Theorem:
A metric space (X, d) is connected iff every continuous function \( f:X\to {\lbrace 0, 1\rbrace} \) is constant.
Proof:
Let (X, d) be a connected metric space and suppose that there exists a continuous function \( f:X\to {\lbrace0, 1\rbrace} \).
Claim: f is constant.
Since, \( {\lbrace0, 1\rbrace} \) is finite w.r.t. usual metric, set \( {\lbrace 0\rbrace} \) and\( \lbrace 1 \rbrace \)are open and closed subsets of \( \lbrace 0, 1 \rbrace \),
\( \therefore f^{-1}\lbrace 0 \rbrace \) and \( f^{-1}\lbrace 1 \rbrace \) both are open and closed subset of X. (\( \because \) f is continuous)
\( f^{-1}{\lbrace 0 \rbrace} \cup f^{-1}{\lbrace 1\rbrace}=X \) and \( f^{-1}{\lbrace 0 \rbrace} \cap f^{-1}{\lbrace 1\rbrace}=\phi \).
Thus, X is union of disjoint sets.
But, by hypothesis,
X is connected and hence, either \( f^{-1}\lbrace 0 \rbrace=\phi \) or \( f^{-1}\lbrace 1 \rbrace=\phi \).
WLOG,
Suppose, \( f^{-1}\lbrace 0 \rbrace=\phi \).
\( \therefore f^{-1}\lbrace 1 \rbrace=X \)
\( \therefore f(X)=1 \)
i.e. f is constant.
Conversely,
Suppose that every continuous function \( f:X\to {\lbrace0, 1\rbrace} \) is constant.
Claim: Metric space (X, d) is connected.
Since, \( f:X\to {\lbrace0, 1\rbrace} \) is constant function,
either \( f(X)=\lbrace 0 \rbrace \) or \( f(X)=\lbrace 1 \rbrace \).
WLOG, suppose that .
Since, \( \lbrace 1 \rbrace \) is closed as well as open subset of \( {\lbrace0, 1\rbrace} \) w.r.t. usual metric and f is continuous function, \( f^{-1}\lbrace 1 \rbrace \) is closed as well as open.
But \( f^{-1}\lbrace 1 \rbrace=X \).
\( \therefore \) X is only nonempty set which is both open as well as closed in X.
\( \therefore \) X is connected.
