0
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ I don't quite agree with this definition of disconnected set, because I think (for good reason) that (ii) should be substituted by $U_1\cap U_2\cap D=\emptyset$. $\endgroup$
    – user239203
    Dec 19 '19 at 16:05
  • $\begingroup$ @Gae.S. Thank you. In this case how can I show $f^{-1}(U_1) \cap f^{-1}(U_2) \cap A \neq \emptyset$ ? $\endgroup$
    – user519955
    Dec 19 '19 at 16:31
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 19 '19 at 17:37
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.