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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.

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  • $\begingroup$ 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
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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.

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