Let $\{0,1\}$ be equipped with the Sierpiński topology $\{\emptyset, \{0,1\},\{1\}\}$, and $\mathbb{R}^d$ with the usual Euclidean topology. Then is the pointwise-convergence (point-open) topology on $C(\mathbb{R}^d,\{0,1\})$ indeed weaker than the compact-open topology?


I have in mind the case where $n>1$.



Comparison of topology of pointwise convergence and compact-open topologies for Sierpiński space
Functions $\mathbb R^d\to \{0,1\}$ are indicator functions $f=I\_{A}$ with $A=f^{-1}(\{1\})$ and continuity with respect to the Sierpiński topology precisely means that $A$ is open in $\mathbb R^d$. The topology of pointwise convergence on $C(\mathbb R^d,\{0,1\})$ is strictly coarser than the compact-open topology: Since the only interesting open set in $\{0,1\}$ is $\{1\}$ the sets $W(K)=\{I\_B: K\subseteq B\}$ ($K$ compact and $B$ open in $\mathbb{R}^d$) form a base of the compact-open topology and the sets $W(E)$ with finite $E$ form base of the [point-open topology](https://math.stackexchange.com/questions/411045/topology-of-pointwise-convergence-and-point-open-topology-are-equivalent) (that name is probably not very common).


To show that the claim one has to find a compact set $K$ such that $W(K)$ does not contain any $W(E)$ for a finite set $E$. This is the case, e.g, for the closed unit ball $K$ in $\mathbb R^d$: For any finite $E$ take a union $B$ of smalls open balls centered at the points of $E$ so that the volume of the $B$ is strictly less than the volume of $K$. Then $I\_B \in W(E) \setminus W(K)$.


