Prove that $\overline{C}$ and $D$ are disjoint

Question

Suppose that $C$ and $D$ form a separation of $B$. Prove that $\overline{C}$ and $D$ are disjoint.

(A separation of $B$ is a pair of disjoint nonempty open subsets of $B$ whose union is $B$.)

My attempt:

Suppose $\overline{C}\cap D\ne\varnothing$.

Let $x\in\overline{C}\cap D$.

$\implies x\in\overline{C}$ and $x\in D$

$\implies x\in\overline{C}$ and $x\notin C$

How do I get a contradiction?

Answer 1

I assume you mean that a separation of $B$ is a pair of two nonempty disjoint subsets both open in $B$ whose union is $B$.

If $U$, $V$ form a separation of $B$, then notice that they are also both closed in $B$, since they are the complements of each other.

Now $\overline{U}\cap V = U \cap V = \emptyset$.

If you don't assume that $U, V$ are both open/closed in $B$, then the statement is false:

Take $B = [0,1]$, $U = \left[0, \frac12\right]$ and $V = \left\langle \frac12, 1\right]$. Clearly $U \cap \overline{V} = \left\{\frac12\right\} \ne \emptyset$.

Answer 2

As I learnt it, a separation of a space $X$ is a pair of sets $A, B$ such that $A \cup B = X$ and $\overline{A} \cap B =\emptyset = A \cap \overline{B}$. The latter condition I call "$A$ and $B$ are separated". The $T_5$ separation axiom is stated as "$T_1$ and any two separated sets can be separated by disjoint open sets", e.g.

The union condition than implies that $A, B$ are both open in $X$ and both closed in $X$: $X = A \cup B \subseteq A \cup \overline{B} \subseteq X$ and as the last union is disjoint: $A = X\setminus \overline{B}$ which is open, as the complement of a closed set. By symmetry, $B$ is also open (as $B = X \setminus \overline{A}$). And as $A$ and $B$ are mutual complements (as $\emptyset = A \cap \overline{B} \supseteq A \cap B$ etc.), $A$ open means $B$ is closed and vice versa.

On the other hand if we write $X = A \cup B$, $A \cap B = \emptyset$, where either both sets are closed or both sets are open, then $A$ and $B$ are separated in the sense above: (if both open) $A = X\setminus B$ and as the latter set is closed (as the complement of $B$) $\overline{A} \subseteq X \setminus B$ and so $\overline{A} \cap B = \emptyset$ and symmetrically $A \cap \overline{B} = \emptyset$. If both are closed, $A = \overline{A}$ and likewise for $B$ so then being separated is trivial from disjointness.

So it's equivalent for a partition $\{A,B\}$ of $X$ to be both open, both closed or separated, and all can be used to define a disconnected space.