Let $\mathcal{F} \subset C[0,1]$ consider $$g(x)= \sup{ \{f(x)\mid f\in \mathcal{F}} \}$$ Can we say $\exists \{f_n\} \subset \mathcal{F} $ s.t. $f_n \rightarrow g$ pointwise?
Sorry if this is a very easy question, I am a bit confused. From the definition we know that for each $x$ there exists a sequence (depending on $x$) $\{f_n^x\}$ s.t. $f_n^x(x)\rightarrow g(x)$. How can we get a sequence which would be "uniform" and would converge for all $x$? Thank you.
It's false : consider $f:[0,1]\to\mathbb{R}$ such that $f(x)=x$ and $h:[0,1]\to\mathbb{R}$ such that $h(x)=1-x$. Then $g=f$ if $x\geq 1/2$ and $g=h$ if $x\leq 1/2$, and there is no sequence in $\mathcal{F}=\{f,h\}$ which verify what you are looking for.
The answer is no. As a counterexample, let $\mathcal F$ be the collection of functions defined over integers $1 \leq k \leq n$ by $$ f_{{n,k}} = \begin{cases} 1 & x \in [(k-1)/n, k/n]\\ [\text{interpolating line}] & x \in [(k-2)/n,(k-1)/n] \cup [k/n,(k+1)/n]\\ 0 & \text{otherwise} \end{cases} $$