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This problem is from Rudin. I am trying to Prove that the closure of a connected set is always connected. Here is my proof.

Let $E$ be a connected set in a space $X$. Suppose to the contrary that the closure of $E$, $\overline{E}$ is not connected. Then there exist two sets $A$ and $B$ such that $\overline{E}=A \cup B$ and $\overline{A}\cap B= \emptyset=A\cap\overline{B}$. $E$ being connected, we know that $A\cup B \neq E$ so there exist $p \in \overline{E} \backslash E$. We also know that $\overline{E}=E\cup E'$ (where $E'$ is the set of the limit points of $E$) so $p$ must be a limit point of $E$ (but not in $E$). Taking the set of all such $p$ we obtain the set $E''=\{p|p\in E', p\not \in E\}$. We find then that $E'' \subset A$ or $E'' \subset B$. Say $E'' \subset A$. Then $E \subset B$ and we see that $A\cap\overline{B}\neq \emptyset$ which is a contradiction. So $\overline{E}$ must be connected.

Can anyone help me critique this proof? I feel uneasy about it but I don't know exactly what is wrong with it.

Thank-you

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    $\begingroup$ Why does $E''\subset A$ or $E''\subset B$? $\endgroup$
    – David P
    Feb 16 '14 at 3:05
  • $\begingroup$ My reasoning was that since $\overline{E}=A\cup B$ and also that $\overline{E}=E\cup E'$, and since $E'' \subset E'$, then $E''$ must be a subset of one of the sets that separate $\overline{E}$ (because certainly $E'' \cup E = \overline{E}$ $\endgroup$
    – illysial
    Feb 16 '14 at 3:12
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    $\begingroup$ The problem (I believe) with your proof is that you never use that $E$ is connected. You say it to imply $E$ does not contain its limit points, but that isn't using the property of connectedness. Take for example $E$ as the disjoint union of two open balls, forgetting your statement of the assumption that $E$ is connected. Then everything you write up until "$E''\subset A$ or $E''\subset B$" is valid, but breaks at this point. $\endgroup$
    – David P
    Feb 16 '14 at 3:28
  • $\begingroup$ I also felt awkward about not using that condition more...but I see your point I think. $\endgroup$
    – illysial
    Feb 16 '14 at 3:44
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Suppose that $E$ is connected. Let $A,B\subseteq X$ be separated sets (that is, $\overline{A}\cap B=A\cap\overline{B}=\varnothing$) such that $\overline{E}=A\cup B$, and suppose that $A\neq\varnothing$. Let us prove that $B=\varnothing$.

Let $a\in A$. Since $A\cap \overline{B}=\varnothing$, there exists a neighborhood $U$ of $a$ such that $U\cap B=\varnothing$. Since $a\in\overline{E}$, then there exists some point $x\in E\cap U$, so $x\not\in B$, hence $x\in E\cap A$. Therefore, $E\cap A\neq\varnothing$.

Notice that $E=(A\cap E)\cup (B\cap E)$, and $A\cap E$ and $B\cap E$ are obviously separated. As $A\cap E\neq\varnothing$, from the previous paragraph, and $E$ is connected, then $B\cap E=\varnothing$.

(See PS below for an alternative end to the proof without the argument by contradiction)

Finally, suppose, in order to obtain a contradiction, that $B\neq\varnothing$, and take $b\in B$. By the same arguments as those used in the second paragraph above, switching $A$ and $B$ and $a$ by $b$, we would conclude that $B\cap E\neq\varnothing$, contradicting what we have just proved.

Therefore, $B=\varnothing$. This proves that $\overline{E}$ is connected.


PS: As $E\subseteq A\cup B$ and $E\cap B=\varnothing$, then $E\subseteq A$, so $\overline{E}\subseteq\overline{A}$. It follows that $$B=B\cap\overline{E}\subseteq B\cap\overline{A}=\varnothing.$$

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    $\begingroup$ I dont follow $U\cap\bar{E}=\varnothing$. Since $a\in A$ and $U$ is a neighborhood of $a$, $a\in U$. When I say neighborhood, I would mean $U\subset A$ and $U$ is open. Using this definition and $A\cup B=\bar{E}$, we have $U\cap\bar{E}=(U\cap A)\cup(U\cap B)=U$, and since $A,U$ are nonempty, we cant have $U\cap\bar{E}=\varnothing$. If $U$ is a open set such that $A\subset U$ and $a\in A$, $U\cap\bar{E} = A$ which again is nonempty since $a\in A$. $\endgroup$
    – dustin
    Mar 17 '15 at 19:21
  • $\begingroup$ Is it the similar proof for "Closure of a connected subset is connected"? $\endgroup$
    – S786
    Apr 23 '15 at 11:50
  • $\begingroup$ @Luiz Corderio I also don't understand the $U\cap\bar{E}=\emptyset$. Can you please explain. $\endgroup$
    – Babai
    Nov 8 '15 at 14:34
  • $\begingroup$ @Babai This was a typo. I meant $U\cap\overline{B}=\varnothing$. $\endgroup$ Nov 9 '15 at 16:20
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    $\begingroup$ @davidharun This proof is for $\color{red}{all}$ general topological spaces which of course include $\mathbb R^k$. $\endgroup$
    – Sam Wong
    Apr 27 at 5:06
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I believe this can be made even more concise: Suppose $\overline{E}=A\cup B$ for disjoint, nonempty, and open $A,B$.

$E$ connected and $E=(A\cap E)\cup (B\cap E)$, so wlog, $A\cap E=E$. Then $B$ is an open set containing a limit point of $E$, and so it must intersect $E\subseteq A$ nontrivially - contradiction, as $A\cap B=\emptyset$.

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  • $\begingroup$ why does B containing a limit point means that it does intersect E nontrivially? $\endgroup$
    – user23657
    Apr 22 at 12:09
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    $\begingroup$ $x$ being a limit point of $A$ means (by definition) that for any open $U$ containing $x$, $A\cap U\neq \emptyset$. Hence, if there exists such an $x$ in $B$, then, since $B$ is open (by assumption), there exists an open neighbourhood $U$ in $B$ that contains $x$. But then $A\cap U\neq \emptyset$ as per the definition above, which means that $A\cap B\neq \emptyset$ because $B$ contains $U$. $\endgroup$
    – mss
    Apr 22 at 17:46
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There is only one part which might not have been explained in detail **

$ E''\subset A$ or $E''\subset B$

**

$A$ and B are separation of $\bar{E}$ implies $A\cap E$ and $B\cap E$ is a separation of $E$(trivial to proof).

$\implies$ $\overline{(A\cap E)}\cap(B\cap E)=\emptyset$ $\implies (\bar{A}\cap\bar{E})\cap B \cap E=\emptyset (\because \bar{X}\cap \bar{Y}\subset \overline{X\cap Y} )\implies \bar{A}\cap B \cap \bar{E}=\emptyset \implies A\cap B\cap\bar{ E}=\emptyset$

(I kept using the fact: $C\cap D=\emptyset $ and $C'\subset C$ then $C'\cap D=\emptyset$)

$A\cap B\cap \bar{E}=\emptyset$ says if $x\in \bar{E}$ and also $x\in A$ then $x\not\in B$ (Similarly, $x\in \bar{E}$ and also $x\in B$ then $x\not\in A$.

Therefore, $E''\subset A$ or $E''\subset B$

It is important to keep using the equivalent definitions of connectedness:

A topological space $X$ is disconnected if

Definition 1: there are two non-empty open sets $A$ and $B$ such that $X=A\cup B$ and $A\cap B=\emptyset$

Definition 2: there are two subsets $A$ and $B$ such that $X=A\cup B,$ $\bar{A}\cap B=\emptyset$ and $A\cap \bar{B}=\emptyset$

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Yes.

For any topological space $X$, let $E$ be the connected set . To prove that $E$ closure is connected, suppose $\mathrm{Cl}(E)$ is disconnected. Then $\exists$ at least two non-empty open sets say $H$ and $K$ in $\mathrm{Cl}(E)$ such that $\mathrm{Cl}(E)= H\cup K$.

Since $H$ and $K$ are open in $\mathrm{Cl}(E)$ and $E$ contained in $\mathrm{Cl}(E)$, so $H\cap E$ and $K \cap E$ are non-empty disjoint open sets in $E$ such that :

$$E= (H \cap E) \cup (K \cap E)$$

Which gives us that $E$ is disconnected, which is a contradiction to the fact that $E$ is connected. Thus, our supposition is wrong. Hence, $\mathrm{Cl}(E)$ is connected.

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  • $\begingroup$ You might wanna use Mathjax to format your answer. It would make your answer easier to read.. $\endgroup$
    – sai-kartik
    Sep 24 '20 at 12:05
  • $\begingroup$ As an example, I have formatted it for you this time. Please go through your answer again to make sure no content from the original has changed. $\endgroup$
    – sai-kartik
    Sep 24 '20 at 12:16

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