
b) $M=\mathcal{B}(R;R)$, $B=\{f\in M:f$ is discontinuos in all points of $R\}$
c) $M=\mathcal{C}_0([a,b];R)$, $C=\{f\in M:\int_a^bf(x)\,dx>0\}$.
In b) I thought about characteristic function of $Q$ like counterexample, but I couldn't nothing. There is a theorem that says if $g$ is descontinuos on point $a$ and is a distance finite of $f$, then $g$ is interior in $\mathcal{B}(X;R)$, idk if this help. In a) and c) I missing something.
 A: If $f \in A$, let $r= f(a)>0$. Show that $B_d(f, \frac{r}{2}) \subseteq A$ so that $f$ is an interior point of $A$.
For c) define $I: M \to \Bbb R$ by $I(f)=\int_a^b f(x)dx$, then $|I(f)| \le d(f,0)(b-a)$ so that $I$ is conrtinuus, and note that $C=I^{-1}[(0,+\infty)]$ is thus open.
The theorem you quote says that if $f$ is discontinuous at some $a \in \Bbb R$ then for some $r>0$ we have that $B_d(f, r)$ only contains functions that are also discontinuous at $a$. I don't think that helps you for b) though.
A: For a) consider the complement and proof that it contains all limits of sequences. For c) do the same as for a).
An alternative proof for a) is the following: For $a\in X$ fixed consider the function $L:M\to\Bbb R:f\mapsto f(a)$. As the metric space $M$ is equiped with the supremum metric, $L$ is continuous. Since $]0,\infty[$ is an open set in $\Bbb R$ the inverse image $A_a$ is open in $M$.
You can make a similar proof for c) by considering the function $L:M\to\Bbb R:f\mapsto \int_a^bf(x){\rm d}x$.

To b): $B$ is not an open set is equivalent to $$B^\complement=\{f\in M: f \text{ is continuous in at least one point of }\Bbb R \}$$ being not closed. If we can find a sequence in $B^\complement$ whose limit is in $B$ then we are done. Such a sequence of functions is given by
$$f_n(x):=x\cdot 1_{\Bbb Q}(x)1_{\Bbb R\setminus[-1/n,1/n]}(x)+1_{\{0\}}(x),
$$
whereas $1_S(x)$ denotes the indicator function of the set $S$ (i.e. one for $x\in S$ and zero otherwise). The limit of this sequence with respect to the supremum metric is
$$f(x):=x\cdot 1_{\Bbb Q}(x)+1_{\{0\}}(x),
$$ which is an element of $B$. Note: without the $1_{\{0\}}$-part the function would be continuous in zero.
