0
$\begingroup$

i have two disjoint connected sets $A,B$ such that $\overline{A}\cap B\neq \emptyset$, how to prove that $A\cup B$ is connected? "Using the continuous function)

let $f: A\cup B\to \{0,1\}$ a continuous map, and we have to prove that it is a constant map.

let $a,b\in A\cup B$ if $a,b\in A$ then $f(a)=f(b)=cst$ because A is connected, and the same thing when $a,b\in B$

now if $a\in A$ and $b\in B$ how to prove that $f(a)=f(b)$?

how to use in this case the fact that $\overline{A}\cap B\neq \emptyset$ ?

$\endgroup$
2
$\begingroup$

Let $p\in \overline A\cap B$. By continuity of $f$, the set $f^{-1}(f(p))$ is an open neighbourhood of $c$, hence intersects $A$. Hence there are some points in $q\in A$ with $f(q)=f(p)$. As $f|_A$ is constant, $f(x)=f(p)$ for all $x\in A$. As $f|_B$ is constant, $f(x)=f(q)$ for all $x\in B$. As $f(p)=f(q)$, we conclude that $f$ is constant.

$\endgroup$
  • $\begingroup$ a neighbourhood of p ? h $\endgroup$ – Poline Sandra Sep 11 '18 at 20:10
  • $\begingroup$ can you tel me what is c? $\endgroup$ – Poline Sandra Sep 11 '18 at 20:17
  • $\begingroup$ where we use the intersection $f^{-1}(f(p))\cap A\neq\emptyset$ please answer me $\endgroup$ – Poline Sandra Sep 12 '18 at 17:52
0
$\begingroup$

Note that by connectedness of $A$ and $B$, and by the fact that restrictions are continuous, the restrictions $f_A$ and $f_B$ are both constant. If $f_A = f_B$, we are done.

Otherwise, without loss of generality suppose that $f_A \equiv 1$ and $f_B \equiv 0$. This is absurd, because taking $x \in \bar{A} \cap B$ guarantees a net in $A$, $(x_\alpha)_{\alpha}$ so that $x_\alpha \to x$, which by continuity would imply

$$ 1 = \lim_\alpha f_A(x_\alpha) = \lim_\alpha f(x_\alpha) = f(x) = 0 $$

since $x \in B$ and $x_\alpha \in A$ for any $\alpha$.

$\endgroup$
0
$\begingroup$

First note that if $f: A \cup B \to \{0,1\}$ is continuous so is $f|A$, and as $A$ is connected $f|A$ has constant value $i_A \in \{0,1\}$ so that $f(x) = i_A$ for all $x \in A$.

Similarly $\exists i_B \in \{0,1\}$ such that for all $x \in B$: $f(x) = i_B$, or $f[B] = \{i_B\}$.

Let $p \in \overline{B} \cap A$. Then by continuity of $f$: $f(p) \in f[\overline{B}] \subseteq \overline{f[B]} = \overline{\{i_B\}} = \{i_B\}$ as $\{0,1\}$ is discrete. So $f(p) = i_B$ but $p \in A$ gives $f(p) =i_A$ so that $i_A = i_B$ and thus $f$ is constant on $A \cup B$, and $A \cup B$ is connected.

$\endgroup$

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.