Prove that a proper subset $E$ of $\Bbb R^n$ is connected $\iff$ it contains exactly two relatively clopen sets.

I researched the meaning of "clopen set". And I reached the result that so as to for a set $A$ be clopen, the set $A$ need to be both closed and open.

I cannot do this proof. Please help me to do this. Thank you

  • $\begingroup$ Could you add your definition of connectedness as well? $\endgroup$ – Henno Brandsma Jul 26 '13 at 12:16

I'll use the term "open" in the following as "open for the subspace topology inherited by $E$".

$\Rightarrow$: Assume that $E$ is connected. This means that the union of two disjoint non-empty open subsets of $E$ is not $E$ itself. So let's have $X$ a clopen set of $E$. $E\setminus X$ is open, then, and $E = X \cup (E\setminus X)$. What can you conclude with the remark I made above?

$\Leftarrow$: you know that $\emptyset$ and $E$ are clopen sets, so the hypothesis implies that there are no other clopen set. What can you get from this?

  • $\begingroup$ For $\Rightarrow$, let $U=X$ and $V=E$\$X$ be relatively open subsets of $E$ then $E=U\cup V$ also, $U\cap V\not= \emptyset$ thus, E is not connected. $\endgroup$ – Bstr Jul 26 '13 at 12:30
  • $\begingroup$ For ⇒, let U=X and $V=E\$$X$ be relatively open subsets of E then $E=U\cup V$ also,$U\cap V\not= \emptyset$ thus, E is not connected. – $\endgroup$ – Bstr Jul 26 '13 at 12:38
  • $\begingroup$ For $\Rightarrow$, let $U=X$ and V= E\X be relatively open two subsets of E. $\endgroup$ – Bstr Jul 26 '13 at 12:40
  • $\begingroup$ Then, $E=U\cup V$ and $U \cap V \not= \emptyset$ $\endgroup$ – Bstr Jul 26 '13 at 12:42
  • $\begingroup$ Thus, E is not connected. $\endgroup$ – Bstr Jul 26 '13 at 12:42

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