Let $S$ be a set, or equivalently the topological space with discrete topology and $2$ two point set with discrete tiopology. $βS$ be the Stone–Čech compactification of $S$. By Tychonoff theorem the topology on $2^S$ is compact with respect to the product topology.  

Is the compact-open topology on $2^{\beta S}$ the product topology or it is something much stronger, say uniform topology? How one can see it.


In particular can $2^{\beta S}$ be a compact in case $S$ is an infinite set. 


Given X, $2^X$ means the topological space with following property:


1. There exists continuous map, an evaluation map $e:2^{X}\times X\to 2$
2. For any compact Hausdorff and zero-dimensional topological space A and continuous map $f:A\times X\to 2$, there exists $\hat{f}:A\to 2^{X}$ such that $f=e\circ (\hat{f}\times id\_{X})$.
The last says that if $A=\{\*\}$ a one point set then $2^X$ as a set is set of continuous maps from $X$ to $2$.



Compact open topology on $2^{\beta S}$ with $S$ a set
Let me try to put some things in order here.


For a topological space $X$ we denote here by $2^X$ the set of all continuous functions $X\to \{0,1\}$. 
Fixing a set $S$ we consider the set $2^{\beta S}$ where $\beta S$ denotes the Stone–Čech compactification of $S$.
For $A,B\subset \beta S$ let me write
$$U(A,B)=\{f : f|\_A=0,~f|\_B=1\} \subset 2^{\beta S}.$$
We consider here three topologies on the same space $2^{\beta S}$. These given as follows:


1) The *compact-open topology*, which basis is given by the sets $U(A,B)$ where $A,B\subset \beta S$ are compact.


2) The *product topology*, which basis is given by the sets $U(A,B)$ where $A,B\subset \beta S$ are finite.


3) The *$2^S$ topology*, which basis is given by the sets $U(A,B)$ where $A,B\subset S\subset \beta S$ are finite.


Let me note that (1) is a natural topology on a space of continuous functions, (2) is a natural topology on a subspace of the set of *all* maps $\beta S\to\{0,1\}$ and (3) is the pull back of the natural topology on $2^S$ under the (bijective) restriction map $2^{\beta S}\to 2^S$. So these are all reasonable.


The question seems to be about the comparison of these topologies.
It is clear from the above description that $(3) \leq (2) \leq (1)$.


If $S$ is finite than they are all the same. Let me assume from now on that $S$ is infinite.
I claim that $(3)\lneq (2) \lneq (1)$.


To see the inequalities, note that for $x\in \beta S-S$ the function $f\mapsto f(x)$ is (2)-continuous but not (3) continuous and that the set of functions which are supported on $S$ is (1) open but not (2) open.


Note that (3) is Hausdorff and compact by Tychonoff, so it also follows that (2) and (1) are not compact, otherwise the corresponding identity map would be a homeomorphism.


