# Compact open topology on $2^{\beta S}$ with $S$ a set

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 $2$.
• What do you mean by $2^{\beta S}$? all maps? only the continuous ones? if all maps, why will you expect any relation with the structure of $\beta S$? if only continuous, note that the restriction map to $2^S$ is a homeomorphism. – Uri Bader Mar 2 '17 at 9:11
• @Uri Bader I want $2^{\beta S}$ to be an exponent of $2$ by $\beta S$. Of course I consider the continuous maps. The restriction is bijection, but how do you see it is an homeomorphism? Thanks for a feedback – Evgeny Kuznetsov Mar 2 '17 at 9:21
• A continuous bijective map between Hausdorff compact spaces is a homeomorphism (because it is closed). – Uri Bader Mar 2 '17 at 9:23
• But how do you know that $2^{\beta S}$ is a compact? Is it clear and I do not know? Compactness is exactly I want to know about $2^{\beta S}$. I guess $2^{\beta S}$ is a compact iff $S$ is finite. – Evgeny Kuznetsov Mar 2 '17 at 9:25
• I am sorry, it is not compact if $S$ is infinite. I got confused. I guess you should edit your question and explain what $2^{\beta S}$ means. – Uri Bader Mar 2 '17 at 9:34

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.

• Thank you @Uri Bader, this is that kind of answer I expected to see when I asked. – Evgeny Kuznetsov Mar 3 '17 at 9:51