$A,B$ are connected subsets of a topological space $X$.

What I've tried:

As $B$ is connected, so is its closure $\overline{B}$. We have:

$\overline{B}\cap A\neq \emptyset$

$\overline{B} \cap B\neq \emptyset$

hence, $\overline{B}\cup A \cup B$ is connected. However, this isn't what I want.

I was thinking of trying to show that $A\cap \overline{B} \neq \emptyset$ implies $A \cap B \neq \emptyset$, though I don't think this would be true if $A$ is closed.

Any hints?

  • 1
    $\begingroup$ Let $A=[0,1]$ and $B=(1,2)$. Then $A \cap \bar{B} = \{1\} \neq \emptyset$ however $A \cap B = \emptyset$. So $A \cap \overline{B} \neq \emptyset$ does not imply $A \cap B \neq \emptyset$. $\endgroup$ – Hetebrij Sep 11 '15 at 17:22

Hint: Use that $X$ is connected $\iff$ the only continuous functions $f:X\to\{0,1\}$ are constant, where $\{0,1\}$ is endowed with the discrete topology.

Consider a continuous function $f :A \cup B \to \{0,1\}$ and restrict it to $A$ and $B$ and use the fact $A$ and $B$ are connected.Can you conclude from here i.e. can you show that $f$ is constant ?

  • $\begingroup$ $f:A\rightarrow {0,1}$ and $f:B\rightarrow {0,1}$ are both constant. Consider $x\in A\cap \overline{B}$. Whatever it maps to, every element in $A$ maps that same element. However, $x$ need not be in $B$, so how can I conclude that every element in $B$ maps to the same thing that every element in $A$ maps to? $\endgroup$ – man_in_green_shirt Sep 11 '15 at 17:43
  • $\begingroup$ Can you show that $f$ on $ \overline B$ is also constant?(Since $f$ is continuous) $\endgroup$ – Arpit Kansal Sep 11 '15 at 17:44
  • $\begingroup$ As $B$ is connected, so is $\overline{B}$. Hence, $f$ on $\overline {B}$ is constant and this proves the claim. Thanks! $\endgroup$ – man_in_green_shirt Sep 11 '15 at 18:09
  • $\begingroup$ However, I haven't used the continuity of $f$ to prove it. The proof I know that $B$ connected implies $\overline{B}$ connected doesn't use continuous functions. How would you show it using the continuity of $f$? $\endgroup$ – man_in_green_shirt Sep 11 '15 at 18:10
  • 1
    $\begingroup$ Well,note that initially $f$ is not defined on $ \overline B$ So You can't use connectedness of $ \overline B$.But for an element b in $ \overline B $ \B choose a sequence $b_n$ in $B$ which converges to $b$,and use continuity of $f$ on $B$ $\endgroup$ – Arpit Kansal Sep 11 '15 at 18:13

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.