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Prove or disprove: The closure of a connected set in $\mathbb{R}$ is always connected.

Response: I don't really have a grasp on this conceptually, but here's an attempt. Proof: Let $X$ be a connected subset of $R$, and let $Z$ be the closure of $X$. Suppose (to get a contradiction) that $Z$ is disconnected. That means there are two nonempty open subsets of $R$ covering $Z$, call them $U$ and $V$, such that $U\cap Z$ and $V\cap Z$ are nonempty while $U\cap V\cap Z$ is empty.

Then $U$ and $V$ cover $X$ also. And since $U\cap V\cap X \subseteq U\cap V\cap Z$, then $U\cap V\cap X$ is empty. Pick points $u\in U\cap Z$ and $v\in V\cap Z$ (which exist since those sets are nonempty). That means $u$ is in the closure of $X$ and $U$ is an open neighborhood of $u$, so $U\cap X$ is nonempty. Similarly, $V\cap X$ is nonempty.

But then the previous paragraph shows that $X$ is disconnected, which is a contradiction, so $Z$ must have actually been connected.

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    $\begingroup$ Have a look here: at.yorku.ca/cgi-bin/… $\endgroup$ – Lucien Nov 8 '12 at 15:39
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    $\begingroup$ In the case of $\mathbb{R}$, a set is connected if and only if it is an interval. $\endgroup$ – Michael Greinecker Nov 11 '12 at 0:39
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You can get started simply with definitions. You will need the definition of connectedness, and you will need to understand that we are using the subspace topology to determine connectedness of subsets.

Suppose $S$ is a connected subset of $X$, and that $O$ and $U$ are disjoint open subsets of $X$ such that $O\cup U\supseteq \overline{S}$. We will show that one of $O$ or $U$ contains $\overline{S}$ and the other does not intersect $\overline{S}$, proving that $\overline{S}$ is connected.

Certainly $O\cup U\supseteq S$, therefore by connectness of $S$ we have that (...?) (Draw conclusions about where $\overline{S}$ must live.)

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Do you know any definition or property of connectivity? A good one is that if $X$ is connected then any continuous $f : X \to \{0, 1\}$ must be constant. Say $\forall x \in X$, $f(x) = 1$. How can you extend this to the closure of $X$ without destroying continuity? Note that this statement is not restricted to $\mathbb{R}$.

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I will prove this through contradiction. Let $\overline{S}$ be disconnected, such that there exists $A, B$ disjoint, closed sets which don't intersect $\overline{S}$. Let $S$ be connected, such that there exists $A, B$ disjoint, closed sets with $S\subseteq A \cup B$, where either $A$ or $B$ intersects $S$. Let $S\subseteq A$, then $\overline{S}\subseteq \overline{A} = A$, which is a contradiction to our statement that $\overline{S}$ is disconnected. Therefore, the closure of a connected set in $\mathbb{R}$ is always connected.

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Here is another proof by contradiction, using the definition of connectedness from Rudin's Principles of Mathematical Anaylsis.

Suppose $S$ is a connected set, where its closure $ \bar{S} $ is not connected. Therefore there exist two sets $A$ and $B$, such that $ \bar{A} \cap B = \bar{B} \cap A = \emptyset $ and $A\cup B = \bar{S} $. Define $ G:= A \cap S $ and $ H:= B\cap S $. Because $G$ is a subset of $A$ and $H$ is a subset of $B$, it is clear that $ \bar{G} \cap H = \bar{H} \cap G = \emptyset $. and $G\cup H = S$. So that $S$ is not connected, contrary to our first assumption. q.e.d.

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You can actually show that a connected subset in R must be singleton or an interval. In both the cases, closure is also connected.

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