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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.

1 Answers 1

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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 $F$.

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    I understand, thanks. But the thing is, $f$ does not have to be in all of the basis elements. I might aswell choose $h(t^*)>0$ for all other points $t^*$ and then I have that $h$ lies in the union of those. – 2017-02-09
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    If $F$ was open and $f \in F$, then there would be some open $U$ such that $f \in U \subset F$. There would be **some** basis element $B$ such that $f \in B \subset U$. This basis element (in fact all) will contain some $h$ that is **not** in $f$. Hence such a $U$ cannot exist. The point is that the weak topology does not 'constrain' open sets enough so that they can be contained in $F$. – 2017-02-09
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    May I ask, if all f with $\int^b_a f(t)dt =0 \forall t \in [a,b]$ would be open too? I could again assert arbitrary values to some $x$ between $x_k$ and $x_{k-1}$ that make the integral not dissapear, right? – 2017-02-09
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    That set would not be open, as you could find $g$ in any basis set containing $f$ that has a non zero integral. – 2017-02-09