Is F open in the pointwise convergence topology? Let $F=\{f\in \mathcal{C}([a,b]): f(t)>0,\ \forall t\in [a,b]\}$
Is F open in the pointwise convergence topology $\mathcal{O}_{ptc}$ on $\mathcal{C}([a,b])$?
Given a basis of $\mathcal{O}_{ptc}$, consisting of the family of all sets 
$\mathcal{B}(x_1,...,x_n;t_1,...t_n;\epsilon_1,...\epsilon_n)=\{f\in\mathcal{C}([a,b]): f(x_i)\in B_{\epsilon_i}(t_i),i=1,...,n\}\\ 
n\in\mathbb{N}, x_1,...,x_n\in[a,b]; t_1,...,t_n\in\mathbb{R};\epsilon>0$
I have to prove or disprove the statement. 
Now my answer (just in words) would be, that I can always chose elements from the basis to "model" every function of F, so that F is necessarily contained in a particular union of elements of the basis and never touches the x-axis.
My solution says, F is NOT open, thus I am quite confused.

Suppose $f \in F$ and let $f \in U$ where $U$ is open. Then there is some basis
element $B$ such that $f \in B \subset U$. 
By choosing a smaller basis element as necessary, we can assume that $B$ has
the form
$B= \{ g | g(x_k) \in (t_k-\epsilon, t_k+\epsilon), k=1,...,n \}$ for some
fixed $t_k, x_k$ and $\epsilon >0$.
Note that the only constraints that membership of $B$ imposes is that $g$
must lie in the interval $(t_k-\epsilon, t_k+\epsilon)$ at $x_k$, for a
finite number of $x_k$.
Choose $t^*$ to be distinct from the $t_k$ and choose
$h$ to be any polynomial such that $h(t_k) = f(x_k)$ and $h(t^*) = -1$. Then
$h \in U$ but clearly $h \notin F$.
Hence $F$ cannot be open since any non empty open set contains elements that are not in $F$.
