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

Question

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

Answer 1

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 $A\cap B=\emptyset$.

Answer 2

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.

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

Answer 3

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

Answer 4

Theorem: $(X, \tau) $ be a topological space. $A\subset X$ connected, then $\overline{A}$ is also connected.

Proof:

A topological space $(Y, \tau') $ is connected iff every continuous map from $(Y, \tau') $ to a two-point discrete space is constant.

Let $f:\overline{A}\to (\{0, 1\},\tau_{\text{discrete}})$ be a continuous map.

Since $A$ is connected, the map $f|_{A} : A\to (\{0, 1\},\tau_{\text{discrete}})$ is constant.

WLOG suppose $f(A) =\{0\}$

Then $f=0$ on $A$ implies $f=0 $ on $\overline{A}$ (see here)

Implies $f(\overline{A})=\{0\}$

Hence the map $f:\overline{A}\to (\{0, 1\},\tau_{\text{discrete}})$ is $\text{constant}$.

Answer 5

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.

Answer 6

Firstly, regarding your question, I don't see why we have

$$\bar{E} = E \cup E'' = A \cup B \implies E'' \subset A \text{ or } E'' \subset B.$$

This assertion need not be true without the connectedness of $E$, let's consider this case in $\mathbb{R}$: $$\bar{E} = [0,1] \cup [2,3], \; E = (0,1) \cup (2,3), \; E'' = \{0, 1, 2, 3\}.$$

$\bar{E}$ is not connected, which satisfies our hypothesis. Now take $A = [0,1]$ and $B = [2,3]$ as subsets of $\bar{E}$.
$A$ and $B$ are nonempty, with $\bar{A} = [0,1]$, $\bar{B} = [2,3]$ and $\bar{A} \cap B = \bar{B} \cap A = \emptyset$. Thus, $A$ and $B$ are separated sets while $E'' \not\subset A$ and $E'' \not\subset B$. So, the simple fact that $\bar{E}$ is not connected (without knowing that $E$ is connected) cannot imply that

$$\bar{E} = A \cup B \implies E'' \subset A \text{ or } E'' \subset B.$$

I think there is also an issue with Babai's answer, which is

$$A \cap B \cap \bar{E} = \emptyset \implies (x \in \bar{E} \text{ and } x \in A \implies x \notin B, \; x \in \bar{E} \text{ and } x \in B \implies x \notin A).$$

I don't think this has any meaning (since $ A \cap B = \emptyset $, there is no need for $ \bar{E} $). Moreover, this does not imply that $ E'' \subset A $ or $ E'' \subset B $, because it’s possible to have limit points in both $ A $ and $ B $. If this is not the case, it must be proven.

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I will present my method in the following paragraphs, hoping that someone will correct me if there is anything wrong.

My method involves proving that

$$ \bar{E} \text{ is not connected } \implies E \text{ is not connected.} $$

If $\bar{E}$ is not connected, then

$$ \exists C,D \subset \bar{E}, C \ne \emptyset, D \ne \emptyset, \text{ such that } \\ C \cup D = \bar{E} = E \cup E'' \land C \cap \bar{D} = D \cap \bar{C} = \emptyset. \\(E'' = E' \setminus E, E' \text{ as the set of limit points of } E) $$

Since $E'' = E' \setminus E$, we have $E \cap E'' = \emptyset$. Therefore,

$$ C \cup D = \bar{E} = E \cup E'' \implies \left( C \setminus E'' \right) \cup \left( D \setminus E'' \right) = E. $$

Let $M = C \setminus E''$ and $N = D \setminus E''$, then $M \cup N = E$.

In order to prove that $E$ is not connected, it suffices to show that $M$ and $N$ are nonempty, separated sets.

We begin by proving that $M$ and $N$ are nonempty. That is, to prove

$$ E \cap M \ne \emptyset \land E \cap N \ne \emptyset. $$

Without loss of generality, suppose that $E \cap M = E \cap \left( C \setminus E'' \right) = \emptyset$. Then

$$ E \cap M = E \cap \left( C \setminus E'' \right) = \emptyset \quad \left( \text{and } \because E \cap E'' = \emptyset \right) \implies E \cap C = \emptyset, $$

and since $ C \cup D = \bar{E} = E \cup E'', $ we have $$ E \subset D \implies E'' \cap C = \emptyset \quad \left( \text{for otherwise } C \cap \bar{D} \ne \emptyset \right). $$

Thus, with $E \cap C = \emptyset$ previously obtained,

$$ E'' \cap C = E \cap C = \emptyset \quad \left( \text{whereas } C \subset \bar{E} = E'' \cup E \right) \implies C = \emptyset, $$

which contradicts the condition given by the hypothesis, i.e., $C$ and $D$ are nonempty.

So, neither $M$ nor $N$ is empty.

It remains to be proven that $M$ and $N$ are separated, and it will be done. Since $M = C \setminus E''$ and $N = D \setminus E''$, we have $M \cap N = \emptyset$.

we have $ C \cap \bar{D} = \emptyset$, so $$ \forall x,\quad x\in M \implies x \in C \setminus E'' \subset C \quad \implies x \notin \bar{D} \implies x \notin \bar{D} \setminus E'' \implies x \notin \bar{N}. $$

Thus, $M \cap \bar{N} = \emptyset $. And similarly, we can obtain $\bar{N} \cap M = \emptyset. $

Therefore, $M$ and $N$ are nonempty separated sets, and since $M \cup N = E$, il follows that $E$ is not connected. $\quad \square$