I’m trying to prove that the set of continuous and bounded functions $C(M,\mathbb{R})$ is closed in the set of the bounded functions $\mathbb{B}(M,\mathbb{R})$ which is equipped with the supremum metric $d_{\infty}(f,g)= \sup_{x \in M} |f(x)-g(x)|$.
I know I can prove it directly: Suppose $f_n$ are continuous, that $f_n\rightarrow f$ uniformly and pick $a\in M$. Fix $\epsilon>0$ and choose $n$ such that $|f(x)-f_n(x)|<\epsilon$ for all $x\in M$. [This is true for $n$ sufficiently large but we need only one]. Pick an open set $U\subset M$ with $a\in U$ such that $x\in U$ implies $|f_n(x)-f_n(a)|<\epsilon$. Then $x\in U$ implies $$ |f(x)-f(a)| \le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)| < 3\epsilon. $$ Hence $f$ is continuous in $a$.
The problem is I can't use sequences nor convergence so I have to try something else. So, I want to prove that the set of discontinuous functions is open but I'm really stuck, I don't even know how to start. I hope you guys can help me out.
Thanks so much in advance.
The set of continuous is indeed closed under uniform convergence and this is a valid proof strategy.
As to your idea, suppose that $f$ is discontinuous. This means that (assuming $M$ is a metric space with metric $d$) that
$$\exists a \in M: \exists \varepsilon>0: \forall \delta >0: \exists a ' \in M: d(a,a')< \delta \land |f(a) - f(a')| \ge \varepsilon$$
Can the $\frac{\varepsilon}{2}$-ball around $f$ contain any continuous function ?
If not, that ball shows $f$ is an interior point of the set of discontinuous functions.