**Theorem** :

**Continuous image of connected set is connected.**

**Proof** :

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Friday, February 3, 2023

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.

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