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A metric space (X, d) is connected iff every continuous function is constant

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.
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