Is a subset open in a topology?

Question

Given are two metrics $d_\infty (f,g) := \sup \{ f(t)-g(t)| , t\in[a,b]\}$ and

$d_1(f,g) := \int_{[a,b]} |f(t)-g(t)|\mathrm{d}t$ on $C([a,b])$.

Let $F = \{ f\in C([a,b]) : f(t)>0, \ \forall t\in [a,b]\}$

Is $F$ open in the topology

$\mathcal{O}(d_1)$ or $\mathcal{O}(d_\infty)$?

If I am correct $\mathcal{O}(d_1)$ means the set of all open set with respect to the metric $d_1$ on $C([a,b])$. Now that means, by intersecting and combining elements of a basis of $\mathcal{O}(d_1)$ i should obtain $F$, if it is in the topology.

Since $f\in C([a,b])$ it is bounded. Thus, $\infty>f(t)>0$ for all $t \in [a,b]$.

I am not sure what to do now. Do I have to find a basis first? I could then intersect all negative valued functions and the zero functions from all C functions and would somehow obtain all positive functions.

Answer 1

Consider a function $f \in F$. This is continuous and so has a minimum $m >0$ on $[a,b]$. If a function $g$ is now close enough in the $\sup$-metric (so for all $t$, $f(t)$ is close to $g(t)$) it will also not touch the $x$-axis and lie in $F$... So $f$ is then an interior point...

For the integral-metric, what does it mean to be close to $f$? We know that $|f(t) - g(t)|$ is bounded by $d_\infty(f,g)$, so the integral over $[a,b]$ of this difference is also bounded: $d_1(f,g) \le (b-a)\cdot d_\infty(f,g)$. But this goes the wrong way, inequality-wise. So take the function $f$ that is constantly $1$ for definiteness. Can you find a function $g$ that is close in integral (all $1$ with a spike down to $0$ which is very steep, say $g$ is linear from $1$ downto $\frac{1}{n}$ and up to $1$ at from $\frac{1}{n}$ to $\frac{2}{n}$ and $1$ onwards. This has difference integral with $f$ of $\frac{1}{n}$ which can be made as small as we like; note that $g \notin F$. So any open ball around $f$ intersects $X \setminus F$. So we have found a non-interior point of $F$ in this metric.