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My text gives the definition that $E$ is disconnected if there exist disjoint open sets $A, B$ such that:

  1. $A \cap E$, $B \cap E$ are nonempty.
  2. $(A \cap E) \cup (B \cap E) = E$.

Then for the $(\implies)$ direction, the text offers the following as proof:

Suppose $E$ is an interval and $E$ is not connected. Then there exist disjoint open sets $A$, and $B$ with $A \cap E \ne \emptyset$, $B \cap E \ne \emptyset$, and $E \subset A \cup B$. Suppose $a \in A \cap E$ and $b \in B \cap E$. Inasmuch as $E$, an open set in $\mathbb R$ is a finite or countable union of open and disjoint intervals, there exist disjoint open intervals $I$ and $J$ such that $a \in I$ and $b \in J$. Suppose $a < b$ and $J = (t, s)$. Since $a < t < b$, $t \in E$. However, $t \notin A \cup B$, creating a contradiction.

Is the part with $E$ being a finite or countable union of open and disjoint intervals necessary? I did it in a much simpler way: Without loss of generality, call the open set on the left $A$, with $a \in A \cap E$, and call the open set on the right $B$, with $b \in B \cap E$. Now, set $c =$ sup$A$; $a < c < b \rightarrow c \in E$. Since $c$ is a boundary point of $A$ and $A^c$, $c \notin A$. But $c \notin B$ either. Otherwise, there is a $\delta$ such that $N(c, \delta) \subset B$, entailing that $A \cap B$ is nonempty, which is not possible. As a result, we have $c \in E$ and $c \notin E$, a contradiction.

Why did the book need to use the fact that $E$ is a union of open, disjoint intervals?

Ivo Terek
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Andy Tam
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  • There is no such thing as "left" and "right". The 2 sets can mixed up in anyway. Of course, your "proof" can still work too and won't require that theorem, but it needs to be fixed. – Gina Jul 12 '14 at 03:36
  • It seems to me that the text’s proof is not completely right. Disconnectedness of $E$ means that there are disjoint open subsets of $E$ with the specified property. So there would be open subsets $A$ and $B$ of $\mathbb R$ with $A\cap B\cap E=\emptyset$, not merely $A\cap B=\emptyset$. – Lubin Jul 12 '14 at 05:28

2 Answers2

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Your proposed proof isn't quite right. The problem is that it's not guaranteed that there is a "left" or "right" open set - this would be true if $A$ and $B$ were intervals, but they need not be. The open sets $A$ and $B$ could interlace.

So it's not actually guaranteed that $\sup A \in E$ or that $\sup B \in E$, precisely because of this issue: If we take smaller and smaller intervals in $A$ growing to the right, and likewise for $B$, then it's certainly possible to have

$$\sup A = \sup B = \sup E \notin E$$

Notice that this is fixed precisely by knowing that an open set can be decomposed into intervals; this actually does allow you to select a "left" interval and a "right" interval, and then study how they don't intersect.

  • So the author avoided my error by using the Hausdorff property. On second look, with the correction, my proof is the same as the one in the text. TY .... – Andy Tam Jul 14 '14 at 16:36
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The book needs the Hausdorff separation property -- that around each of two points there are disjoint open sets. This property is normally covered in Topology, but your book is importing the idea via $E$ being a finite or countable union of open and disjoint intervals so it can pick out two disjoint intervals as needed.

Eric Towers
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