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Friday, June 9, 2023
HomeConnectednessA metric space (X, d) is disconnected iff there exists a non-empty...

# 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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