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Question: Equip $\{0,1\}$ with the Sierpiński topology $\{\{1\},\{0,1\},\emptyset\}$, 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?

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  • $\begingroup$ What is $I_B$? Is it not just $B$? Is you space Hausdorff? $\endgroup$
    – Andrej Bauer
    Commented Feb 1, 2021 at 21:25
  • $\begingroup$ Do you realize that by writing $N_K$ instead of just $N$ you are needlesly bringing in the axiom of choice? Just say "for every compact $K \subseteq B$ there is some $N \in \mathbb{N}$ such that $K \subseteq B_N$". (I am saying this because you used the tag constructive-mathematics so maybe you care.) $\endgroup$
    – Andrej Bauer
    Commented Feb 1, 2021 at 21:28
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    $\begingroup$ Since $X$ is locally compact the compact-open topology on $C(X,\mathbb{S})$ coincides with the Scott topology induced by the pointwise ordering. For this reason its basically never going to be Hausdorff (in any case $\mathbb{S}$ is a retract of $C(X,\mathbb{S})$ whenever $X$ is nonempty). I don't think sequences are particular useful here. $\endgroup$
    – Tyrone
    Commented Feb 1, 2021 at 22:13
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    $\begingroup$ Another way to go is to identify $C(X,\mathbb{S})$ with the set of closed subsets $CL(X)$ of $X$. Then the compact-open topology becomes the cocompact topology, and because of $X$ agrees with the upper-Kuratoski topology. This is the topology determined by the following convergence structure: a filter $\mathcal{F}$ on $CL(X)$ upper-Kuratowski converges to $A_0$ if $\bigcap_{F\in\mathcal{F}}cl(\bigcup_{A\in F}A)\subseteq A_0$. People have spent some time writing about this stuff, so it might be useful to you to check out some of their papers. $\endgroup$
    – Tyrone
    Commented Feb 1, 2021 at 22:14
  • $\begingroup$ @Tyrone Doesn't the Upper-Kurtatowksi convergence coincides with the set-theoretic limit superior if each $A \in F$ is compact? Or have I missed something? $\endgroup$
    – ABIM
    Commented Feb 2, 2021 at 15:13

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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.

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