4
$\begingroup$

This is a proof by contradiction, unsing the definition of connectedness from Rudin, Principles of Mathematical Anaylsis. Can someone please check if it is any good?

Suppose S is a connected set, where its closure $ \bar{S} $ is not connected. Therefore there exist two nonempty sets A and B, such that $ \bar{A} \cap B = \bar{B} \cap A = \emptyset $ and $ A \cup B = \bar{S}$. Define $ G:= A \cap S $ and $ H:= B \cap S $. Because G is a subset of A and H is a subset of B, it is clear that $ \bar{G} \cap H = \bar{H} \cap G = \emptyset $. and $A\cup B = S$. So that S is not connected, contrary to our first assumption. q.e.d.

Thank you

$\endgroup$
9
  • 1
    $\begingroup$ You seem to have changed your $S$ into an $E$ at the end there. $\endgroup$
    – user259242
    Nov 4 '16 at 11:57
  • 1
    $\begingroup$ $G$ and $H$ are the same set. This is probably not what you want. $\endgroup$
    – user259242
    Nov 4 '16 at 11:59
  • 1
    $\begingroup$ Shouldn't $A\cap\overline{S}$ and $B\cap\overline{S}$ be assumed to be nonempty, for the contradiction? $\endgroup$ Nov 4 '16 at 12:01
  • 1
    $\begingroup$ Something is not quite right. You have not used the assumption that $\overline{S}$ is not connected. $\endgroup$
    – user259242
    Nov 4 '16 at 12:02
  • 3
    $\begingroup$ The current version is correct. $\endgroup$ Nov 4 '16 at 14:59
3
$\begingroup$

The contradiction is that $G\cup H = S$ where G and H are separated, not $A\cup B=S$

$\endgroup$
3
  • 1
    $\begingroup$ Can you be sure that $G$ and $H$ are both not empty? $\endgroup$
    – 311411
    Apr 30 at 19:39
  • 1
    $\begingroup$ Yes, if $x \in S$ then $x \in \overline{S}$, and then $x \in A $ or $x \in B$. If $x \in A$ then $x \in A \cap S$, which implies that $A \cap S \neq \emptyset$. Now, since $A \cup B = \overline{S}$ with $A \cap B = \emptyset$, then $B = \overline{S} - A$. $B \cap S = (\overline{S} - A) \cap S = \overline{S} \cap A^{c} \cap S = S - A = \emptyset$ if and only if $S = A$, but $\overline{A} \cap B = \overline{S} \cap B = \emptyset$, which implies that $B = \emptyset$ due to $B \subset \overline{S}$, but $B$ is a non emptyset by definition, which is a contradiction. So, $B \cap S \neq 0$ too. $\endgroup$
    – DIEGO R.
    May 1 at 14:56
  • $\begingroup$ That works well, thank you. Another proof, organized slightly differently, is the one I wrote here: math.stackexchange.com/a/4122663/688046 $\endgroup$
    – 311411
    May 8 at 15:28

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.