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Show that $\mathbb{R}^n$ cannot be written as a countable union of proper subspaces

Ok so I know I have to use Baire's Cathegory Theorem here. And I've done the following, lets suppose on the contrary that $\mathbb{R}^{n}=\bigcup_{n\in \mathbb{N}}S_n$ Such that $\forall n \in \mathbb{N}, S_n$ is a proper subspace. Now if I show that for every $n \in \mathbb{N}$, $S_{n}$ is closed and $int(S_{n})=\emptyset$ then Baire would imply $int(\mathbb{R}^n)=\emptyset$ which is a contradiction.

Now I was able to prove that for every $n \in \mathbb{N}$, $int(S_{n})=\emptyset$.

However, when I am trying to prove that $S_{n}$ is closed, I have the following inconvenient. $S_{n}$ is closed iff $\mathbb{R}^n-S_{n}$ is open. Therefore, if $\mathbb{R}^n-S_{n}$, is open given $x\in X$there exists $\epsilon>0$ such that $B(x,\epsilon)\subset \mathbb{R}^n-S_{n}$. Now given $a \in S_{n}$, we can consider $\frac{x-a}{\|x-a\|}\frac{\epsilon}{2}$ which satisfies $\frac{x-a}{\|x-a\|}\frac{\epsilon}{2} \in B(x,\epsilon)\subset \mathbb{R}^n-S_{n} $ Which implies $a \in \mathbb{R}^n-S_{n}$ which is a contradiction, and therefore $\mathbb{R}^n-S_{n}$ is not open.

So with this proof I can't use Baire! I don't know if this proof that shows that $S_{n}$ is not closed is ok and therefore my strategy to prove the exercise is wrong or if this latter proof is incorrect!

Suggestions to solve the problem in a different way are also more than welcome!

Joaquin Liniado
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  • See this post: http://math.stackexchange.com/questions/134477/countable-union-of-proper-subspaces?rq=1 – Peter Haine May 16 '15 at 02:04
  • You don't need any measure or topology for this. The argument [I gave here](https://math.stackexchange.com/a/60719/11619) shows this for any uncountable field in place of $\Bbb{R}$. – Jyrki Lahtonen May 18 '20 at 20:08

2 Answers2

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I assume "subspaces" means "$\mathbb{R}$-vector subspaces" rather than "topological subspaces".

Anyway, no, you don't need Baire's Category (sp!) Theorem. Let us generalize a bit: Let $K$ be an uncountable field. We claim that if $n \in \mathbb{N}$, then an $n$-dimensional $K$-vector space cannot be written as a countable union of proper subspaces.

Proof. We proceed by induction over $n$.

We need two induction bases: the cases $n=0$ and $n=1$. The case $n=0$ is vacuously true, since a $0$-dimensional $K$-vector space has no proper subspaces to begin with but is not the union of an empty collection of subspaces (indeed, that latter union is $\emptyset$, which is not the same as a $0$-dimensional $K$-vector space). The case $n=1$ is also true, since a $1$-dimensional $K$-vector space has only one proper subspace (namely, $0$), which is not enough to cover the whole space.

Now to the induction step. Let $N > 1$ be a positive integer. Assume that we already have shown that

(1) an $N-1$-dimensional $K$-vector space cannot be written as a countable union of proper subspaces.

Now, we need to show that an $N$-dimensional $K$-vector space cannot be written as a countable union of proper subspaces. Indeed, assume the contrary. Then, there exists an $N$-dimensional $K$-vector space $V$ which can be written as a countable union of proper subspaces. This vector space $V$ can therefore be written as a countable union of hyperplanes (since every proper subspace is contained in a hyperplane). Let $\mathfrak{H}$ be the set of these hyperplanes. Thus, $\mathfrak{H}$ is a countable set, and $V$ is the union of the hyperplanes $G \in \mathfrak{H}$.

Choose a hyperplane $H$ of $V$ which is not in $\mathfrak{H}$ (this clearly exists since $V$ has uncountably many hyperplanes (because $V$ is a $K$-vector space of dimension $N > 1$, and because $K$ is uncountable) whereas $\mathfrak{H}$ is only a countable set). Then, $G \cap H$ is a proper subspace of $H$ for every $G \in \mathfrak{H}$; the union of these subspaces $G \cap H$ is the whole $H$ (since the union of the hyperplanes $G \in \mathfrak{H}$ is $V$). So the $K$-vector space $H$ is written as a countable union of proper subspaces (namely, of its proper subspaces $G \cap H$ for $G \in \mathfrak{H}$). But this contradicts (1), since $H$ is $N-1$-dimensional (because $H$ is a hyperplane in the $N$-dimensional space $V$). This is the desired contradiction, and so the induction step is proven.

darij grinberg
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The result of this question was used here: Uncountable subset $X$ of $\mathbb{R^{n}}$ with property that every subset of $X$ with $n$ elements is a basis of $\mathbb{R}^{n}$.

The argument can be reversed!

Consider the set $A = \{(1, x, \ldots, x^{n-1}) : x > 0\}$. It has the following properties:

  • It is uncountable (because $\mathbb R_{>0}$ is uncountable)
  • Every $n$ distinct elements of $A$ form a basis (because Vandermonde determinants are nonzero).

By the second property, $A$ contains at most $n-1$ points on every hyperplane. Therefore if $\mathbb R = \bigcup_{n \in \mathbb N} S_n$ is a countable union of proper subspaces, then $A = \bigcup_{n \in \mathbb N} (A \cap S_n)$ is a countable union of finite sets, so that $A$ is countable. This is a contradiction!

Bart Michels
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