For which $\mathcal{F} \subset C[0,1]$ does there exist a sequence converging pointwise to the supremum? This question is following this one.
Take a subset $\mathcal{F} \subset C[0,1]$ and consider 
$$g(x)= \sup{ \{f(x)\mid f\in \mathcal{F}} \}.$$ Added: $g$ may take infinite values.
What property(ies) should have $\mathcal{F}$ in order that it exists a sequence $(f_n)$ of elements of $\mathcal{F}$ that converges pointwise to $g$?
I know that the question is open. Answers can be in the form of examples. Please don't shoot to quick to close it because it is an open question!
 A: As the comments hint at, it suffices to assume that $\mathcal{F}$ is upper-directed, i.e. for $f,g \in \mathcal{F}$, there should be some $h \in \mathcal{F}$ with $h \geq \max \{f, g\}$.
Indeed, it is well-known that $C([0,1])$ is separable when equipped with the sup-norm. Hence, so is any subset, in particular so is $\mathcal{F}$. Thus, let $\{g_n \mid n \in \Bbb{N}\} \subset \mathcal{F}$ be dense with respect to the sup-norm.
Using the directedness of $\mathcal{F}$, we find (inductively) for each $n \in \Bbb{N}$ some $h_n \in \mathcal{F}$ with $h_n \geq \max \{h_{n-1}, g_1, \dots, g_n\}$. In particular, $(h_n)_n$ is a non-decreasing sequence, so that $h := \lim_n h_n$ exists. Since $h_n \in \mathcal{F}$, we trivially have $h_n \leq g$ for all $n$ and thus $h \leq g$.
It thus suffices to show $g \leq h$, for which it suffices to show $f \leq h$ for each $f \in \mathcal{F}$. But by construction, we have $g_n \leq h$ for each $n$. Since $\{g_n \mid n \in \Bbb{N}\} \subset \mathcal{F}$ is dense, we easily get $f \leq h$ for each $f \in \mathcal{F}$ as desired.
Thus, in the end (assuming directedness) everything boiled down to a separability statement.
