Friday, February 3, 2023
HomeConnectednessContinuous 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.

RELATED ARTICLES

### Dimension theorem of a quotient space: If W be a subspace of a finite dimensional vector space V over ​( mathbb{R} )​ then ​(...

$${}$$