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.
