Vtyjkyuk

I’m trying to prove that the set of continuous and bounded functions $C(M,mathbbR)$ is closed in the set of the bounded functions $mathbbB(M,mathbbR)$ 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 $f_nrightarrow f$ uniformly and pick $ain M$. Fix $epsilon>0$ and choose $n$ such that $|f(x)-f_n(x)|<epsilon$ for all $xin M$. [This is true for $n$ sufficiently large but we need only one]. Pick an open set $Usubset M$ with $ain U$ such that $xin U$ implies $|f_n(x)-f_n(a)|<epsilon$. Then $xin U$ implies
$$
|f(x)-f(a)| le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)| < 3epsilon.
$$
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 $fracvarepsilon2$-ball around $f$ contain any continuous function ?

If not, that ball shows $f$ is an interior point of the set of discontinuous functions.