The formulation of the problem states:

Show that $\mathbb{R}$ is not a countable, disjoint union of proper, closed sets. Can this be generalized to $\mathbb{R}^n$?

There are many questions very similar to this, and they all use the Baire category theorem. can open set in $\mathbb{R}^n$ be written as the union of countable disjoint closed sets?, Is $[0,1]$ a countable disjoint union of closed sets?, and Question about nowhere dense condition in Baire category theorem application are some (I posted the last question, but it was more proof-completion/verification/explanation oriented, so I'm posing a direct question this time). The second question is the only one with a complete argument; unfortunately, it uses compactness, which we don't have in the case of $\mathbb{R}^n$. The first question is a bit more general, as it pertains to every open subset of $\mathbb{R}^n$, and the answer to the first question outlines the idea of a sufficient proof for my problem, but doesn't fill in the details, which I'm having trouble filling in. The third question has some blanks filled in (by me; the reason I'm posting this question separately at all is because I suspect that there might be a different strategy than the one I attempted in the previous question), but one key element is missing.

I'd appreciate any new (or "old", but more complete than mine) idea towards solving this problem.

  • $\begingroup$ What is a "proper closed set"? Do you mean "closed proper subset"? Or what? $\endgroup$ – Rob Arthan Sep 1 '17 at 22:30
  • $\begingroup$ A set $A \subset \mathbb{R}^n$ which is closed, i.e. $\mathbb{R} ^n\setminus A$ is open in the standard topology and which is not equal to $\mathbb{R}^n$. $\endgroup$ – Matija Sreckovic Sep 1 '17 at 22:43
  • $\begingroup$ I knew what you meant, but it isn't standard terminology. $\endgroup$ – Rob Arthan Sep 1 '17 at 22:57
  • $\begingroup$ Could you tell me a better way to phrase this so as to avoid confusion? I'll edit it immediately. $\endgroup$ – Matija Sreckovic Sep 1 '17 at 23:45

Here's a fun trick:

First, note that the usual argument for $[0, 1]$ shows a slightly stronger result: that we can't write $[0, 1]$ as the disjoint union of countably many closed sets, at least two of which are nonempty. (We certainly can't do better than $2$ - take $C_0=[0, 1], C_{i+1}=\emptyset$).

Now we'll show the same for $\mathbb{R}^n$. Suppose $\mathbb{R}^n$ can be written as a disjoint union of countably many closed sets $C_i$ ($i\in\mathbb{N}$), at least two of which are nonempty; say, $C_0, C_1\not=\emptyset$. Pick $x\in C_0, y\in C_1$, and let $L\subset \mathbb{R}^n$ be the line segment joining $x$ and $y$. Note that $L$ with the subspace topology is homeomorphic to $[0, 1]$. Now we have that $L$ is the disjoint union of the closed (with respect to the subspace topology on $L$) sets $D_i$, where $D_i=C_i\cap L$, and at least two of the $D_i$s are nonempty (namely, we know that at least $D_0, D_1\not=\emptyset$). So the argument that $[0, 1]$ can't be written as the disjoint union of countably many closed sets, at least two of which are nonempty, kicks in, and we're done.

Note that this argument applies to any path-connected space.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.