# Math Help - Proof of disconnected sets

1. ## Proof of disconnected sets

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!

2. this proof may have been done in munkres or sheldon davis topology books. I don't remember how it goes, sorry.

3. Originally Posted by EricaMae
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.
Imagine that $S$ is disconnected and closed. But since $S$ is disconnected it means there exists $U,V$, open sets in the topological space so that $S\subseteq U\cap V$ and $U\cap V = \emptyset$. Now let $A = S\cap U$ and $B = S\cap V$ then $A,B$ are closed and $A\cup B = S$ (if we think of this geometrically).

4. This is great, but there is one thing I'm a little confused about.

S is contained in U intersect V, but U intersect V is empty? I'm not sure how to draw this geometrically.

5. Originally Posted by EricaMae
This is great, but there is one thing I'm a little confused about. S is contained in U intersect V, but U intersect V is empty? I'm not sure how to draw this geometrically.

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 $S \cap U\,\& \,S \cap V$ are both closed. That is easy because S is closed.
And $\left( {S \cap U} \right) \cap \left( {S \cap V} \right) = \emptyset$ again that is easy because $U \cap S \cap V = \emptyset$.
All that is left is $S \subseteq U \cup V\quad \Rightarrow \quad \left( {S \cap U} \right) \cup \left( {S \cap V} \right) = S$.

6. Looks like a recap of some basic definition is order here.

To say that $C$ is a disconnected subspace of $Y$ means that there exist subsets $S,T$ which are open in $Y$ such that $U=S\cap C$ and $V=T\cap C$ are disjoint and $C=U\cup V.$

Points to note:

• $U=S\cap C$ and $V=T\cap C$ are disjoint, but $S$ and $T$ themselves need not be disjoint.

• $S$ and $T$ are open in $Y$ but $U=S\cap C$ and $V=T\cap C$ need not be open in $Y$. However, $U$ and $V$ are open in $C$ considered as a subspace of $Y.$

To show that $U$ is closed, show that $U^c=Y\setminus U$ is open.

Let $x\in U^c.$

Either $x\in C$ or $x\in C^c.$

If $x\in C^c,$ then $C^c$ is open and $x\in C^c\subseteq S^c\cup C^c=U^c.$

If $x\in C=U\cup V,$ then $x\in V$ since $x\notin U.$

Thus $x\in T\cap C\subseteq T$. Note that $(S\cap C)\cap(T\cap C)=\O$ $\Rightarrow$ $T\subseteq(S\cap C)^c=U^c.$ Hence $x\in T\subseteq U^c$ and $T$ is open.

In either case, there exists an open set $E$ such that $x\in E\subseteq U^c.$ This shows that $U^c$ is open; hence $U$ is closed.

The proof of the closedness of $V$ is similar.