**Question:**
Equip $\{0,1\}$ with the [Sierpiński topology $\{\{1\},\{0,1\},\emptyset\}$](https://ncatlab.org/nlab/show/Sierpinski+space), let $X$ be a compact metric space, and equip $C(X,\{0,1\})$ with the compact-open topology. Let $\{B\_n\}\_{n=1}^{\infty}$ be a sequence of open subsets of $X$ and $B$ also be open in $X$. Concretely, what does it mean for $I\_{B\_n}$ converges to $I\_B$ in this (compact-open) sense?


**Answer?:**
Am I correct in understanding that this means, for every compact subset $K\subseteq B$, there is some $N\in \mathbb{N}$ such that $K\subseteq B\_{N}$?


*Or am I overlooking something?*



Convergence in compact-open topology on the Sierpiński space
Yes, the answer is not correct. Let Y the space $\{0,1\}$ with the Sierpinsky-topology. The compact-open topology is generated by the set of $T(K,U) := \{f \in C(X,Y) \colon f(K) \subset U\}$ with $K \in \mathcal{K}(X)$ (compact subsets) and $U$ open in $Y$. W.l.o.g. $U = \{1\}$. Since $C(X,Y) = \{1\_B \colon B \text{ open in } X\}$ ($1\_B$ the indicator function) we get $T(K,\{1\}) = \{1\_B \colon K \subset B, B \text{ open }\}$. It follows that $\{T(K,\{1\}) \colon K \in \mathcal{K}(X)\}$ is a base (not only a subbase) for the compact-open topology.


Now $1\_{B\_n} \to 1\_B$ (for an arbitrary net) is equivalent to $1\_{B\_n} \in T(K,\{1\})$ finally in $T(K,\{1\})$ for each (neighbourhood) $T(K,\{1\})$ of $1\_B$, i.e. $K \subset B\_n$ finally for each $K \subset B$. For an (arbitrary) counterexample let $X = [0,1]$, $B = [0,1/2)$ and $B\_n = \emptyset$ if $n$ is uneven and $B\_n = [0,1/2-1/n)$ if $n$ is even.


