Let $E$ be a subset of a metric space $X$, and say $E$ is connected if and only if it is not disconnected. I have the following two definitions of disconnected:

I) $E$ is disconnected if there exist two nonempty subsets $A$, $B$ of $X$ such that:

  1. $E = A \cup B$
  2. $A \cap \overline{B} = \emptyset$ and $\overline{A} \cap B = \emptyset$

II) $E$ is disconnected if there exist two open (relative to the metric space $X$) subsets $A$, $B$ of $X$ such that:

  1. $E \subset A \cup B$
  2. $E \cap A \neq \emptyset$ and $E \cap B \neq \emptyset$
  3. $E \cap A \cap B = \emptyset$

Is there a concise way to prove that these definitions are equivalent?

Here are the definitions I'm using: A neighborhood of radius $r > 0$ around a point $p \in X$ is the set $N_r(p) = \{q \in X \; | \; d(p, q) < r\}$. A limit point of a subset $E$ of $X$ is a point $p$ such that every neighborhood of $p$ contains at least one other point of $E$. A subset $E$ of $X$ is closed if it contains all of its limit points. The closure of $E$, written $\overline{E}$, is the set $E$ together with its limit points. A subset $E$ of $X$ is open if, for every point $p \in E$, there is some neighborhood of $p$ which is contained in $E$.

Assume we have already proved that every closure is closed, and that a set is closed if and only if its complement is open.


I$\implies$ II: Suppose we have such sets $A,B$. The sets we use are $E\setminus \bar{A}$ and $E\setminus \bar{B}$, which are open. Then $A\subseteq E\setminus \bar{B}$ and $B\subseteq E\setminus \bar{A}$, so $E\setminus \bar{A}\cup E\setminus \bar{B}=E$ thus 1 is satisfied. Neither $E\cap A$ nor $E\cap B$ are empty since $A,B\subset E$, so $E\setminus \bar{B}$ and $E\setminus \bar{A}$ are nonempty thus 2 is satisfied. Furthermore, $$E\cap(E\setminus \bar{A})\cap (E\setminus \bar{B})=E\setminus(\bar A\cup\bar B) = E\setminus E=\emptyset$$ thus 3 is satisfied.

II$\implies$I: Suppose we have such sets $A,B$. The sets we use are $E\cap A$ and $E\cap B$. Then $E=(A\cap E)\cup (B\cap E)$ so 1 is satisfied, and $(E\cap A)\subseteq E\setminus E\cap B$ so since $E\cap A$ is open in $E$ we have $(E\cap A)\subseteq E\setminus \overline{E\cap B}$ thus $(E\cap A)\cap \overline{E\cap B}=\emptyset$, and the other intersection is similar, so 2 is satisfied.

  • $\begingroup$ You know that $E \backslash \overline{A}$ is open relative to $E$ (since you are taking the complement of the closed set $\overline{A}$ in $E$), but how do you know that $E \backslash \overline{A}$ is also open relative to the embedding metric space $X$? $\endgroup$ – jamaicanworm Mar 1 '12 at 22:37
  • $\begingroup$ Maybe you can take $X \backslash \overline{A}$ instead? $\endgroup$ – jamaicanworm Mar 1 '12 at 22:50
  • 1
    $\begingroup$ @jamaicanworm Sure, you can take $X\setminus \bar A$. $\endgroup$ – Alex Becker Mar 2 '12 at 4:26

Assume firstly: If $E$ is disconnected as Definition-I sense,

There exists $A$ and $B$ some subsets of $X$, such that $$E= A\cup B\text{ and } A\cap \bar{B}=\phi,\bar{A}\cap B=\phi$$ These condition implies that $A$ and $B$ are open subset in $X$, as $A$ and $B$ are subset of $X$ and $\bar{A},\bar{B}$ are closed sets and complement of closed set is open.

Hence we have two non empty open set $A$ and $B$ which will satisfy condition-II (verify it). Hence $E$ is disconnected in sense of definition-II.

Conversely, if we have given condition of definition-II:

That is we have given $A$ and $B$ open set satisfying the definition-II, Take $A'= E\cap A$ and $B'= E\cap B$, $A'$ and $B'$ will satisfy the condition of definition-I (verify it).

Hence $E$ is disconnected in sense of definition-1.

  • 1
    $\begingroup$ Unrelated $\LaTeX$ note: $\overline$ looks better than \bar! $\endgroup$ – user21436 Mar 1 '12 at 7:23
  • $\begingroup$ @KannappanSampath, yeah i think so.. :) $\endgroup$ – zapkm Mar 1 '12 at 7:42
  • $\begingroup$ In the first part, how do you know $A$ and $B$ are open? All we know is that $\overline{A}$ and $\overline{B}$ are closed, but that says nothing about their complements... $\endgroup$ – jamaicanworm Mar 1 '12 at 15:08
  • $\begingroup$ @jamaicanworm, As $\bar{A}$ is closed and $B\cap \bar{A}=\phi$.... We know that complement of closed set is open... $\endgroup$ – zapkm Mar 1 '12 at 15:17

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