Monday, December 23, 2024

Continuous image of connected set is connected

Table of Contents

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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