# Theorem :

## A metric space (X, d) is disconnected iff there exist a non-empty proper subset of X which is both open and closed.

# Proof :

Let (X, d) be a metric space and suppose that it is disconnected.

__Claim__: $\exists$ a non-empty proper subset of X which is both open & closed.

Since, X is disconnected by definition, $\exists$ non-empty sets A & B such that $X=A\cup B$, $\bar{A}\cap B=\phi$, $\bar{B}\cap A=\phi$.

Since, $A\neq\phi$,$B\neq\phi$ and $X=A\cup B$, $A\cap B=\phi$,

$\implies A=X\backslash B$.

$\therefore$ A is non-empty proper subset of X.

Also, $B=(\bar{A})^C$, $A=(\bar{B})^C$.

Clearly, A and B are open subset of X.

(Since, complement of closed set is open)

$\therefore A=X\setminus B$ is closed subset of X.

Thus, A is closed as well as open subset of X.

i.e. $\exists$ a non-empty proper set A of X which is both open and closed.

Conversely,

Suppose that metric space (X, d) has non-empty proper subset which is both open and closed.

__Claim__: X is disconnected.

Suppose A is non-empty proper subset of X which is both open and closed.

Let $B=X\setminus A$

$\therefore B\neq\phi$, $X=A\cup B$ and $A\cap B=\phi$

Since, A is closed and open subset of X, B is open as well as closed.

$\therefore \bar{A}=A$ & $\bar{B}=B$

$\therefore \bar{A}\cap B=A\cap B=\phi$

$A\cap \bar{B}=A\cap B=\phi$

Thus, $X=A\cup B$, $A\neq\phi$, $B\neq\phi$, $\bar{A}\cap B=\phi$, $\bar{B}\cap A=\phi$

$\therefore$ X is disconnected.

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