Friday, February 3, 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: ​$$\exists$$​ 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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