# Continuous Image of a Connected set

Our definition for disconnectedness is below.

$$D\subseteq X$$ where $$X$$ is a metric space is a disconnected set if followings satisfy.

1. There exists two open sets $$U_1,U_2$$ such that $$D\subseteq U_1\cup U_2$$

2. $$U_1\cap U_2=\emptyset$$

3. $$U_1\cap D\neq \emptyset$$ and $$D\cap U_2 \neq \emptyset$$

Let $$f:X\to Y$$ is continuous where $$(X,d_1)$$ and $$(Y,d_2)$$ are metric spaces and $$A\subseteq X$$ is connected. Show that $$f(A)$$ is connected.

I was trying to show via contrapositive. Let $$f(A)$$ is disconnected. Therefore,

1. There exists two open sets $$U_1,U_2$$ such that $$f(A)\subseteq U_1\cup U_2$$
2. $$U_1\cap U_2=\emptyset$$
3. $$U_1\cap f(A)\neq \emptyset$$ and $$f(A)\cap U_2 \neq \emptyset$$

I know $$A\subseteq f^{-1}(U_1) \cup f^{-1}(U_2)$$ but I couldn't show $$A\cap f^{-1}(U_1) \neq \emptyset$$ and $$A\cap f^{-1}(U_2) \neq \emptyset$$ only for contradicting with connectedness of $$A$$. How can I show it?

Thanks for any help.

• I don't quite agree with this definition of disconnected set, because I think (for good reason) that (ii) should be substituted by $U_1\cap U_2\cap D=\emptyset$.
– user239203
Dec 19 '19 at 16:05
• @Gae.S. Thank you. In this case how can I show $f^{-1}(U_1) \cap f^{-1}(U_2) \cap A \neq \emptyset$ ? Dec 19 '19 at 16:31
• If $x \in f^{-1}[U_1] \cap f^{-1}[U_2] \cap A$, then $f(x) \in A \cap U_1 \cap U_2$ by definitions, and this is what we know does not happen. Dec 19 '19 at 17:37

Your hypothesis (3) that $$U_1 \cap f(A) \neq \emptyset$$ means for some $$x \in A$$ we have $$f(x) \in U_1$$. Another way of writing this is that $$x \in f^{-1}(U_1)$$. In other words, we have $$x \in A \cap f^{-1}(U_1)$$.
since $$U_1 \cap f^{-1}(A) \not = \emptyset$$ there is some $$y \in U_1 \cap f^{-1}(A)$$.
So $$f^{-1}(y) \in f^{-1}(U_1) \cap A$$. which implies $$f^{-1}(U_1) \cap A \not= \emptyset$$
If $$U_1, U_2$$, disconnect $$f(A)$$ then $$f^{-1}(U_1), f^{-1}(U_2)$$ disconnect $$A$$.
As to your question, $$3)\implies A\cap f^{-1}(U_i)\ne\emptyset,i=1,2$$