# Alternative proof that the continuous image of a connected set is connected

I have written a proof of the proposition (that the continuous image of a connected set is connected) slightly different from the ones on here so far (of which I am aware there are many). Is the following correct?

I have a continuous function $$f:X \rightarrow Y$$, where $$X$$ and $$Y$$ are metric spaces such that $$X$$ is connected. I want to show that $$f(X)$$ is connected as a metric space, with the metric induced by $$Y$$.

Suppose, for a contradiction, that $$f(X)$$ is not connected. That is, there exists open sets $$A,B$$ such that $$f(X)=A\cup B$$ and $$A \cap B = \emptyset$$. Note that $$A,B\subseteq Y$$, and so, by the continuity of $$f:X \rightarrow Y$$, $$f^{-1}(A)$$ and $$f^{-1}(B)$$ are both open in $$X$$. Additionally, $$f^{-1}(A)\cap f^{-1}(B)= \emptyset$$ (since, otherwise, $$\exists x^*$$ such that $$x^*\in f^{-1}(A) \cap f^{-1}(B)$$ and so $$f(x^*)\in A \cap B$$, contradicting the assumed disjointedness of $$A$$ and $$B$$). Finally, we show that $$X=f^{-1}(A)\cup f^{-1}(B)$$.

$$w\in f^{-1}(A)\cup f^{-1}(B) \iff f(w) \in A \cup B = X \iff w\in X$$.

So $$X=f^{-1}(A)\cup f^{-1}(B)$$, and is not connected. A contradiction.

• This is correct, and never uses the fact that either $X$ or $Y$ is a metric space, so you've actually proved the proposition for topological spaces in general. Oct 18, 2022 at 0:17
• This looks like the “standard proof” to me. Oct 18, 2022 at 0:20
• A minor point: you seem to be misusing $:=$. That's usually used to introduce a definition; in contrast, if you're merely asserting equality of two preexisting objects, which is the case everywhere here, then you should use plain $=$. Oct 18, 2022 at 0:22
• $A$ and $B$ are open sets in $f(X),$ not $Y.$ You will need a little more care. Oct 18, 2022 at 0:52
• @RobertShore No, it is not correct. $A\cup B=f(X)$ means we are talking about open sets in $f(X),$ not in $Y.$ Oct 18, 2022 at 0:57

This is fine. I'll just make one small comment that it's totally unnecessary to phrase this as a proof by contradiction. This is a proof by contrapositive: you've shown that if $$f(X)$$ is not connected then $$X$$ is not connected.