# The closure of a connected set in a topological space is connected

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

• Why does $E''\subset A$ or $E''\subset B$? Feb 16, 2014 at 3:05
• 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 $E'' \cup E = \overline{E}$ Feb 16, 2014 at 3:12
• 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. Feb 16, 2014 at 3:28
• I also felt awkward about not using that condition more...but I see your point I think. Feb 16, 2014 at 3:44

## 4 Answers

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 $$B=B\cap\overline{E}\subseteq B\cap\overline{A}=\varnothing.$$

• Is it the similar proof for "Closure of a connected subset is connected"?
– S786
Apr 23, 2015 at 11:50
• In the proof made by Luiz Cordeiro above, I don't follow why $a∈\bar{E}\Rightarrow , \exists x∈E∩U$. Could anyone clarify?
– I000
May 25, 2018 at 20:28
• Is it because: $a \in \bar{E}=E \cup E^{'}$ where $E^{'}$ is the set with all limit points of $E$. If $a\in E$, $E\cup U\neq\emptyset$ since $a\in U$. If $a\in E^{'}$, by the definition of limit point, $\forall\varepsilon>0, \exists y \in N_{\varepsilon}(a)$ such that $a\neq y$ and $y \in E$. ?
– I000
May 25, 2018 at 20:52
• @I_. Yes, even though you're using notation for metric spaces. This is standard: $\overline{E}=\left\{x\in X:\forall U^{open}\ni x, U\cap E\neq\varnothing\right\}$. May 25, 2018 at 22:13
• @davidharun This proof is for $\color{red}{all}$ general topological spaces which of course include $\mathbb R^k$. Apr 27, 2021 at 5:06

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 $$A\cap B=\emptyset$$.

• why does B containing a limit point means that it does intersect E nontrivially? Apr 22, 2021 at 12:09
• $x$ 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 $U$.
– mss
Apr 22, 2021 at 17:46

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 $A\cap \bar{B}=\emptyset$

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.

• You might wanna use Mathjax to format your answer. It would make your answer easier to read.. Sep 24, 2020 at 12:05
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