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Continuous image of connected set is connected

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.

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