this proof may have been done in munkres or sheldon davis topology books. I don't remember how it goes, sorry.
I have to show that given any topological space Y, if a closed subset of Y is disconnected then it must be the union of two disjoint closed sets.
i.e. C= U union V where U and V are closed nonempty sets, and their intersection is empty.
My proof so far:
Let C be a closed subset of Y. Assume C is disconnected. Then there exists two nonempty subsets U,V such that U does not equal C and V does not equal C. Im not sure if I can assume that U= C(V) (complement of V) and V= C(U). If so that would imply that they are disjoint. So I need to show that if U is relatively closed that the complement is relatively open. This implies that if V is relatively open and closed than U is also relatively open and closed. But this is where I get confused because I have to show that C=U union V where U and V are closed disjoint sets. I'm stuck.
thanks!
There are a couple of mistakes(typos?) in the TPH’s explanation.
It should read that U & V are open and have nonempty intersections with S; S, U &V have an empty intersection; and S is a subset of U union V.
Now you need to show that are both closed. That is easy because S is closed.
And again that is easy because .
All that is left is .
Looks like a recap of some basic definition is order here.
To say that is a disconnected subspace of means that there exist subsets which are open in such that and are disjoint and
Points to note:
- and are disjoint, but and themselves need not be disjoint.
- and are open in but and need not be open in . However, and are open in considered as a subspace of
To show that is closed, show that is open.
Let
Either or
If then is open and
If then since
Thus . Note that Hence and is open.
In either case, there exists an open set such that This shows that is open; hence is closed.
The proof of the closedness of is similar.