I looked at previous answers to this question but they appear to take a slightly different approach that I couldn't comprehend.

$C$ is connected.

We would like to prove that $\bar{C} = C \cup C'$ is connected.

My attempt:

Suppose $\bar{C}$ is disconnected. Then $\exists$ open, disjoint subsets of $(X,d)$ $U$ and $V$, such that $\bar{C}$ is entirely contained in $U \cup V$ and $\bar{C} \cap U \neq \emptyset$ and $\bar{C} \cap V \neq \emptyset$.

That is to say, $C \cup C'$ is entirely contained in $U \cup V$, where $C'$ is the set of limit points of $C$. Equivalently, $C$ must be contained in $U \cup V$. Therefore, $\exists$ two open disjoint subsets such that $C$ is contained in their union. But this is a contradiction to $C$ being connected. Hence, $\bar{C}$ is connected.

Is this proof adequate?

  • $\begingroup$ You went too quickly at the end. While $C$ must be contained in $U\cup V$, that is not enough to prove $C$ is not connected. E.g. you haven't ruled out the possibility that $U$ contains $C$ and $V$ contains $C'$. $\endgroup$ – hardmath Mar 16 '15 at 0:48

You have to argue that $C \cap U \ne \emptyset$ and $C \cap V \ne \emptyset$. Maybe this is clear enough. (Any $x \in \overline{C} \cap U$ is either in $C \cap U$, or it's a boundary point of $C$, which also forces $U$ to intersect $C$)

Another way to look at this is the equivalent definition that $C$ is connected if and only if every continuous map $$f : C \rightarrow \{0,1\} \; \; (\mathrm{discrete} \; \mathrm{topology})$$ is constant, and a function $f : \overline{C} \rightarrow \{0,1\}$ is constant if and only if $f|_C$ is constant.


I think you might have some of the definitions mixed up. A set $C$ of a metric space is disconnected if there exist nonempty sets $A,B \subset X$ such that $C=A\cup B$, $A \cap \overline B=\overline A \cap B = \emptyset$ (i.e. $A$ and $B$ are separated). This is equivalent to another definition of connectedness: $C$ is disconnected if there exist nonempty sets $E,F \subset X$ open relative to $C$ such that $C=E \cup F$ and $E \cap F=\emptyset$.

I'll prove that the closure of a connected set $C$ is connected using the first definition. Suppose $\overline C$ is not connected, then there exist nonempty sets $A,B \subset X$ such that $\overline C=A\cup B$, $A \cap \overline B=\overline A \cap B = \emptyset$. Since $C \subset \overline C$ we get $C=C \cap \overline C=C\cap (A \cup B)=(C\cap A)\cup (C \cap B)$. Put $G=C \cap A$ and $H=C \cap B$.

We need to show that $G$ and $H$ are nonempty. If $G$ were empty we would have $C \subset A^c$. Take $x \in C$, then $x \in \overline C=A\cup B$, and since $x \not \in A$ we get $x \in B$. But then $C \subset B$, which gives $\overline C \subset \overline B$, hence $A \cap \overline B \supset A \cap \overline C =A \cap \overline{(A \cup B)}=A\cap (\overline A \cup \overline B)=(A \cap \overline A) \cup (A \cap \overline B)=A \neq \emptyset$, which contradicts that $A$ and $B$ are separated. Hence $G$ is nonempty. The same reasoning shows that $H$ is nonempty.

Also note that: $G \cap \overline H\subset (C \cap A) \cap (\overline C \cap \overline B)\subset (\overline C \cap A) \cap (\overline C \cap \overline B)= \overline C \cap (A \cap \overline B)=\emptyset$. The same reasoning shows that $\overline G \cap H = \emptyset$. Hence $G$ and $H$ are separated. But then $C$ is not connected, which is absurd.


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