There is a theorem saying that every open set in $R$ can be written as a countable union of disjoint open intervals. And we can also show any open set in $R^n$ can be written as the union of countable many bounded open intervals (see here).

I understand the proofs of the above two statements, but it feels strange that $R^n$ does not share the same property as $R$. Is there any explanation that why an open set in $R^n$ might not be written as countable union of disjoint open intervals?

  • 2
    $\begingroup$ What do you mean by open intervals in $\mathbb R^n$ when $n>1$? $\endgroup$
    – user99914
    Jun 28, 2015 at 3:25
  • $\begingroup$ FYI, a much better answer was given by Noah Schweber in response to a "duplicate" version of this question: math.stackexchange.com/a/1900272/169852 $\endgroup$
    – user169852
    Aug 22, 2016 at 23:10

2 Answers 2


Yes. $\:$ For integers $n$ that are greater than $1\hspace{-0.02 in}$, $\mathbf{R}^n$'s usual topology is not an order topology.
Specifically, $\mathbf{R}^n$ will have no cut points, whereas every point of $\mathbf{R}$ is a cut point.


$\Bbb R$ has many topological properties that $\Bbb R^2$, for example, doesn't share. Another example is $\Bbb R\setminus\{0\}$ is not connected, while $\Bbb R^2\setminus\{(0,0)\}$ is connencted.

  • $\begingroup$ Hi, sorry if it's a trivial question but I didn't really understand how connectedness plays the role. Can you please explain a bit more? $\endgroup$ Jul 7, 2022 at 23:07

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