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**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__: â€‹\( \exists \)â€‹Â a non-empty proper subset of X which is both open & closed.

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

Since, â€‹\( A\neq\phi \)â€‹,â€‹\( B\neq\phi \)â€‹ and â€‹\( X=A\cup B \)â€‹, â€‹\( A\cap B=\phi \)â€‹,

â€‹\( \implies A=X\backslash B \)â€‹.

â€‹\( \therefore \)â€‹Â A is non-empty proper subset of X.

Also, â€‹\( B=(\bar{A})^C \)â€‹, â€‹\( A=(\bar{B})^C \)â€‹.

Clearly, A and B are open subset of X.

(Since, complement of closed set is open)

â€‹\( \therefore \)â€‹ â€‹\( A=X\setminus B \)â€‹ is closed subset of X.

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

i.e. â€‹\( \exists \)â€‹ 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 â€‹\( B=X\setminus A \)â€‹

â€‹\( \therefore B\neq\phi \)â€‹, â€‹\( X=A\cup B \)â€‹ and â€‹\( A\cap B=\phi \)â€‹

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

â€‹\( \therefore \bar{A}=A \)â€‹ & â€‹\( \bar{B}=B \)â€‹

â€‹\( \therefore \bar{A}\cap B=A\cap B=\phi \)â€‹Â

Â Â â€‹\( A\cap \bar{B}=A\cap B=\phi \)â€‹

Thus, â€‹\( X=A\cup B \)â€‹, â€‹\( A\neq\phi \)â€‹, â€‹\( B\neq\phi \)â€‹, â€‹\( \bar{A}\cap B=\phi \)â€‹, â€‹\( \bar{B}\cap A=\phi \)â€‹

â€‹\( \therefore \)â€‹Â X is disconnected.

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