# 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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