Continuous Image of a Connected set

Question

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.

Answer 1

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 $x \in A \cap f^{-1}(U_1)$.

Answer 2

For your third case

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 $f^{-1}(U_1) \cap A \not= \emptyset$

Answer 3

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 $