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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 $overline{A}cap B= emptyset=Acapoverline{B}$. $E$ being connected, we know that $Acup B neq E$ so there exist $p in overline{E} backslash E$. We also know that $overline{E}=Ecup E'$ (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 $E''={p|pin E', pnot in E}$. We find then that $E'' subset A$ or $E'' subset B$. Say $E'' subset A$. Then $E subset B$ and we see that $Acapoverline{B}neq emptyset$ 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

Suppose that $E$ is connected. Let $A,Bsubseteq X$ be separated sets (that is, $overline{A}cap B=Acapoverline{B}=varnothing$) such that $overline{E}=Acup B$, and suppose that $Aneqvarnothing$. Let us prove that $B=varnothing$.

Let $ain A$. Since $Acap overline{B}=varnothing$, there exists a neighborhood $U$ of $a$ such that $Ucap B=varnothing$. Since $ainoverline{E}$, then there exists some point $xin Ecap U$, so $xnotin B$, hence $xin Ecap A$. Therefore, $Ecap Aneqvarnothing$.

Notice that $E=(Acap E)cup (Bcap E)$, and $Acap E$ and $Bcap E$ are obviously separated. As $Acap Eneqvarnothing$, from the previous paragraph, and $E$ is connected, then $Bcap E=varnothing$.

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

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

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

PS: As $Esubseteq Acup B$ and $Ecap B=varnothing$, then $Esubseteq A$, so $overline{E}subseteqoverline{A}$. It follows that
$$B=Bcapoverline{E}subseteq Bcapoverline{A}=varnothing.$$

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 $Acap E$ and $Bcap E$ is a separation of $E$(trivial to proof).

$implies$ $overline{(Acap E)}cap(Bcap E)=emptyset$ $implies (bar{A}capbar{E})cap B cap E=emptyset (because bar{X}cap bar{Y}subset overline{Xcap Y} )implies bar{A}cap B cap bar{E}=emptyset implies Acap Bcapbar{ E}=emptyset$

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

$Acap Bcap bar{E}=emptyset$ says if $xin bar{E}$ and also $xin A$ then $xnotin B$ (Similarly, $xin bar{E}$ and also $xin B$ then $xnotin A$.

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 $X=Acup B$ and $Acap B=emptyset$

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