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

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 math.stackexchange.com/questions/10760 or the answers at mathoverflow.net/questions/26 – Bart Michels Sep 2 '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 2 '19 at 15:42
• @PeterFranek It means that its dimension is less than $n$. – amsmath Sep 2 '19 at 15:43
• user658532, not even a countable number of such sets could fill $\mathbb R^n$ because they have measure zero. – amsmath Sep 2 '19 at 15:45
• – Jyrki Lahtonen Sep 2 '19 at 15:49

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.