Prove that $\mathbb R^n$ is not the union of finitely many its proper subspaces.

Question

I am reading An Introduction to Algebraic Topology by Rotman. After proving

Theorem 2.7: For every $k\geq 0$, euclidean space$\mathbb R^n$ contains $k$ points in general position,

the book remarked: There are other proofs of this theorem using induction on k. The key geometric observation needed is that $\mathbb R^n$ is not the union of only finitely many proper affine subsets . I want to prove that observation.

See https://math.stackexchange.com/questions/10760 or the answers at https://mathoverflow.net/questions/26 — Bart Michels, Sep 02 '19 at 15:42
I'm not sure if "proper affine subset" is a standard well-known definition... could you include what exactly it means, pls (affine subspace, maybe)? — Peter Franek, Sep 02 '19 at 15:42
user658532, not even a countable number of such sets could fill $\mathbb R^n$ because they have measure zero. — amsmath, Sep 02 '19 at 15:45
See also here. The same holds for $\Bbb{Q}$ without any measure theory or topology (though a countable union will obviously work in that case :-) — Jyrki Lahtonen, Sep 02 '19 at 15:49
Possible duplicate of If a field $F$ is such that $|F|\ge n-1$ why is $V$ a vector space over $F$ not equal to the union of $n$ proper subspaces of $V$ — Jyrki Lahtonen, Sep 02 '19 at 15:49

score 1 · Answer 1 · answered Sep 02 '19 at 16:40

Every affine proper subspace of $\mathbb{R}^n$ i contained in an affine hyperplane.

Now, let $C=\{(t,t^2,t^3,\ldots,t^n),\,t \in \mathbb{R}^n\}$.

Since every nonzero polynomial with degree at most $n$ has at most $n$ roots, an affine hyperplane of $\mathbb{R}^n$ contains at most $n$ points in $C$.

So if you have a covering of $\mathbb{R}^n$ by proper affine subsets, you need at least $|\mathbb{R}|$ of them.

Prove that $\mathbb R^n$ is not the union of finitely many its proper subspaces.

1 Answers1