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HomeConnectednessContinuous image of connected set is connected

# Proof :

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

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