Theorem :
Continuous image of connected set is connected.
Proof :
Let \( (X, d) \) and \( (Y, d’) \) be any two metric spaces and \( f:X\to Y \) be continuous function.
Claim: f(X) is connected.
Suppose, if possible that f(X) is not connected.
So, there exists non-empty disjoint open subsets A and B such that \( f(X)=A\cup B, \bar{A}\cap B=\phi \) and \( \bar{B} \cap A=\phi \).
\( \therefore f^{-1}(f(X))=f^{-1}(A\cup B) \)
\( X=f^{-1}(A\cup B) \)
\( X=f^{-1}(A)\cup f^{-1}(B) \)
Also,
\( f^{-1}(A)\cap f^{-1}(B)=\phi \)
Since, A and B are open subsets of Y and \( f:X\to Y \) is continuous ,
\( f^{-1}(A) \) and \( f^{-1}(B) \) are open subsets of X.
\( X=f^{-1}(A)\cup f^{-1}(B) \)
\( f^{-1}(A)\cap f^{-1}(B)=\phi \)
\( \therefore \) X is disconnected, which is contradiction to X is connected.
\( \therefore \) Our assumption is wrong.
\( \therefore \) f(X) is connected.
Thus, continuous image of connected set is connected.
