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Theorem. If a set $E \subset \mathbb{R}$ is disconnected, then there exist $x, y \in E$, and some $z \in \mathbb{R} \setminus E$ with $x < z < y$.

I want to prove this theorem using the following definition of disconnectedness:

Definition. A subset $E$ of a metric space $X$ is disconnected if there exist $U, V \subset X$ such that:

  1. $U$ and $V$ are open relative to $X$ (meaning every point of $U / V$ has some neighborhood which is completely contained in $U / V$)
  2. $E \subset U \cup V$
  3. $E \cap U \neq \emptyset$ and $E \cap V \neq \emptyset$
  4. $E \cap U \cap V = \emptyset$

Here's what I have so far:

There must be some $x \in E \cap U$, and some $y \in E \cap V$ (since they are nonempty), and we assume without loss of generality that $x < y$. Now let $z = $???

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    Tricky problem-a good one for a take home problem set in either analysis or point-set topology.2012-03-05

1 Answers 1

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First, for simplicitly, define U'=E\cap U and V'=E\cap V.

Set w=\sup( [x,y] \cap U') and then v=\inf( [w,y] \cap V').

Finally take $z=\frac{w+v}{2}$. We must prove that that this $z$ can be in neither U' nor V'.

If $v\ne w$, this is immediate, because then the entire interval $(w,v)$ must be disjoint from both U' and V'.

On the other hand, if $v=w$, then $z=v=w$, and every neighborhood of $z$ must contain both points from U' and points from V'. Because U' is closed in $E$ and U'\cap V'=\varnothing, this means that $z$ cannot be in U', and similarly z\not\in V'.

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    I mean $[w,y]$ as written. Notice that later in the argument I depend on $v \ge w$ -- otherwise $(w,v)$ in the $w\ne v$ case wouldn't be an interval.2012-03-05