Friday, February 3, 2023
HomeConnectednessA metric space (X, d) is connected iff every continuous function is...

# 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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