Union of proper subspaces of the same dimension of vector space is not a subspace

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I have a series of very related questions and thoughts on each... Would appreciate insight!!


  1. Let $E$ be an real inner-product space of dimension $n \geq 2$. Show that $E$ is not the union of finitely many hyperplanes of dimension $n-1$.

Proof: $E \cong \mathbb{R}^n$ and so we can topologize $E$ so that it is homeomorphic to $\mathbb{R}^n$. The finitely many hyperplanes, by the same argument, will be homemorphic to a finite product of $\mathbb{R}^{n-1}$. Thus $E$ is not the union of finitely many hyperplanes of dimension n-1 because their natural typologies inherited from real space are not equivalent.

remark: I feel a bit sketch about this proof, I mean doesn't the topology we put on $E$ depend on a choice of basis? Does this proof work? What is the regular way to prove this?


  1. Suppose that $E$ is a finite dimensional vector space over an infinite field. Suppose $U_1,U_2,....,U_n$ are proper subspaces of $E$ of the same dimension. Show that the set-theoretic union $\bigcup_{i=1}^nU_i$ is not a subspace. In particular, it is a proper subset of $E$.

To show that it is not a subspace, would I try to find two elements such that adding them together is not in this union? How would I go about doing this?

Could I prove that it is a proper subset using topological arguements like the first part?

Also, what is the counter example where this isn't true over a finite field? Thanks!!

Also, why do we have to assume that $U_i$'s are all the same dimension?

Maybe I should have split this post up into multiple different posts.. I kept having more questions as I typed it up..

Thanks!!